BUSINESS STATISTICS›PROBLEM SOLVING›UNIT 01

Unit 01 — How to Solve Anything

Five topics. Each one opens with how to recognise that a question belongs to it, then the master method — every formula, the fixed routine, the traps — then questions worked in full, then every exercise Levin sets on that topic, solved, with the textbook page number against each one.

TOPICS: 5 QUESTIONS: 174 (96 FROM LEVIN) TEXTBOOK: LEVIN §2.1–§3.6, pp. 17–111 EXAM SLOT: Q3 (a) — 5 MARKS PAIRS WITH: Unit 01 notes

◆ How to use this page tomorrow morning

Start with the ◎ topic ID card — it tells you what an unseen question on this topic looks like, so you can name the method before you touch the numbers. Then read the master method once; it is the only part you have to hold in your head. Then work down the worked questions with a pen: the left column of every grid tells you why the line exists, the right column shows what to write. Finish on the textbook bank — every exercise Levin sets on that topic, so nothing your lecturer can lift from the book is unseen.

The exam does not ask you to draw a pie chart. It hands you a table of six statistics for two service channels and asks what they mean — so Topic 5 is the one that is actually worth 5 marks. Topics 1–4 are what make Topic 5 readable.

THE SPINE OF THE WHOLE UNIT

Everything in Unit 1 is one pipeline, and every question is a cut across it at some point:

Pipeline Spine

RAW DATA $\to$ CLASSIFY IT $\to$ TABULATE IT $\to$ PICTURE IT $\to$ SUMMARISE IT $\to$ READ IT

If you are ever stuck on a Unit 1 question, ask "which arrow am I on?" — the answer names the formula.

Topic 1 of 5 · identification questions · 2–5 marks

Foundations — Descriptive vs Inferential, Population vs Sample, and Data Types

No arithmetic in this topic. Every question is a labelling question, and every one of them is answered by the same three-step reflex: who is being described → what is being measured → what kind of number came out. Getting the label right is also what decides your table and your chart in Topics 2 and 3, so a mistake here propagates.

How to know a question belongs to this topic

It asks you to

Name something, not compute something. Label · classify · identify · comment · justify. If the question has no numbers to combine, you are almost certainly here.

Give-away words

is this a sample or a population? is this raw data? identify the variable discrete or continuous? qualitative or quantitative? comment on… what type of classification primary or secondary

What it hands you

A sentence or a scenario — sometimes a small table you are not asked to compute from. The data is there to be described, not processed.

What a full answer contains

The label + the reason + the consequence. "It is a sample, because only 30 of the 500 looms were measured, so any conclusion about all 500 is an inference and carries sampling error."

Don't confuse it with

If the question…	then it is…
asks you to build a table from a list of numbers	Topic 2 — frequency distributions
asks which graph and why	Topic 3 — charts (but it starts by classifying the data type, so this topic feeds it)
gives a table and asks for an average	Topic 4 or 5
asks whether a conclusion is justified by the evidence	Still here — that is Levin's five tests for data

Master Method — Topic 1

The definitions you are actually tested on

$N$ Population size — all the elements you want a conclusion about.

$n$ Sample size — the subset you actually measured.

$\mu$ Population mean — a parameter.

$\bar{x}$ Sample mean — a statistic.

$\sigma$ Population standard deviation — parameter.

$s$ Sample standard deviation — statistic.

$p$ Population proportion — parameter.

$\hat{p}$ Sample proportion — statistic.

The rule that generates all eight: Greek letter ⇒ population ⇒ parameter. Roman letter ⇒ sample ⇒ statistic. deck slide 3

The routine — apply it in this order, every time

Find the group. Read the sentence and ask: does the number describe everyone (population) or some of them (sample)? The give-away words are "all / every / entire / census" versus "randomly chosen / surveyed / a sample of".
Name the number. Computed on a population it is a parameter; computed on a sample it is a statistic. Write the correct symbol next to it — markers give the mark for the symbol.
Name the activity. If the sentence only reports what was measured, it is descriptive. If it extends the finding to a wider group — "therefore", "we estimate", "it is likely that all" — it is inferential.
Classify the variable. Countable → discrete. Measured on a scale → continuous. A category → qualitative (nominal if unordered, rank/ordinal if ordered).
Chain to the chart. Qualitative → bar or pie. Quantitative → histogram, ogive or scatter. This single link earns the "justify your choice" mark.

Pattern recognition — what the wording is telling you

Pattern	Wording cue in the question	What to write
Descriptive vs inferential	"…the average of the 200 surveyed was…" vs "…so the average of all customers is about…"	Report only ⇒ descriptive. Generalises beyond the measured group ⇒ inferential.
Parameter vs statistic	"all 5,000 employees" vs "a random sample of 80"	Give the symbol: $\mu,\sigma,p$ vs $\bar{x},s,\hat{p}$.
Discrete vs continuous	"number of…" vs "time / weight / length / temperature"	"Number of" is always discrete; anything on a measuring instrument is continuous.
Nominal vs rank	"gender, brand, region" vs "1st, 2nd, 3rd; poor→excellent"	Ask: does reordering the categories lose information? Yes ⇒ rank/ordinal.
Primary vs secondary	"conducted its own survey" vs "obtained from the RBI bulletin"	Collected first-hand ⇒ primary. Already existed ⇒ secondary, "not necessarily in the form actually required".
Method of classification	by state / by month / by income slab / by gender	Geographical · chronological · variable (quantitative) · qualitative.

How each of your sources says it

Source	Wording it uses
Deck	"A parameter is a descriptive value for a population; a statistic is a descriptive value for a sample." Data tree: quantitative (discrete / continuous) vs qualitative (nominal / rank) slides 3, 8–10
Class notes	Descriptive → predictive → prescriptive analysis tree notes p. 1
Levin	Ch. 1–2 — subdivisions within statistics, then raw data and arrangement pp. 13–18
Exam	Never asked directly. It is assumed vocabulary inside Q3 (a) and inside every sampling question in Units 5–6.

Traps

Calling a sample figure a parameter because the sample was large. Size is irrelevant — only who was measured matters.
Calling "number of children" continuous because it is a number. If you cannot have 2.4 of it, it is discrete.
Treating a rank as a quantity. The gap between rank 1 and 2 is not the same as between 2 and 3, so you may not average ranks and call it a mean of the attribute.
Saying "secondary data is unreliable". The examinable objection is narrower: it may exist in a form that does not fit your question slide 7.

Section A — Ten questions, fully explained read the WHY column, copy the ANSWER column

A1Descriptive vs inferential

Q. Label each statement descriptive or inferential, and say why.
(i) Of the 400 invoices audited this month, 38 contained an error.
(ii) Based on those 400 invoices, roughly 9.5% of all invoices raised this year contain an error.
(iii) The mean waiting time of the 60 customers observed on Monday was 6.2 minutes.
(iv) The branch should add a second counter, because waiting time is likely to exceed 8 minutes at month-end.

Why this method: the split is not about how clever the sentence is — it is about whether the sentence stays inside the group that was measured. Draw a mental box around the measured group; if the claim leaks out of the box, it is an inference.

Why we do itWhat to write

Statement (i) counts only the 400 audited and claims nothing beyond them.

(i) Descriptive — a simple count of the measured set.

Statement (ii) uses the same 400 to make a claim about all invoices this year — the claim leaves the box.

(ii) Inferential — a sample result generalised to the population.

Statement (iii) summarises the 60 observed customers and stops there.

(iii) Descriptive — a summary measure, $\bar{x}=6.2$ min.

Statement (iv) predicts an unobserved future and recommends an action on that basis.

(iv) Inferential — prediction plus prescription from sample evidence.

Close with the general rule so the marker sees you know the principle, not just the four answers.

Descriptive summarises what was measured; inferential generalises to what was not.

Marker's note: a one-line reason per statement is what earns the marks — four bare labels typically score half.

A2Population / sample / parameter / statistic

Q. A bank has 12,400 savings accounts. An auditor draws 250 of them at random and finds a mean balance of ₹48,300 with a standard deviation of ₹9,150. The bank's own core system reports that the mean balance over all accounts is ₹47,100. Identify the population, the sample, and every parameter and statistic, with symbols.

Why this method: two means are floating in the same paragraph. The only thing that separates them is who they were computed on — so sort by group first, then attach symbols.

Why we do itWhat to write

Name the group the conclusion is about. "A bank has 12,400 accounts" — all of them.

Population = all 12,400 savings accounts, $N = 12{,}400$

Name the measured subset.

Sample = the 250 audited accounts, $n = 250$

₹48,300 and ₹9,150 came from the 250 → sample → statistics, Roman letters.

$\bar{x} = ₹48{,}300$ · $s = ₹9{,}150$ — statistics

₹47,100 came from the core system, i.e. from all accounts → population → parameter, Greek letter.

$\mu = ₹47{,}100$ — a parameter

Name the variable and its type, since the next question always asks for the chart.

Variable = account balance — quantitative, continuous.

The observation the examiner is fishing for.

$\bar{x} \ne \mu$: a sample statistic is expected to differ from the parameter — that gap is sampling error, not an audit error.

Marker's note: the last line is the bridge into Units 5–6. Saying it here costs one sentence and shows the examiner you know why sampling distributions exist.

A3Symbols

Q. Fill in the correct symbol and the word parameter or statistic: (i) the proportion of defectives in a shipment of 5,000 units, all of which were inspected; (ii) the proportion of defectives in a random check of 200 of them; (iii) the SD of the marks of all 90 students in a class; (iv) the mean of a random sample of 15 of those students.

Why this method: one question, one rule — Greek for everybody, Roman for a subset. The words "all / every" versus "random sample" are the switch.

Why we do itWhat to write

"All of which were inspected" = a census of the shipment ⇒ population proportion.

(i) $p$ — parameter

"A random check of 200" = a subset ⇒ sample proportion.

(ii) $\hat{p}$ — statistic

"All 90 students" — the class is the population here, even though 90 is small.

(iii) $\sigma$ — parameter

A sample drawn from that class ⇒ sample mean.

(iv) $\bar{x}$ — statistic

Point out the trap in (iii) explicitly.

Population size does not decide the label — coverage does. 90 out of 90 is a population; 15 out of 90 is a sample.

A4Data types

Q. Classify each variable as quantitative-discrete, quantitative-continuous, qualitative-nominal or qualitative-rank: (a) number of flights landing per hour; (b) thickness of a steel sheet; (c) blood group; (d) position in a sales contest; (e) monthly salary; (f) number of defective bulbs in a carton; (g) customer satisfaction on Poor→Excellent; (h) region of a dealership.

Why this method: two yes/no questions settle every item — is it a number or a category? then can it take an in-between value / does the order carry meaning?

Why we do itWhat to write

Counts. You cannot land 3.4 flights or find 7.5 defective bulbs.

(a), (f) — quantitative, discrete

Read off a measuring scale; any value in an interval is possible.

(b), (e) — quantitative, continuous

Categories with no natural order — A, B, AB, O is not a ladder; nor is North, South, East, West.

(c), (h) — qualitative, nominal

Categories whose order carries information — reordering them destroys meaning.

(d), (g) — qualitative, rank (ordinal)

The follow-up mark: chain each type to its chart.

Discrete & continuous → histogram / ogive. Nominal & rank → bar graph / pie chart.

Watch (e): salary is continuous in principle even though it is paid in whole rupees — the underlying quantity is measured, not counted. Say "continuous (recorded to the nearest rupee)" and you are safe either way.

A5Primary vs secondary

Q. A retail chain wants to know why footfall fell in Q3. It (i) commissions a 900-shopper exit survey, and (ii) downloads district-level consumer-spending data from a government portal. Classify each source, and state one limitation of the second that is specific to this problem.

Why this method: the classification is easy; the mark sits in the limitation, and the deck gives you the exact phrasing to use.

Why we do itWhat to write

The survey did not exist before; it is collected first-hand for this question.

(i) Primary data — collected for the first time from the primary source, for this specific purpose.

The portal data already existed, published by someone else for their own purpose.

(ii) Secondary data — already available in an identifiable published source.

State the limitation in the deck's own words, then make it concrete for this chain.

It is "not necessarily in the form actually required": it is district-level, so it cannot say why this chain's footfall fell — only what the district spent.

One sentence balancing the two, which is the usual last mark.

Secondary data is cheaper and faster and good for context; primary data is fit for purpose but costly. Use secondary to frame the question, primary to answer it.

A6Classification method

Q. Name the method of classification used in each table, and for the last one state whether the qualitative classification is dichotomous or manifold.
(i) Sales by state · (ii) Production by month, 2020–2025 · (iii) Employees by income slab ₹20k–30k, ₹30k–40k, … · (iv) Employees split male/female, then each split employed/unemployed.

Why this method: the four names come straight from the deck; the only judgement is on the last one, where you count how many attributes are being used at once.

Why we do itWhat to write

The classifying characteristic is a place.

(i) Geographical (spatial) classification.

The classifying characteristic is time.

(ii) Chronological (temporal) classification.

The characteristic is measurable and has been cut into class intervals.

(iii) Classification by variable — continuous, so class intervals are used.

The characteristic is an attribute, and two attributes are applied one after the other.

(iv) Qualitative, and manifold — a dichotomy is a single two-fold split; here a second attribute is nested inside the first, giving four groups.

Give the count, because that is the check the examiner applies.

Manifold with two dichotomous attributes ⇒ $2\times2 = 4$ classes: male-employed, male-unemployed, female-employed, female-unemployed.

A7Type → table → chart

Q. For each dataset name the data type, the tabular summary and the graph, with a one-line justification. (a) Preferred payment mode of 500 shoppers; (b) delivery times of 240 orders; (c) market share of six brands; (d) advertising spend and sales for 36 months; (e) "how many orders were delivered in under 30 minutes?"

Why this method: this is the deck's closing decision tree read out loud. Type decides the table; the table decides the chart. Never jump straight to the chart.

Why we do itWhat to write

Categories, unordered ⇒ qualitative nominal ⇒ counts per category ⇒ separated bars.

(a) Qualitative nominal · frequency + relative frequency distribution · bar graph (bars separated — the categories are distinct).

Measured on a continuous scale ⇒ class intervals ⇒ touching bars.

(b) Quantitative continuous · grouped frequency distribution · histogram (bars touch — the classes are adjacent intervals).

Parts of one whole ⇒ proportions ⇒ sectors of a circle.

Two quantitative variables measured on the same unit ⇒ paired points.

(d) Two quantitative variables · paired data table · scatter diagram (shows the relationship between the two).

The question is "how many below a value" — that is the definition of a cumulative reading.

(e) Quantitative, cumulative question · cumulative ("less than") frequency distribution · ogive, read straight off at 30 minutes.

This is the highest-yield question in Topic 1 — it is the only Unit 1 shape that has ever appeared as a stand-alone sub-part in this style of paper, and the justification clause is where every mark is.

A8Theory · 5-mark skeleton

Q. "A sample is preferred to a census." Give five reasons, with an example of one situation where a sample is the only option.

Why this method: a pure theory question — the marks are for the number of distinct reasons, so answer as a list, one line each, never as a paragraph.

Why we do itWhat to write

Reason 1 — the one every marker expects first.

Cost. Measuring 12,000 units costs far more than measuring 300.

Reason 2 — pairs with cost.

Time. A census takes so long the answer can arrive after the decision is due.

Reason 3 — the counter-intuitive one, worth stating.

Accuracy. A small, carefully supervised sample can be more accurate than a huge census with untrained enumerators.

Reason 4 — feasibility.

Practicality. Some populations are infinite or constantly changing (every future unit off a production line).

Reason 5 — the example the question asks for.

Destructive testing. Testing the burn-life of every bulb or the breaking strength of every cable leaves nothing to sell — a sample is the only option.

Close by naming the price you pay, or the answer reads one-sided.

The cost of sampling is sampling error: $\bar{x}$ will differ from $\mu$. Units 5–6 exist to measure exactly how much.

A9Scales of measurement

Q. State the scale of measurement (nominal, ordinal, interval, ratio) for: (a) employee ID number; (b) finishing position in a race; (c) temperature in °C; (d) annual turnover in ₹; and say for each whether computing a mean is meaningful.

Why this method: work up the ladder — each scale adds one property: labels → order → equal gaps → a true zero. The mean becomes meaningful only from "equal gaps" upward.

Why we do itWhat to write

An ID is a label that happens to be written as a number; arithmetic on it is meaningless.

(a) Nominal. Mean: not meaningful — the "average employee ID" describes nobody.

Order is real, but the gaps are not equal — 1st to 2nd may be 0.1 s, 2nd to 3rd 4 s.

(b) Ordinal. Mean: not strictly meaningful; use the median instead.

Equal gaps (1 °C is the same step everywhere), but 0 °C is a convention, not an absence of heat.

Equal gaps and a genuine zero (₹0 means no turnover).

(d) Ratio. Mean: meaningful, and so are ratios — ₹40 lakh is twice ₹20 lakh.

One line the examiner can tick.

Ladder: nominal ⊂ ordinal ⊂ interval ⊂ ratio. Mean needs interval or ratio; ordinal gets the median; nominal gets the mode.

Source note: the four scales are Levin's framing rather than your deck's — the deck stops at nominal/rank slide 10. If the question uses the words "nominal/rank", answer in those; if it says "scale of measurement", use all four.

A10Full mini-case

Q. A hospital treated 3,800 post-operative patients last year. An administrator records the length of stay of 200 randomly chosen patients; the mean is 7.7 days and the SD is 4.6 days. She reports: "the typical post-operative stay in this hospital is about a week." Identify: population, sample, variable and its type, the symbol for each number given, whether her report is descriptive or inferential, and the tabular and graphical summary you would build.

Why this method: this is your own class's Georgia-hospital data notes pp. 3–4 turned into a labelling question. Run the five-step routine straight down the page — it answers all six parts in order.

Why we do itWhat to write

Step 1 — the group the conclusion is about.

Population: all 3,800 post-operative patients, $N=3{,}800$. Sample: the 200 recorded, $n=200$.

Step 2 — name the numbers. Both came from the 200.

$\bar{x}=7.7$ days, $s=4.6$ days — both statistics.

Step 3 — the report speaks about "this hospital", i.e. all patients, not just the 200.

The report is inferential — it generalises $\bar{x}$ to $\mu$.

Step 4 — classify the variable. Days of stay are counted in whole days but the underlying quantity is duration.

Variable = length of stay. Quantitative; recorded as discrete whole days, treated as continuous once grouped into class intervals.

Step 5 — chain to the summaries.

Table: grouped frequency distribution (1–3, 4–6, 7–9, …) plus a cumulative column. Graph: histogram, and an ogive to read off "how many stayed under 10 days".

Add the caveat that lifts the answer.

"About a week" is defensible only because the sample was random; $\bar{x}=7.7$ is an estimate of $\mu$ carrying sampling error of order $s/\sqrt{n}=4.6/\sqrt{200}\approx0.33$ days.

Marker's note: the final line borrows one formula from Unit 5. It costs a line and reliably converts a 4 into a 5.

Section B — Solve a problem ten fresh questions · one or two lines of reasoning, then the answer

Do them on paper first. Each answer carries only the reason it exists — the full method is in Section A.

B1Descriptive vs inferential

Q. "In a taste test, 62 of 100 panellists preferred Brand X; the company concludes that a majority of Indian consumers prefer Brand X." Which part is descriptive, which is inferential?

Solution

Why: split the sentence at the point where the claim stops describing the 100 people who were actually measured.

Counts inside the measured group.

"62 of 100 preferred Brand X" — descriptive.

Extends the finding to a population never measured.

"A majority of Indian consumers prefer Brand X" — inferential.

B2Symbols

Q. A factory has 6,000 workers. HR measures absenteeism for every worker and gets a mean of 4.2 days; a consultant samples 150 workers and gets 4.6 days with SD 1.8. Write all four numbers with their correct symbols and labels.

Solution

Why: "every worker" is a census ⇒ Greek; "samples 150" ⇒ Roman.

Whole workforce measured.

$N=6{,}000$, $\mu=4.2$ days — parameter.

Subset measured.

$n=150$, $\bar{x}=4.6$ days, $s=1.8$ days — statistics.

B3Data types

Q. Classify: (a) number of complaints per shift; (b) call duration in seconds; (c) preferred language; (d) hotel star rating; (e) diameter of a bearing.

Solution

Why: number or category first, then countable-vs-measured / ordered-vs-unordered.

Counts.

(a) quantitative discrete.

Read off a scale.

(b), (e) quantitative continuous.

Unordered categories.

Ordered categories — 5★ outranks 3★.

(d) qualitative rank / ordinal.

B4Type → chart

Q. Which graph for: (a) share of five loan products in a bank's portfolio; (b) distribution of 300 loan sizes; (c) relationship between loan size and default rate?

Solution

Why: shares of a whole → pie; one continuous variable's distribution → histogram; two quantitative variables → scatter.

Parts of one portfolio.

(a) Pie chart from a percent frequency distribution.

One continuous variable, grouped into classes.

(b) Histogram (bars touch).

Two quantitative variables paired per loan.

B5Primary vs secondary

Q. Classify: (i) a company's own employee-engagement survey; (ii) industry salary benchmarks bought from a consultancy. Give one limitation of (ii).

Solution

Why: collected first-hand for this purpose vs already existing for someone else's purpose.

Fresh collection.

(i) Primary.

Pre-existing, published for a general audience.

(ii) Secondary. Limitation: it is "not necessarily in the form actually required" — the grades, regions or job families may not match this company's own structure.

B6Classification

Q. Name the method of classification: (i) rainfall by district; (ii) exports by quarter; (iii) students by marks band; (iv) staff by union member / non-member.

Solution

Why: ask what the classifying characteristic is — place, time, a measurable quantity, or an attribute.

Place / time.

(i) Geographical · (ii) Chronological.

Measurable quantity / two-fold attribute.

(iii) By variable (class intervals) · (iv) Qualitative, dichotomous.

B7Trap

Q. A manager says: "We surveyed 4,000 of our 4,500 customers, so 4,000 is a big enough number to call the mean a parameter." Is she right?

Solution

Why: the parameter/statistic label depends on coverage, never on size.

4,000 out of 4,500 is still a subset.

No. It is a sample mean, $\bar{x}$ — a statistic. (It is a very precise one: with $n/N=0.89$ the finite population correction shrinks its standard error almost to zero — but that changes its precision, not its name.)

B8Theory

Q. Give four management decisions in which statistics is used, and name the statistical input each needs.

Solution

Why: the deck lists the decisions slide 6; the marks come from pairing each with an input, not just naming it.

Two operational decisions.

Plant location — demand and cost data by region. Plant size — forecast demand distribution.

Two functional decisions.

Quality control — sample defect proportions and control limits. Marketing — consumer-preference survey proportions.

B9Scales

Q. A survey codes satisfaction as 1 = very dissatisfied … 5 = very satisfied and reports a mean of 3.9. Is the mean legitimate? What would a purist compute instead?

Solution

Why: the codes are ordinal — the order is real but the gaps are not guaranteed equal.

Strictly ordinal ⇒ the median is the defensible average; in practice Likert means are reported anyway.

Strictly, no — a mean assumes equal gaps between 1–2 and 4–5. A purist reports the median and the mode. In business practice the mean is reported, and the paper's own Q3 (a) does exactly that — so quote the mean and the median together.

B10Mini-case

Q. An e-commerce firm ships 90,000 orders a month. It tracks delivery time for 500 randomly selected orders: mean 38.4 minutes, SD 11.2, and 12% took over an hour. Identify the population, sample, variable and type; label each number; and name the table and chart you would build to answer "what fraction is delivered within 45 minutes?"

Solution

Why: run the five-step routine; the last part is a cumulative question, which fixes both the table and the chart.

Group and subset.

Population = all 90,000 orders ($N=90{,}000$); sample = 500 orders ($n=500$).

All three numbers came from the 500.

$\bar{x}=38.4$ min, $s=11.2$ min, $\hat{p}=0.12$ — all statistics.

Time is measured on a scale.

Variable = delivery time — quantitative, continuous.

"Within 45 minutes" is a below-a-value question.

Cumulative ("less than") frequency distribution, drawn as an ogive; read the height at 45 minutes and divide by 500.

Textbook bank — every Levin exercise on this topic §2.1 and §2.2, solved, with the page each one is on

8 exercises Levin §2.1 How can we arrange data? p. 17 · §2.2 Examples of raw data p. 18 All 8 are discussion questions — this section sets none that compute.

THE FIVE TESTS FOR DATA — three of the eight exercises below are just this list applied

5 Tests For Data

1. Source Bias: Where did data come from? Is source biased?
2. Consistency: Does it support or contradict other evidence?
3. Completeness: Is critical evidence missing?
4. Sample Size: How many observations? Are all groups represented?
5. Logic: Is the conclusion supported by data?

Levin p. 15. Learn the five headings; any "comment on this conclusion" question is answered by walking down them.

2-1Sample or populationLevin Ex. 2-1 · p. 17

Q. When asked what they would use if they were marooned on an island with only one choice for a pain reliever, more doctors chose Bayer than Tylenol, Bufferin or Advil. Is this conclusion drawn from a sample or a population?

Why: ask whether every member of the group the claim is about was asked. "Doctors" is a population of hundreds of thousands; nobody polled all of them.

AnswerA sample. The claim is about doctors in general, but only a subset was surveyed — so "more doctors chose Bayer" is an inference from a sample, not a measured fact about the population. Worth adding: the advertisement does not state the sample size or who paid for the survey, which is test 1 and test 4 of the five tests.

2-2Sample or populationLevin Ex. 2-2 · p. 17

Q. Twenty-five percent of the cars sold in the United States in 1996 were manufactured in Japan. Is this conclusion drawn from a sample or a population?

Why: same test, opposite answer — car sales are registered, so every single sale in the period is on record.

AnswerA population. Every car sold in the US in 1996 is counted in registration records, so 25% is a parameter computed on the whole population — not an estimate. Contrast with 2-1: identical wording, different answer, and the only thing that decides it is whether the whole group was measured.

2-3Five tests for dataLevin Ex. 2-3 · p. 17

Q. An electronics firm introduced a new amplifier; warranty cards indicate 10,000 have been sold. The president, upset after reading three letters of complaint, told the production manager to implement costly control measures immediately. Comment on the president's reaction using the five tests for data.

Why: a "comment on this reaction" question is the five tests, walked in order. Answer under the test headings and the marks fall out.

TestApplied to this case

1. Source biased?

Complaint letters are self-selected. Only angry customers write in; satisfied ones never do. This is the most biased sample possible.

2. Supports or contradicts other evidence?

No other evidence was consulted — no return rates, no service-centre data, no quality-control test results.

3. Evidence missing?

Yes — the 9,997 customers who did not write. Their silence is the bulk of the information and it was ignored entirely.

4. How many observations?

Three out of 10,000 — a defect rate of 0.03%, which for most consumer electronics is excellent, not alarming.

5. Conclusion logical?

No. The data do not support spending on costly controls. The president should first sample units from the line, or examine warranty-return rates, before acting.

2-4Five tests for dataLevin Ex. 2-4 · p. 17

Q. "Germany will remain ever divided," said Walter Ulbricht after the Berlin Wall was built in 1961. By 1989 the wall was destroyed. Give some reasons for Ulbricht's incorrect prediction.

Why: the same five tests, applied to a forecast rather than a measurement — the point being that a prediction is only as good as the data behind it.

Answer Test 1 — biased source. Ulbricht headed the East German state; he had a direct interest in the conclusion that division was permanent, so his own judgement was not a neutral source.
Test 3 — missing evidence. He could not observe the political and economic pressures building inside East Germany, nor the future changes in Soviet policy.
Test 5 — illogical conclusion. He extrapolated an indefinite future from a single point in time. No dataset that stops in 1961 can support a claim containing the word "ever".
The transferable lesson: a forecast that extends far beyond the range of the data is unsupported, no matter how good the data is.

2-5Five tests for dataLevin Ex. 2-5 · p. 17

Q. Discuss the chapter-opening problem in terms of the five tests for data. (Dalmon Carpet Company: the production manager samples the daily output, in yards, of 30 of his 500 carpet looms and draws a conclusion about the average production of all 500.)

Why: the same checklist, this time on a well-designed study — so the answer is mostly "passes", which is what makes it a useful contrast with 2-3.

Answer 1. Source. The data are the firm's own production measurements, taken for internal use — no incentive to bias them. Passes.
2. Other evidence. They can be checked against previous days' output and against loom specifications. Passes, if he does so.
3. Missing evidence. One day only. Output may vary by day of week, shift or yarn batch, and none of that is captured. A genuine weakness.
4. Observations. 30 looms out of 500 is a reasonable fraction — but only if the 30 were chosen at random. If he always samples the same accessible looms, the sample is not representative. Conditional.
5. Logical. Yes — concluding about average output of 500 looms from a random sample of 30 is exactly what inferential statistics is for, provided he reports it as an estimate and not as the exact figure.

2-6Raw dataLevin Ex. 2-6 · p. 18

Q. Look at the data in Table 2-1 (20 pairs of high-school and college grade-point averages). Why do these data need further arranging? Can you form any conclusions from the data as they exist now?

Why: this question is the reason Topics 2 and 3 exist — it asks you to state what raw data cannot do.

AnswerThe table is raw data: 20 pairs listed in the order the students happened to be recorded. In that form you can read individual values but you cannot see where the values cluster, how spread out they are, or whether the two variables move together. Almost no conclusion can be drawn as it stands — beyond noting the rough range of GPAs. Arranging it (a data array, then a frequency distribution, then a scatter diagram for the pairing) makes the centre, the spread and the relationship visible.

2-7Raw dataLevin Ex. 2-7 · p. 18

Q. A marketing manager receives a monthly report listing sales of one product by state for the previous month. Is this an example of raw data?

Why: raw data is data that has not been processed by statistical methods. Ask: has anything been summarised, or only listed?

AnswerYes — it is raw data. The report is a listing, not a summary: no totals, no averages, no distribution. Sorting or grouping by state is only classification (geographical classification), which arranges the observations without reducing them to a summary measure. Compare 2-8, where the same kind of report is not raw.

2-8Raw dataLevin Ex. 2-8 · p. 18

Q. A production manager receives a monthly quality-control report giving the reject rate for the line (rejects per 100 units), the machine causing the greatest number of rejects, and the average cost of repairing rejected units. Is this an example of raw data?

Why: the same test as 2-7 with the opposite answer — look for words that name a computed quantity: rate, greatest, average.

AnswerNo — this is processed, not raw. All three items are summary measures computed from raw data: a rate (rejects ÷ units × 100), a maximum (which machine caused the most), and a mean (average repair cost). The raw data would be the individual inspection records, one per unit. The pairing of 2-7 and 2-8 is the whole point of the section: a list is raw, a summary is not.

Topic 2 of 5 · table-building questions · 3–5 marks

Frequency, Relative, Percent and Cumulative Distributions

This is the workshop of Unit 1. Everything downstream — the histogram, the ogive, the mean, the median, the mode — is computed off a table you build here. Build the table wrong and every later mark is lost, so the routine below is worth more than it looks.

How to know a question belongs to this topic

It asks you to

Build or read a table. Turn raw numbers into classes, or turn a given table into relative, percent or cumulative form.

Give-away words

construct a frequency distribution arrange these data in an array use intervals of 6 how many are less than… relative frequency class boundaries closed / open-ended classification

What it hands you

A list of raw values to be grouped, or a table to be extended with a column. Never a mean or a standard deviation — those belong to Topics 4 and 5.

The three checks that earn marks

$\Sigma f = n$ · relative frequencies sum to 1.00 · percent frequencies sum to 100. Write all three.

Don't confuse it with

If the question…	then…
asks which graph to draw	Topic 3 — charts
asks for an average from the table	Topic 4 or 5
asks you to group, tally or cumulate	Here

Master Method — Topic 2

Formula bank

Range

\[ R = L - S \qquad \text{(largest value } - \text{ smallest value)} \]

Number of classes
(Sturges' rule)

\[ k = 1 + 3.322\,\log_{10} n \]

Class width

\[ h = \frac{R}{k} \quad\longrightarrow\quad \text{round \textbf{up} to a convenient number} \]

Class mark
(midpoint)

\[ x_i = \frac{\text{lower limit} + \text{upper limit}}{2} \]

Relative & percent
frequency

\[ \text{rel} = \frac{f_i}{n} \qquad \text{pct} = \frac{f_i}{n}\times 100 \qquad \Sigma\,\text{rel}=1.00,\;\; \Sigma\,\text{pct}=100 \]

Inclusive → exclusive
(class boundaries)

\[ d = \frac{\text{next lower limit} - \text{this upper limit}}{2},\qquad \text{boundaries} = (\text{lower}-d,\;\text{upper}+d) \]

Cumulative
frequency

\[ \text{“less than”: } cf_i=\sum_{j\le i} f_j \qquad \text{“more than”: } cf_i = n-\sum_{j

Frequency density
(unequal widths)

\[ \text{density} = \frac{f_i}{h_i} \qquad\text{or}\qquad f_i^{\,\text{adj}} = f_i \times \frac{h_{\min}}{h_i} \]

Sturges' rule gives a starting point, not a law. Round $k$ to a whole number and $h$ to something readable — 5, 10, 20 — then check the classes cover the smallest and largest value.

The routine — building a grouped table from raw data

Range. Write $S$, $L$ and $R=L-S$ explicitly. Markers look for this line.
Decide $k$, then $h$. Sturges for $k$, then $h=R/k$ rounded up. State the rounding: "$h=5.67$, taken as 6".
Write the classes. Start at or just below $S$, keep the width constant, keep them non-overlapping and exhaustive.
Tally. One pass through the data, one tally mark each. Then total the frequencies and check $\Sigma f = n$ before going further.
Add the columns the question asks for — midpoint $x_i$, relative $f/n$, percent, "less than" $cf$, "more than" $cf$. Never add columns you were not asked for; do add the total row.
Check the totals. $\Sigma f = n$, $\Sigma\text{rel}=1.00$, $\Sigma\text{pct}=100$, last "less than" $cf = n$, first "more than" $cf = n$.

Pattern recognition

Pattern	Wording cue	What to do
Build from raw data	"The following are the … of 30 …. Construct a frequency distribution."	Range → Sturges → width → tally. Show all four lines.
Inclusive series	classes written 10–19, 20–29 — a gap between them	Convert to boundaries 9.5–19.5, 19.5–29.5 before median, mode, histogram or ogive. The mean does not need it.
Cumulative	"how many are less than / at least / below…"	Build the "less than" (or "more than") $cf$ column and read it at the class boundary.
Unequal widths	classes 0–10, 10–20, 20–40, 40–70	Do not plot $f$. Plot $f/h$ (frequency density), or rescale to a common width.
Open-end class	"below 10", "70 and above"	Give the open class the same width as its neighbour, state the assumption in words, then proceed.
Missing frequency	"$N = 50$ and the mean is 25; find the missing frequencies"	Two equations: $\Sigma f = N$ and $\Sigma f x = \bar{x}N$. Solve simultaneously.
Crosstabulation	two qualitative variables in one table	Add row totals, column totals, grand total; then row-% or column-% depending on which comparison the question wants.

How each of your sources says it

Source	What it gives you
Deck	The Marada Inn table — frequency, relative and percent side by side, and the rule that relative frequencies sum to 1.00 slides 25–30
Class notes	The inclusive-series conversion written out as 0.5–3.5, 3.5–6.5, … with the red rule beside it notes p. 3
Levin	§2.3–2.4 — how to construct one, and the warning about too many classes (the same 20 figures shown in 6 and in 12 classes) pp. 18–37
Exam	Never asked as a stand-alone build, but the table is the object Q3 (a) hands you already built — you must be able to read every column of it.

Traps

Overlapping classes. "20–30, 30–40" is only legal if you state the convention (lower limit included, upper excluded). Otherwise a value of exactly 30 has two homes.
Forgetting to check $\Sigma f = n$. One mis-tally invalidates the mean, median, mode and every chart.
Plotting raw $f$ on unequal widths. A wide class then looks far more common than it is.
Using class limits instead of boundaries on an ogive or a histogram of an inclusive series — the bars end up with gaps between them.
Too many classes. Levin's own example: 12 classes on 20 observations produces mostly-empty cells and hides the pattern that 6 classes reveal.

Section A — Ten questions, fully explained every table shown in full, every check performed

A1Qualitative FD

Q. Twenty-five customers rated a service as Poor, Average, Good or Excellent. The responses were:

Raw Qualitative Ratings

E  G  A  G  E  A  P  G  G  E
A  G  P  G  E  G  A  G  E  G
A  P  G  E  G          (P=Poor, A=Average, G=Good, E=Excellent)

Construct the frequency, relative frequency and percent frequency distributions, and state two things a manager should conclude.

Why this method: the data is qualitative, so the classes are the categories themselves — no range, no width, no Sturges. Tally, then divide by $n$ twice.

Why we do itWhat to write

Tally each category once through the list, then check the total equals $n$.

$f$: Poor 3, Average 5, Good 11, Excellent 6. $\Sigma f = 25 = n$ ✓

Relative frequency is the same count expressed as a fraction of the whole, $f/n$.

\[ \frac{3}{25}=0.12,\quad \frac{5}{25}=0.20,\quad \frac{11}{25}=0.44,\quad \frac{6}{25}=0.24 \]

Percent frequency is just that × 100 — no new work.

12%, 20%, 44%, 24%

Rating	Frequency f	Relative f/n	Percent
Poor	3	0.12	12
Average	5	0.20	20
Good	11	0.44	44
Excellent	6	0.24	24
Total	25	1.00	100

Answer — the reading 68% of customers rated the service Good or Excellent, which should please the manager. But twice as many said Poor (3) as the gap to a full quarter suggests is acceptable, and 12% dissatisfied on a 25-person sample is roughly one customer in eight walking away unhappy — that should concern him.

Marker's note: the deck's own reading of the Marada Inn table has exactly this shape — a number, what it means for the business, and whether it is good or bad news slide 35. Copy that three-part sentence.

A2Build from raw data

Q. The daily number of transactions at a branch over 30 days:

Raw Transaction Data

42  55  61  38  47  52  66  58  45  49
63  57  41  53  60  48  72  50  44  56
39  51  68  46  59  54  43  62  57  50

Construct a grouped frequency distribution using Sturges' rule, add relative, percent and "less than" cumulative columns, and comment on the shape.

Why this method: raw quantitative data must be compressed before it can be read. Sturges fixes how many classes; the range divided by that fixes how wide. Everything after is bookkeeping.

Why we do itWhat to write

Step 1 — the spread of the data decides the total width the classes must cover.

\[ S=38,\quad L=72,\quad R = 72-38 = 34 \]

Step 2 — Sturges turns $n$ into a sensible number of classes.

\[ k = 1+3.322\log_{10}30 = 1+3.322(1.4771)=5.91 \approx 6 \]

Step 3 — width, rounded up so the classes definitely cover $L$.

\[ h = \frac{R}{k} = \frac{34}{6} = 5.67 \;\longrightarrow\; h = 6 \]

Step 4 — start at $S=38$; six classes of width 6 reach 74, which covers 72. Convention: lower limit included, upper excluded.

38–44, 44–50, 50–56, 56–62, 62–68, 68–74

Step 5 — tally, then check the total before doing anything else.

$f = 5,\,6,\,7,\,7,\,3,\,2$ → $\Sigma f = 30 = n$ ✓

Class	Midpoint x	f	Relative	Percent	“Less than” cf
38–44	41	5	0.167	16.7	5
44–50	47	6	0.200	20.0	11
50–56	53	7	0.233	23.3	18
56–62	59	7	0.233	23.3	25
62–68	65	3	0.100	10.0	28
68–74	71	2	0.067	6.7	30
Total	—	30	1.000	100	—

Answer The distribution is mildly positively skewed — the bulk of the days (18 of 30, i.e. 60%) sit between 44 and 62, with a thin right tail of five busy days above 62 and none of comparable weight on the left. The modal class is a tie between 50–56 and 56–62.

Cross-check worth knowing: the grouped mean from this table is $\Sigma fx/\Sigma f = 1608/30 = 53.6$, while the true mean of the 30 raw values is $1586/30 = 52.87$. The 0.73 gap is grouping error — the price of replacing every value by its class midpoint. Mentioning it is a free extra mark.

A3Inclusive → exclusive

Q. Marks of 60 students are given as an inclusive series. Convert it to class boundaries, give the midpoints, and state which later calculations need the conversion.

Marks	10–19	20–29	30–39	40–49	50–59
No. of students	6	14	22	12	6

Why this method: the classes have a gap — nothing can be recorded between 19 and 20. A histogram or an ogive needs a continuous axis, so the gap must be closed by pushing each limit half-way into it.

Why we do itWhat to write

Measure the gap between one class's upper limit and the next class's lower limit, and halve it.

\[ d = \frac{20-19}{2} = 0.5 \]

Push every lower limit down by $d$ and every upper limit up by $d$. The classes now touch.

9.5–19.5, 19.5–29.5, 29.5–39.5, 39.5–49.5, 49.5–59.5

Midpoints — compute them from the original limits; the conversion does not move them.

\[ x=\frac{10+19}{2}=14.5,\;24.5,\;34.5,\;44.5,\;54.5 \]

Confirm the midpoint is unchanged from the boundaries, which is the whole reason the mean is unaffected.

\[ \frac{9.5+19.5}{2}=14.5 \quad\checkmark \text{ same value} \]

State the rule your lecturer wrote in red.

Conversion is NOT required for the mean. It IS required for the median, the mode, the histogram and the ogive.

Source: "For normal calculations, an inclusive series need NOT be converted to an exclusive series for calculating MEAN. However for calculating MEDIAN & MODE conversion is REQUIRED." class notes p. 3, in red

A4Relative & percent

Q. A call centre logged 50 calls in the classes below. Build the relative and percent frequency distributions and state what fraction of calls lasted between 20 and 40 minutes.

Duration (min)	0–10	10–20	20–30	30–40	40–50
No. of calls	5	12	20	9	4

Why this method: a relative frequency is a proportion, so questions phrased as "what fraction / what percentage" are answered by adding relative frequencies, never by re-counting.

Why we do itWhat to write

Confirm $n$ first — every division below uses it.

\[ n = 5+12+20+9+4 = 50 \]

Divide each frequency by $n$.

\[ 0.10,\;0.24,\;0.40,\;0.18,\;0.08 \qquad \Sigma = 1.00\;\checkmark \]

×100 for percents.

\[ 10\%,\;24\%,\;40\%,\;18\%,\;8\% \qquad \Sigma = 100\;\checkmark \]

"Between 20 and 40" spans two classes — add their relative frequencies.

\[ 0.40 + 0.18 = \mathbf{0.58} = \mathbf{58\%} \text{ of calls} \]

Marker's note: always write the two check-sums (1.00 and 100). They cost one line and they catch an arithmetic slip before it reaches the answer.

A5Cumulative both ways

Q. For the same call-duration data, construct both the "less than" and the "more than" cumulative frequency distributions, and verify they are consistent.

Why this method: a cumulative table answers "how many below / above a value" without re-reading the data. The two versions must always add back to $n$ at every boundary — that is the built-in check.

Why we do itWhat to write

"Less than" accumulates downwards and is quoted at the upper boundary of each class.

\[ cf_{<}: \; 5,\;17,\;37,\;46,\;50 \]

"More than" accumulates upwards and is quoted at the lower boundary of each class.

\[ cf_{>}: \; 50,\;45,\;33,\;13,\;4 \]

Consistency check at the same boundary — the two must total $n$.

At 20: $17 + 33 = 50 = n$ ✓ At 30: $37 + 13 = 50 = n$ ✓

Boundary	Less than 10	Less than 20	Less than 30	Less than 40	Less than 50
cf (less than)	5	17	37	46	50
Boundary	0 or more	10 or more	20 or more	30 or more	40 or more
cf (more than)	50	45	33	13	4

AnswerThe last "less than" value and the first "more than" value both equal $n=50$, and every pair at a shared boundary sums to 50 — the two distributions are consistent.

A6Reading a cumulative table

Q. Using the cumulative table from A5, answer: (i) how many calls lasted under 30 minutes, and what percentage is that? (ii) what percentage lasted 30 minutes or more? (iii) estimate how many lasted under 25 minutes.

Why this method: (i) and (ii) are direct table look-ups at a class boundary. (iii) falls inside a class, so it needs linear interpolation — the same arithmetic the median formula performs.

Why we do itWhat to write

30 is a class boundary, so read the "less than" column directly, then convert to a percentage.

\[ cf_{<30}=37 \quad\Rightarrow\quad \frac{37}{50}\times100 = \mathbf{74\%} \]

"30 or more" is the complement — read the "more than" column, or subtract.

\[ 50-37 = 13 \quad\Rightarrow\quad \frac{13}{50}\times100 = \mathbf{26\%} \]

25 sits half-way through the 20–30 class. Assume the 20 calls in that class are spread evenly across it, and take the matching fraction.

\[ cf_{<25} \approx cf_{<20} + \frac{25-20}{10}\times f_{20\text{–}30} = 17 + 0.5(20) \]

State the estimate and flag the assumption — that flag is a mark.

\[ cf_{<25} \approx \mathbf{27 \text{ calls}} \;(54\%) \] — assuming a uniform spread within the class, the same assumption the ogive and the median formula make.

A7Unequal class widths

Q. Seventy-seven claims were grouped as below. A colleague plots the raw frequencies as a histogram and concludes that "20–40 is by far the most common band". Is he right? Show the correct heights.

Claim (₹’000)	0–10	10–20	20–40	40–70
No. of claims	12	18	32	15

Why this method: in a histogram the area of a bar represents frequency, not its height. When widths differ, plotting $f$ inflates the wide classes — so plot $f/h$ instead.

Why we do itWhat to write

Write the width of each class first — the differing widths are the whole problem.

$h = 10,\;10,\;20,\;30$

Divide each frequency by its own width to get frequency per unit of the variable.

\[ \frac{12}{10}=1.2,\quad \frac{18}{10}=1.8,\quad \frac{32}{20}=1.6,\quad \frac{15}{30}=0.5 \]

Equivalent presentation: rescale to the smallest width, $h_{\min}=10$.

\[ f^{\text{adj}} = 12,\;18,\;16,\;5 \quad \text{(per 10 units)} \]

Answer the colleague's claim directly.

No. 20–40 holds the most claims only because it is twice as wide. Per ₹10,000 of claim size the densest band is 10–20 (1.8), ahead of 20–40 (1.6).

Marker's note: the sentence "in a histogram area, not height, represents frequency" is what the mark is for. State it before you compute anything.

A8Open-end classes

Q. Monthly household expenditure was reported as: below ₹10,000 → 6 households; 10–20 → 14; 20–30 → 24; 30–40 → 11; ₹40,000 and above → 5. How do you handle the two open classes in order to compute a mean, and what mean do you get?

Why this method: a class with no lower or no upper limit has no midpoint, and the grouped mean needs one. The standard convention is to give the open class the width of its neighbour and say so.

Why we do itWhat to write

State the assumption before using it — an unstated assumption loses the mark even if the arithmetic is right.

Assume each open class has the same width (10) as its neighbour: "below 10" ⇒ 0–10; "40 and above" ⇒ 40–50.

Now every class has a midpoint.

$x = 5,\;15,\;25,\;35,\;45$

Build $f x$ and total both columns.

\[ fx = 30,\;210,\;600,\;385,\;225 \qquad \Sigma f = 60,\;\; \Sigma fx = 1450 \]

Apply the grouped mean formula.

\[ \bar{x} = \frac{\Sigma f x}{\Sigma f} = \frac{1450}{60} = \mathbf{₹24{,}167 \text{ per month}} \]

The examinable caveat: the mean is only as good as the assumption. If a few households in the top class actually spend ₹2 lakh, the true mean is far higher — which is exactly why the median is preferred for open-end income data. Say that and you have pre-answered Topic 5.

A9Missing frequencies

Q. The distribution below has 50 observations and a mean of 25. Find the two missing frequencies $f_1$ and $f_2$.

Class	0–10	10–20	20–30	30–40	40–50
Frequency	4	f₁	16	f₂	6

Why this method: two unknowns need two equations, and the question has handed you exactly two facts — the total $N$ and the mean. Write both as equations and solve.

Why we do itWhat to write

Equation 1 comes from $\Sigma f = N$.

\[ 4+f_1+16+f_2+6 = 50 \;\Rightarrow\; f_1+f_2 = 24 \]

Equation 2 comes from the mean. Midpoints first, then $\Sigma fx = \bar{x}\,N$.

\[ x = 5,15,25,35,45 \] \[ 20+15f_1+400+35f_2+270 = 25(50)=1250 \]

Tidy it into a clean linear equation.

\[ 15f_1+35f_2 = 560 \;\Rightarrow\; 3f_1+7f_2 = 112 \]

Substitute $f_1 = 24-f_2$ from equation 1.

\[ 3(24-f_2)+7f_2 = 112 \;\Rightarrow\; 72+4f_2 = 112 \;\Rightarrow\; f_2 = 10 \]

Back-substitute, then verify — a missing-frequency answer is only safe once it is checked.

\[ f_1 = 14,\quad f_2 = 10 \] Check: $\Sigma f = 4+14+16+10+6 = 50$ ✓ $\Sigma fx = 20+210+400+350+270 = 1250$, $1250/50 = 25$ ✓

A10Crosstabulation

Q. 300 customers were classified by channel and by satisfaction. Complete the crosstabulation with totals, compute row percentages, and say which channel performs better.

Channel	Satisfied	Neutral	Dissatisfied
Online	90	30	30
In-store	80	40	30

Why this method: a crosstab summarises two qualitative variables at once. Raw counts are only comparable across rows when the row totals are equal — so convert to percentages of the row before you compare channels.

Why we do itWhat to write

Add the margins — row totals, column totals, grand total. The grand total must match $n$.

Rows: 150 and 150. Columns: 170, 70, 60. Grand total $=300=n$ ✓

The question compares channels, so divide each cell by its row total.

\[ \text{Online: } \tfrac{90}{150}=60\%,\; \tfrac{30}{150}=20\%,\; \tfrac{30}{150}=20\% \] \[ \text{In-store: } \tfrac{80}{150}=53.3\%,\; \tfrac{40}{150}=26.7\%,\; \tfrac{30}{150}=20\% \]

Compare like with like and name the driver of the difference.

Online performs better: 60% satisfied vs 53.3%. Dissatisfaction is identical at 20% in both — the whole difference sits in the Neutral block, which is 6.7 points larger in-store.

Channel	Satisfied	Neutral	Dissatisfied	Total
Online	90 (60.0%)	30 (20.0%)	30 (20.0%)	150 (100%)
In-store	80 (53.3%)	40 (26.7%)	30 (20.0%)	150 (100%)
Total	170	70	60	300

Marker's note: "dissatisfaction is identical, the gap is in Neutral" is the insight sentence. A table without a sentence like that scores the arithmetic marks only.

Section B — Solve a problem ten fresh tables · one or two lines of reasoning, then the answer

Build the table on paper before you open anything.

B1Sturges

Q. 80 observations range from 12 to 92. How many classes does Sturges' rule suggest, and what width would you use?

Solution

Why: $k$ comes from $n$ alone; $h$ comes from the range divided by $k$, rounded up to something readable.

Sturges on $n=80$.

\[ k = 1+3.322\log_{10}80 = 1+3.322(1.9031)=7.32 \approx 7 \]

Range ÷ classes, rounded up to a round number.

\[ R = 92-12 = 80,\quad h = \tfrac{80}{7} = 11.4 \;\rightarrow\; \mathbf{h=12} \text{ (or 10 with 8 classes)} \]

B2Boundaries

Q. Convert to class boundaries and give the midpoints: 5–9, 10–14, 15–19, 20–24.

Solution

Why: the gap between 9 and 10 is 1, so each limit moves by half of it.

Half the gap.

\[ d = \tfrac{10-9}{2}=0.5 \]

Shift the limits; midpoints are unchanged.

Boundaries 4.5–9.5, 9.5–14.5, 14.5–19.5, 19.5–24.5. Midpoints $x = 7,\,12,\,17,\,22$.

B3Relative frequency

Q. Frequencies 8, 22, 40, 20, 10. Give the relative and percent frequency distributions, and the percentage in the top two classes.

Solution

Why: divide by $n$; "top two classes" means add the last two relative frequencies.

Total first.

\[ n = 8+22+40+20+10 = 100 \]

With $n=100$ the percents are the frequencies themselves.

rel = 0.08, 0.22, 0.40, 0.20, 0.10 (Σ = 1.00); pct = 8, 22, 40, 20, 10. Top two = 30%.

B4Cumulative

Q. Classes 0–20, 20–40, 40–60, 60–80, 80–100 with frequencies 7, 18, 30, 15, 10. Build the "less than" cf and state how many are below 60 and what percentage is 60 or more.

Solution

Why: accumulate downwards; "60 or more" is $n$ minus the cf at 60.

Running totals.

\[ cf_{<}: 7,\;25,\;55,\;70,\;80 \qquad n = 80 \]

Read at 60, then complement.

Below 60 = 55. 60 or more $= 80-55 = 25$, i.e. $\tfrac{25}{80}\times100 = \mathbf{31.25\%}$.

B5Interpolation

Q. Using B4's table, estimate how many observations fall below 50.

Solution

Why: 50 is half-way through the 40–60 class, so take half of that class's frequency on top of the cf at 40.

Uniform spread inside the class.

\[ cf_{<50} \approx 25 + \frac{50-40}{20}\times 30 = 25+15 = \mathbf{40} \] — exactly half the data, so 50 is also the estimated median.

B6Unequal widths

Q. Classes 0–5 (f = 10), 5–10 (f = 15), 10–20 (f = 24), 20–40 (f = 20). Which class is genuinely the densest?

Solution

Why: compare $f/h$, not $f$ — the widths are 5, 5, 10, 20.

Frequency per unit.

\[ \tfrac{10}{5}=2.0,\quad \tfrac{15}{5}=\mathbf{3.0},\quad \tfrac{24}{10}=2.4,\quad \tfrac{20}{20}=1.0 \] Densest is 5–10, even though 10–20 has more observations.

B7Open end

Q. Ages: under 20 → 8; 20–30 → 22; 30–40 → 30; 40–50 → 25; 50 and over → 15. State the midpoints you would use and compute the mean.

Solution

Why: give each open class its neighbour's width (10), then it is an ordinary grouped mean.

Assume 10–20 and 50–60.

$x = 15,\;25,\;35,\;45,\;55$

Grouped mean.

\[ \Sigma f = 100,\quad \Sigma fx = 120+550+1050+1125+825 = 3670 \] \[ \bar{x} = \frac{3670}{100} = \mathbf{36.7 \text{ years}} \]

B8Missing frequency

Q. Classes 0–20, 20–40, 40–60, 60–80 with frequencies 6, f, 10, 4. The mean is 38. Find f.

Solution

Why: one unknown, one fact (the mean) — so one equation, $\Sigma fx = \bar{x}\,\Sigma f$.

Midpoints, then both sides in terms of f.

\[ x = 10,30,50,70 \] \[ 60+30f+500+280 = 38(20+f) \]

Solve and verify.

\[ 840+30f = 760+38f \;\Rightarrow\; 8f = 80 \;\Rightarrow\; \mathbf{f = 10} \] Check: Σf = 30, Σfx = 1140, 1140/30 = 38 ✓

B9Cumulative → simple

Q. Convert this "less than" table into a simple frequency distribution: <10 → 8, <20 → 25, <30 → 52, <40 → 68, <50 → 75.

Solution

Why: each class frequency is the difference between consecutive cumulative totals.

Successive differences; the first class is the first cf itself.

\[ f = 8,\; 17,\; 27,\; 16,\; 7 \qquad \Sigma f = 75 \;\checkmark \] Classes 0–10, 10–20, 20–30, 30–40, 40–50.

B10Crosstab

Q. Of 200 loan applicants: 120 salaried, of whom 84 were approved; 80 self-employed, of whom 44 were approved. Build the crosstab with totals and compare approval rates.

Solution

Why: the groups are different sizes, so compare row percentages, not counts.

Fill in the rejections by subtraction, then the margins.

Salaried 84 / 36 / 120 · Self-employed 44 / 36 / 80 · Totals 128 / 72 / 200 ✓

Row percentages.

\[ \tfrac{84}{120}=\mathbf{70\%} \text{ vs } \tfrac{44}{80}=\mathbf{55\%} \] Salaried applicants are approved at a 15-point higher rate — note that the count of rejections is identical (36 each), which is why percentages, not counts, must be compared.

Textbook bank — every Levin exercise on this topic §2.3 data arrays · §2.4 constructing a frequency distribution

29 exercises Levin §2.3 pp. 18–23 · §2.4 pp. 24–38 Section A above already works six of these shapes in full.

Sort or tally on paper first — that is the examinable part.

2-9Data arrayLevin Ex. 2-9 · p. 23

Q. Service tickets at 20 stores: 823 648 321 634 752 669 427 555 904 586 722 360 468 847 641 217 588 349 308 766. Arrange in a data array and say how many stores are not breaking even (under 475) and how many earn a bonus (over 725).

Solution

Why: a data array is just the values in order — and once sorted, both counts can be read straight off without re-scanning.

Sort ascending.

217 308 321 349 360 427 468 | 555 586 588 634 641 648 669 722 | 752 766 823 847 904

Count each side of the two thresholds.

Below 475: 217, 308, 321, 349, 360, 427, 468 ⇒ 7 stores not breaking even
Above 725: 752, 766, 823, 847, 904 ⇒ 5 stores earn a bonus

2-10Read the arrayLevin Ex. 2-10 · p. 23

Q. Using the same data, how many stores belong on a "watch list" of 550–650 service actions?

Solution

Why: the array from 2-9 is already sorted, so this is a read, not a recount — which is exactly why you build the array first.

Values between 550 and 650 inclusive.

555, 586, 588, 634, 641, 648 ⇒ 6 stores on the watch list.

2-11Build + interpretLevin Ex. 2-11 · p. 23

Q. Mechanic hours: 4.3 2.7 3.8 2.2 3.4 3.1 4.5 2.6 5.5 3.2 6.6 2.0 4.4 2.1 3.3 6.3 6.7 5.9 4.1 3.7. Construct a frequency distribution with 1.0-hour intervals. What does it say about productivity, and is more than 6.0 hours a major or minor problem?

Solution

Classes of width 1.0 from 2.0.

2.0–2.9 → 5 · 3.0–3.9 → 6 · 4.0–4.9 → 4 · 5.0–5.9 → 2 · 6.0–6.9 → 3 (Σf = 20 ✓)

Describe the shape.

The distribution is right-skewed: over half the jobs (11 of 20) finish in under 4 hours, with a thin tail of long jobs.

Answer the management question with a number.

Over 6.0 hours: 6.6, 6.3, 6.7 ⇒ 3 of 20 = 15%.
A minor problem, but not negligible — one job in seven runs long. Worth investigating whether those three were unusually difficult transmissions or a particular mechanic, before treating it as a performance issue.

2-12Two class widthsLevin Ex. 2-12 · p. 24

Q. Speeds of 45 motorists. Build relative frequency distributions with 5 and with 11 equal intervals. (a) Do Orange County motorists match the national claim that no more than 10% exceed 55 mph? (b) Which distribution did you use? (c) What proportion drive between 36 and 59 mph?

Solution

Range 15 to 69, so $R=54$. Five intervals of width 11; eleven of width 5.

5 intervals: $f=3,\,4,\,12,\,18,\,8$ (Σ = 45 ✓)
11 intervals: $f=3,0,2,2,5,3,11,3,8,5,3$ (Σ = 45 ✓)

(a) Count how many exceed 55.

13 of 45 = 28.9%, far above the national 10%. No — Orange County motorists do not follow the national pattern; nearly three times as many speed.

(b) Which distribution answered it.

The 11-interval one. With only 5 wide intervals, 55 falls inside a class and you cannot separate those above it from those below without going back to the raw data.

30 of 45 = 66.7% drive in the safest band. Again the finer distribution is needed, since both limits fall inside classes of the 5-interval version.

The lesson the question is built around: more classes preserve more detail. Choose the class width to match the questions you need to answer, not by habit.

2-15Array + countsLevin Ex. 2-15 · p. 25

Q. Pollution ratings (ppm): 37.2 51.7 68.4 54.2 49.9 33.4 39.8 52.7 60.0 46.1 38.5 49.1. (a) Array highest to lowest. (b) Count in each 10-ppm band. (c) How many exceed 45.0? (d) Largest gap between consecutive samples?

Solution

(a) Descending array.

68.4 60.0 54.2 52.7 51.7 49.9 49.1 46.1 39.8 38.5 37.2 33.4

(b) Band counts.

30.0–39.9 → 4 · 40.0–49.9 → 3 · 50.0–59.9 → 3 · 60.0–69.9 → 2 (Σ = 12 ✓)

46.1, 49.1, 49.9, 51.7, 52.7, 54.2, 60.0, 68.4 ⇒ 8 of 12 samples are excessively polluted — two-thirds.

(d) Largest consecutive gap in the array.

Between 60.0 and 68.4 ⇒ a gap of 8.4 ppm, isolating 68.4 as a possible outlier worth investigating separately.

2-17Goal checkLevin Ex. 2-17 · p. 33

Q. Burgers discarded per shift: 2 16 4 12 19 29 24 7 19 22 14 8 24 31 18 20 16 6. Build frequency and relative frequency distributions with intervals of 5. (a) Is the goal of ≥75% of shifts wasting no more than 16 met? (b) What percentage waste 21 or fewer?

Solution

Classes of width 5 from 0.

0–4 → 2 · 5–9 → 3 · 10–14 → 2 · 15–19 → 5 · 20–24 → 4 · 25–29 → 1 · 30–34 → 1 (Σ = 18 ✓)
Relative: 0.111, 0.167, 0.111, 0.278, 0.222, 0.056, 0.056

(a) Count shifts wasting ≤ 16.

9 of 18 = 50%. The goal is NOT met — 75% was required.
Note the class 15–19 straddles 16, so the frequency distribution alone cannot answer this: you must go back to the raw data. Say so — that is the examinable point.

(b) Shifts wasting ≤ 21.

13 of 18 = 72.2%. The relative frequency distribution gives this directly once the classes align, which is why it is more useful than raw counts for percentage questions.

2-19Negative valuesLevin Ex. 2-19 · p. 34

Q. CPI percentage changes in 30 communities. (a) Array lowest to highest. (b) Frequency distribution with classes −0.5 to −0.2, −0.1 to 0.2, 0.3 to 0.6, 0.7 to 1.0. (c) How many had no change or an increase under 1.0%? (d) Discrete or continuous?

Solution

(a) Sorted.

−0.5 −0.4 −0.3 −0.3 −0.3 −0.1 0.0 0.0 0.0 0.1 0.1 0.1 0.1 0.2 0.2 0.2 0.2 0.3 0.3 0.3 0.4 0.4 0.4 0.5 0.5 0.5 0.5 0.7 0.7 1.0

(b) Tally into the four given classes.

−0.5 to −0.2 → 5 · −0.1 to 0.2 → 12 · 0.3 to 0.6 → 10 · 0.7 to 1.0 → 3 (Σ = 30 ✓)

Values from 0.0 up to but excluding 1.0 ⇒ 23 communities.

(d) The data type.

Continuous. A percentage change can in principle take any value in an interval — it is recorded to one decimal place, but that is a rounding convention, not a restriction on the underlying quantity.

2-21 / 2-22Closed vs open-endedLevin Ex. 2-21, 2-22 · p. 35

Q. Ages of 30 video-recorder buyers: 26 37 40 18 14 45 32 68 31 37 20 32 15 27 46 44 62 58 30 42 22 26 44 41 34 55 50 63 29 22. Build a 6-category closed classification, then a 5-category open-ended one with relative frequencies. What does each add?

Solution

Raw data alone: note the range first.

Ages run 14 to 68 — a very broad market, which is itself the first conclusion.

6-category closed classification, width 10.

10–19 → 3 · 20–29 → 7 · 30–39 → 7 · 40–49 → 7 · 50–59 → 3 · 60–69 → 3 (Σ = 30 ✓)
Relative: 0.100, 0.233, 0.233, 0.233, 0.100, 0.100

What the grouping reveals.

The market is remarkably flat between 20 and 49 — three equal blocks of 7 — with thinner tails below 20 and above 50. Symphony is not selling to a single age group.

5-category open-ended version, and what it adds.

e.g. under 25 → 6 · 25–34 → 8 · 35–44 → 7 · 45–54 → 4 · 55 and over → 5.

Open ends handle the sparse tails without leaving nearly-empty classes, and the relative frequencies let Symphony compare this market with a different-sized one directly. The cost is that the exact ages in the open classes are lost — so a mean computed from it needs an assumption.

2-23Too many classesLevin Ex. 2-23 · p. 35

Q. John Lyon has 17 intervals of purchase amounts and says this is cumbersome. Can you simplify without losing too much value?

Solution

AnswerYes — combine adjacent intervals into wider ones. Levin's guidance is 6 to 15 classes; 17 is above that, and with 17 two-number labels the table is hard to read.

How: merge in pairs or triples to give roughly 6 classes — for example $0–6.99, $7–12.99, $13–18.99, $19–24.99, $25–36.99, $37 and over. The frequencies simply add.

What is lost, and why it does not matter here: you can no longer distinguish a $19 purchase from a $21 one. But John's decisions — stocking, staffing, checkout design — turn on broad spending bands, not on £2 differences. Detail that no decision depends on is not value.
Keep the last class open-ended ("$37 and over") since the upper tail is thin — that alone removes several near-empty rows.

2-24 / 2-28 / 2-29Too few classesLevin Ex. 2-24, 2-28, 2-29 · pp. 35–36

Q. (2-24) A run-time distribution has only the midpoints 25, 35, 45 — can the coach get enough information? (2-28) Aircraft noise is grouped with midpoints at 100 and 130 dB; is that adequate for residents suing the airport? (2-29) Would the airport's lawyer approve of those midpoints?

Solution

Why: the mirror image of 2-23 — here there are too few classes, and the three parts show that "adequate" depends on who is asking.

(2-24) Three midpoints means three classes of width 10.

No. Three classes hide almost everything — a runner at 21 minutes and one at 29 land in the same class, yet for a coach that difference is the entire point. Six to eight classes of 2–3 minutes would let him group athletes by training need.

(2-28) Two midpoints, 100 and 130, means a class width of 30 dB.

Not adequate for the residents. The class centred on 130 spans roughly 115–145 dB, which lumps "loud" together with "almost deafening" (over 140). The residents need to show how often readings exceed 140, and this grouping cannot show that.

(2-29) The same table, the opposite interest.

Yes, the airport's lawyer would happily approve it — precisely because it conceals the extremes. Wide classes hide the worst readings inside an average-sounding class.

The lesson: class width is not a neutral technical choice. It determines what a table can and cannot prove, and an interested party may choose it deliberately.

2-25Find the errorsLevin Ex. 2-25 · p. 35

Q. Barney's french-fry waste distribution: 0.0–3.9 → 37, 4.0–7.9 → 46, 8.0–11.9 → 23, 12.0–16.9 → 27, 17.0–25.9 → 7, 26.0–40.9 → 0, total 180. List and explain the errors.

Solution

Why: a checklist question — run through the requirements a frequency distribution must satisfy and name each failure.

The rule brokenWhere and why it matters

1. Frequencies do not sum to the stated total.

$37+46+23+27+7+0 = \mathbf{140}$, not the 180 printed. One of the two figures is wrong, and every relative frequency computed from this table is therefore wrong.

2. Class widths are unequal.

Widths are 4, 4, 4, 5, 9, 15. A histogram of raw frequencies would badly misrepresent the wide upper classes, and no simple class-width formula applies.

3. An empty final class.

26.0–40.9 has frequency 0. Either drop it, or merge it with the class below — an empty class at the end contributes nothing and stretches the axis.

4. Gaps between classes.

The series is inclusive (3.9 then 4.0), so for a histogram or ogive the limits must first be converted to boundaries (3.95, 7.95, …). As printed, the classes do not touch.

2-26 / 2-27 / 2-32Classification typeLevin Ex. 2-26, 2-27, 2-32 · pp. 36–37

Q. (2-26) Build a discrete closed classification for marital status, and a 3-category open-ended one. (2-27) For stock listings, classify by industry / closing price / price change as quantitative or qualitative, continuous or discrete, open or closed. (2-32) For ice-cream flavours and amounts sold, answer the same questions.

Solution

(2-26) Marital status is a category, not a measurement.

Closed, discrete: Single · Married · Divorced · Widowed · Separated.
3-category open-ended: Single · Married · Other — the open class absorbs every remaining possibility.

(2-27) Three different variables from one listing.

(a) Industry — qualitative, discrete, closed (a fixed list of industries).
(b) Closing price — quantitative, continuous, closed (bounded below by 0).
(c) Price change — quantitative, continuous, open-ended at both ends, since there is no fixed limit either way.

And the last part: if the change were reported only as "higher / lower / unchanged", it becomes qualitative — specifically ordinal, since the three states have a natural order.

(2-32) Ice cream.

(a) Flavour — qualitative, discrete, closed (55 known flavours).
(b) Amount sold — recorded to the nearest half gallon, so discrete as recorded but continuous underlying; closed, since no more than 20 gallons of a flavour is ever used in a day.
(c) Together the data are both — flavour is qualitative, amount is quantitative.
(d) What Martin should do: record the amount to a finer precision than the half gallon, and record stockouts as well as sales — a flavour that sells out understates its true demand, and no amount of sales data will reveal that.

2-30Two class countsLevin Ex. 2-30 · p. 36

Q. FAA waiting times for 50 applications, running 22 to 51 days. (a) Build 10 equal closed intervals — which holds the most? (b) Build 5 — which holds the most? (c) What would a relative frequency distribution add?

Solution

Range first.

$R = 51-22 = 29$. Ten classes ⇒ width 3; five classes ⇒ width 6.

(a) Ten classes of width 3.

22–24 → 3 · 25–27 → 3 · 28–30 → 6 · 31–33 → 12 · 34–36 → 8 · 37–39 → 6 · 40–42 → 5 · 43–45 → 4 · 46–48 → 2 · 49–51 → 1 (Σ = 50 ✓)
Most populous: 31–33 days.

(b) Five classes of width 6.

22–27 → 6 · 28–33 → 18 · 34–39 → 14 · 40–45 → 9 · 46–51 → 3 (Σ = 50 ✓)
Most populous: 28–33 days.

They convert counts into probabilities the president can act on: 18/50 = 36% of applications are ruled on in 28–33 days, and 48/50 = 96% within 45 days. That is a planning statement; a raw count of 18 is not.

Both groupings agree that the ruling is most likely in the low 30s — a good sign that the pattern is real and not an artefact of where the class boundaries happened to fall.

2-31Choosing class marksLevin Ex. 2-31 · p. 37

Q. Cars sold by 40 salespeople, values 5 to 15. (a) What class marks would you want, based on frequency? (b) Build a frequency and relative frequency distribution with evenly spaced intervals at least two cars wide. (c) Which answer better identifies the unsatisfactory group (under 7 cars)?

Solution

(a) Look at where the data actually concentrates.

Counts: 5 → 7, 6 → 4, 7 → 4, 8 → 8, 9 → 3, 10 → 6, 11 → 2, 12 → 2, 13 → 1, 14 → 2, 15 → 1.
The data cluster at 5, 8 and 10, so class marks near those values keep the structure visible.

(b) Width-2 classes.

5–6 → 11 · 7–8 → 12 · 9–10 → 9 · 11–12 → 4 · 13–14 → 3 · 15–16 → 1 (Σ = 40 ✓)
Class marks 5.5, 7.5, 9.5, 11.5, 13.5, 15.5
Relative: 0.275, 0.300, 0.225, 0.100, 0.075, 0.025

The raw data (a), not the grouped distribution. "Fewer than seven cars" is 11 salespeople — but the class 7–8 straddles the cutoff, so the grouped table cannot separate them.

The general point: if a decision threshold falls inside a class, the grouping has destroyed exactly the information you need. Set class boundaries at the thresholds that matter.

2-33Class marks → boundariesLevin Ex. 2-33 · p. 37

Q. Doug's six class marks for daily passengers are 84.5, 104.5, 124.5, 144.5, 164.5, 184.5; for cars they are 26.5, 34.5, 42.5, 50.5, 58.5, 66.5. What are the upper and lower boundaries of each set of classes?

Solution

Why: the reverse of finding a midpoint. The gap between consecutive marks is the class width; each boundary sits half a width either side of its mark.

People: consecutive marks differ by 20, so $h=20$ and each boundary is 10 from its mark.

74.5–94.5 · 94.5–114.5 · 114.5–134.5 · 134.5–154.5 · 154.5–174.5 · 174.5–194.5

Cars: marks differ by 8, so $h=8$ and each boundary is 4 from its mark.

22.5–30.5 · 30.5–38.5 · 38.5–46.5 · 46.5–54.5 · 54.5–62.5 · 62.5–70.5

The check: consecutive classes must touch — 94.5 ends one and begins the next. If your boundaries leave a gap or overlap, the width was computed wrongly.

2-13 / 2-14 / 2-16 / 2-18 / 2-20Array & distribution drillsLevin Ex. 2-13, 2-14, 2-16, 2-18, 2-20 · pp. 24–34

Q. Five further exercises on the same two skills: 2-13 array concrete-strength data and answer three threshold questions; 2-14 array cable-TV data and draw a conclusion; 2-16 examine SAT differentials against GPA spreads; 2-18 build a relative frequency distribution with 4.0 lb/sq in intervals; 2-20 build frequency and relative frequency distributions for "before" and "after" survey responses.

Solution

Why: all five are the same two moves — sort into an array, then either read thresholds off it or tally into classes and divide by $n$. The data tables sit in the printed pages; the method is identical every time.

ExerciseWhat to do

2-13 concrete strength

Array highest to lowest. Then (a) count values below 2,500 lb/sq in; (b) count those in [2,497, 2,504]; (c) list repeated values with their counts — repeats are visible instantly in an array and invisible in raw data.

2-14 cable channels vs viewing hours

Array the pairs by channels. The conclusion to look for: as channels purchased rise, hours watched tend to fall — a negative association, which is counter-intuitive and therefore the point of the exercise.

2-16 SAT differential vs GPA spread

Two derived variables — compute each student's differential and spread first, then array or plot the pairs. This is the setup for a scatter diagram (Topic 3).

2-18 4.0 lb/sq in intervals

Tally into classes of width 4.0, divide each frequency by $n$, check $\Sigma\text{rel}=1.00$. Conclusion: describe where the strengths concentrate relative to the 2,500 specification.

2-20 Baggit before/after

Two distributions on the same classes (1–2, 3–4, 5–6, 7–8, 9–10) so they can be compared. Why relative frequencies are essential here: only they let you compare the shift in the "after" responses against the "before" ones on equal terms — and the shift is the whole purpose of the survey.

SC 2-1 / SC 2-2 / SC 2-3 / SC 2-4Self-check setLevin SC 2-1 to SC 2-4 · pp. 26–38

Q. SC 2-1 compares a 7-interval and a 13-interval relative frequency distribution of participants' ages. SC 2-2 builds data arrays of GPA pairs. SC 2-3 builds distributions of delivery times with 6-day intervals. SC 2-4 asks what is wrong with a temperature distribution.

Solution

SC 2-1 — which grouping answers the question.

Both distributions show about 90% of participants are older than 50, so the programme is not in compliance. Both are equally easy to use for that. But the 13-interval version is better for estimating the proportion in the 40s, because it has a distinct 45–49 class where the 7-interval version lumps all of 40–49 together.

SC 2-2 — two arrays of the same paired data.

Array once by high-school GPA and once by college GPA. The conclusion: high GPAs at one level tend to go with high GPAs at the other, though with clear exceptions — an association, not a rule.

SC 2-3 — delivery times, 6-day intervals.

1–6 → 4 · 7–12 → 8 · 13–18 → 4 · 19–24 → 4 (Σ = 20 ✓); relative 0.20, 0.40, 0.20, 0.20.
(a) Order processing is concentrated in 7–12 days with a worrying 20% taking 19 days or more.
(b) No — the goal is "half within 10 days", but 10 falls inside the 7–12 class, so the distribution cannot resolve it. (From the raw data it is 8 of 20 = 40%, so the goal is missed.)
(c) Relative frequencies allow comparison with a different-sized sample and express results as proportions of the whole.

SC 2-4 — the faulty temperature distribution.

Three faults: not all-inclusive (500 °C and 541–549 °C are omitted, which is why only 360 of 365 days appear); closed on the high end, discarding everything above 600 °C; and the classes are not mutually exclusive — 530 °C and 580 °C each fall in two intervals at once.

Topic 3 of 5 · choose-and-construct questions · 3–5 marks

Charts — Choosing One, Constructing It, and Reading It Back

Two kinds of question live here. Choosing ("which graph would you use, and why?") is answered from the decision table and is nearly free marks. Constructing is really arithmetic in disguise — a pie chart is a multiplication by 360, a histogram is a frequency-density calculation, an ogive is a cumulative table plotted at boundaries.

How to know a question belongs to this topic

It asks you to

Draw or read a picture — choose a chart, construct one, or extract a value from one.

Give-away words

construct a histogram construct an ogive which graphical method describe the general shape about what proportion the approximate middle value more-than cumulative

The ogive is the workhorse

Nine of Levin's twelve exercises here ask for an ogive, because it answers "what proportion is below X?" and "what is the middle value?" by reading off a curve. If the question says about or approximately, it wants a graphical reading.

Don't confuse it with

If the question…	then…
asks only for a table	Topic 2
asks for the median by formula	Topic 5
asks for the median off a curve	Here — same answer, different route

Master Method — Topic 3

Formula bank

Pie sector angle

\[ \theta_i = \frac{f_i}{n}\times 360^\circ = \text{(relative frequency)}\times 360^\circ \qquad \Sigma\theta_i = 360^\circ \]

Angle → count

\[ f_i = \frac{\theta_i}{360^\circ}\times n \]

Histogram bar height
(equal widths)

\[ \text{height} = f_i \qquad \text{bars drawn on class \textbf{boundaries}, touching} \]

Histogram bar height
(unequal widths)

\[ \text{height} = \frac{f_i}{h_i} \qquad\text{because \textbf{area}, not height, represents frequency} \]

Ogive points

\[ \text{“less than”: plot } (\text{upper boundary},\, cf) \qquad \text{“more than”: plot } (\text{lower boundary},\, cf) \]

Median off an ogive

\[ \text{read the } x \text{ where the “less than” curve crosses } cf=\frac{n}{2} \]

The decision table — the one page to memorise

Data	Tabular methods	Graphical methods
Qualitative nominal or rank	Frequency · relative frequency · percent frequency · crosstabulation	Bar graph (bars separated) · Pie chart
Quantitative discrete or continuous	Frequency · relative · cumulative · cumulative relative · stem-and-leaf	Histogram (bars touch) · Ogive · Scatter diagram

This is the deck's closing slide slide 49. Every "which chart?" mark in this unit is somewhere on it.

The routine — answering any "which chart and why" question

Name the data type in the words of the decision table — "qualitative, nominal" or "quantitative, continuous". Never skip to the chart.
Name the table that type produces. The chart is drawn from the table, not from the raw data.
Name the chart, and add the one distinguishing detail: bars separated (bar graph) / bars touching (histogram) / angle = relative × 360° (pie) / plotted at upper boundaries (ogive).
Justify in one clause — "because the categories are distinct", "because the classes are adjacent intervals on a continuous scale", "because these are shares of one whole".
If the question asks you to read the chart, always add the interpretive sentence: what the pattern means for the business, and whether that is good or bad.

Pattern recognition

Pattern	Wording cue	What to do
Choose the chart	"which graphical method would you use and why"	Type → table → chart → one-clause justification.
Pie construction	"represent by a pie chart / find the central angles"	Relative frequency × 360°; show the angles adding to 360°.
Pie read-back	"a sector of 108° out of 720 respondents"	$\theta/360$ gives the proportion; × $n$ gives the count.
Histogram	"draw a histogram of the following distribution"	Convert inclusive limits to boundaries first, then check whether widths are equal.
Ogive	"draw the cumulative frequency curve and find the median"	cf at upper boundaries → smooth curve → read $x$ at $n/2$ → confirm with the median formula.
Stem-and-leaf	"construct a stem-and-leaf display"	Leading digits = stems, last digit = leaves, leaves in rank order, state the leaf unit.
Scatter	"comment on the relationship"	Name the direction (positive / negative / none) and add "the relationship is not perfect — the points do not lie on a straight line".
Critique a chart	"is this graph misleading?"	Check: does the $y$ axis start at zero? are the widths equal? is the area proportional? is the sample stated?

Traps

Histogram for qualitative data. Ratings, brands and regions get a bar graph, with gaps between the bars.
Drawing a histogram of an inclusive series on the raw limits — you get gaps. Convert to boundaries first.
Plotting an ogive at midpoints. Cumulative frequency is complete only at the class boundary.
Reading causation off a scatter diagram. The deck's own wording is "associated with", never "caused by" slide 45.
Losing the leaf unit. "8 | 6" is 86 if the leaf unit is 1 and 8.6 if it is 0.1 slide 39.

Section A — Ten questions, fully explained including the two charts you may actually be asked to draw

A1Choose the chart

Q. Name the data type, the tabular summary and the graph, with a one-line justification, for: (a) number of employees in each of six departments; (b) monthly salaries of 200 employees; (c) share of five brands in a market; (d) years of experience vs annual bonus for 45 staff; (e) exam grades A/B/C/D of 300 students; (f) "how many staff earn less than ₹50,000?"

Why this method: six sub-parts, one routine. Type first, table second, chart third, justification fourth — write it as four short clauses per item and the marker can tick straight down.

Why we do itWhat to write

Departments are unordered labels; the counts are shares of a workforce but the question asks for a comparison, not a whole.

(a) Qualitative nominal · frequency distribution · bar graph — the six departments are distinct categories, so the bars are separated.

Salary is measured on a continuous scale, so it must be grouped into intervals.

(b) Quantitative continuous · grouped frequency distribution · histogram — the classes are adjacent intervals, so the bars touch.

Market share is explicitly parts of one whole.

(c) Qualitative nominal · percent frequency distribution · pie chart — each share of $r$ occupies $r\times360^\circ$.

Two quantitative variables recorded on the same person.

(d) Two quantitative variables · paired data table · scatter diagram — it displays the relationship between the two.

Grades are ordered categories, so order must be preserved on the axis.

(e) Qualitative rank · frequency + percent frequency · bar graph with the bars in grade order — a pie would lose the ordering.

A "below a value" question is a cumulative question by definition.

(f) Quantitative, cumulative · “less than” cumulative frequency distribution · ogive — read the height directly above ₹50,000.

Marker's note: part (e) is the discriminator. Most answers say "pie chart" for grades; the better answer keeps the rank order visible, which only a bar graph does.

A2Pie construction

Q. Twenty-five customers rated a service Poor 3, Average 5, Good 11, Excellent 6. Construct the pie chart — give relative frequencies, percentages and central angles.

Why this method: a pie chart is a percent frequency distribution wrapped into a circle. Compute the proportion once, then reuse it for both the percentage and the angle.

Why we do itWhat to write

Proportion of the whole, computed once per category.

\[ \frac{f}{n} = \frac{3}{25},\;\frac{5}{25},\;\frac{11}{25},\;\frac{6}{25} = 0.12,\;0.20,\;0.44,\;0.24 \]

The circle is 360°, so a share of $r$ claims $r$ of those 360 degrees.

\[ \theta = 0.12(360)=43.2^\circ,\;\; 0.20(360)=72^\circ,\;\; 0.44(360)=158.4^\circ,\;\; 0.24(360)=86.4^\circ \]

The check that must always be written.

\[ 43.2+72+158.4+86.4 = 360^\circ \;\checkmark \]

Rating	f	Relative	Percent	Angle
Poor	3	0.12	12%	43.2°
Average	5	0.20	20%	72.0°
Good	11	0.44	44%	158.4°
Excellent	6	0.24	24%	86.4°
Total	25	1.00	100%	360.0°

A3Pie read-back

Q. A pie chart of 720 survey respondents shows a sector of 108° for "prefers online banking". How many respondents is that, and what percentage? If a second sector is 25% of the circle, what is its angle?

Why this method: the pie formula runs in both directions. Going backwards, the angle over 360 is the proportion; multiply by $n$ for the count.

Why we do itWhat to write

Turn the angle into a proportion of the whole circle.

\[ \frac{\theta}{360} = \frac{108}{360} = 0.30 \;=\; \mathbf{30\%} \]

Apply that proportion to the number of respondents.

\[ f = 0.30 \times 720 = \mathbf{216 \text{ respondents}} \]

Forward direction for the second sector.

\[ \theta = 0.25 \times 360 = \mathbf{90^\circ} \quad (=180\text{ respondents}) \]

A4Bar vs histogram

Q. A student draws a chart with touching bars for "number of employees by department" and a chart with gaps for "salary distribution in ₹10,000 bands". Both are wrong. Explain each error and state the rule.

Why this method: the gap between bars is not decoration — it encodes whether the horizontal axis is a set of categories or a continuous number line.

Why we do itWhat to write

Departments are separate categories; touching bars falsely imply that one department "flows into" the next.

Error 1: a bar graph was needed. Qualitative data ⇒ bars must be separated.

Salary bands are adjacent intervals on one continuous scale; a gap falsely implies no one earns the boundary value.

Error 2: a histogram was needed. Quantitative data in class intervals ⇒ bars must touch, drawn on class boundaries.

State the rule in one memorable line.

The gap encodes the data type: gaps = categories (bar graph), no gaps = a continuous scale (histogram).

A5Histogram, unequal widths

Q. Draw the histogram for: 0–10 (f = 12), 10–20 (f = 18), 20–40 (f = 32), 40–70 (f = 15). State the height of each bar and explain your choice.

Why this method: a histogram represents frequency by area. When the widths differ, keeping height = frequency multiplies the wide classes by their extra width and distorts the picture.

Why we do itWhat to write

Write the widths — this is the line that justifies everything after it.

$h = 10,\;10,\;20,\;30$ — unequal, so raw frequencies must not be plotted.

Height = frequency density = frequency per unit of the variable.

\[ \frac{12}{10}=1.2,\quad \frac{18}{10}=1.8,\quad \frac{32}{20}=1.6,\quad \frac{15}{30}=0.5 \]

Verify the areas reproduce the original frequencies — that is the definition being satisfied.

\[ 1.2(10)=12,\; 1.8(10)=18,\; 1.6(20)=32,\; 0.5(30)=15 \;\checkmark \]

Say what the corrected picture shows.

The tallest bar is 10–20, not 20–40. The 20–40 class only looks dominant because it is twice as wide.

Heights are frequency densities. Each bar's area equals its class frequency.

A6Ogive & median

Q. Call durations: 0–10 (5), 10–20 (12), 20–30 (20), 30–40 (9), 40–50 (4). Draw the "less than" ogive, read the median off it, and confirm the reading with the median formula.

Why this method: an ogive is the cumulative table drawn as a curve. Because the curve climbs from 0 to $n$, the point where it crosses $n/2$ is by definition the value below which half the data lies — the median.

Why we do itWhat to write

Build the cf column; cumulative frequency is complete only at the upper boundary of each class.

Points to plot: (10, 5), (20, 17), (30, 37), (40, 46), (50, 50). Start the curve at (0, 0).

Locate the half-way height on the vertical axis.

\[ \frac{n}{2} = \frac{50}{2} = 25 \]

Draw across at cf = 25 to the curve, then down to the $x$ axis.

Reading ≈ 24 minutes.

Confirm algebraically. The median class is the one whose cf first exceeds $n/2$ — here 20–30, since $cf$ jumps 17 → 37.

\[ M = L + \frac{\frac{n}{2}-cf_{\text{prev}}}{f_m}\times h = 20 + \frac{25-17}{20}\times 10 \]

The two agree, which is the point of the exercise.

\[ M = 20 + 4 = \mathbf{24 \text{ minutes}} \quad \text{— identical to the graphical reading} \]

The "less than" ogive. Cross at $cf = n/2 = 25$, drop to the axis, read the median.

A7Stem-and-leaf

Q. Construct a stem-and-leaf display for these 20 service times (minutes), then read off the median and the mode.

Service Time Data

52  57  61  63  64  68  68  70  71  73
74  74  76  79  82  85  86  91  94  97

Why this method: a stem-and-leaf is a histogram that keeps the actual data values — which is why you can read the median straight off it, something a histogram cannot do.

Why we do itWhat to write

Split each value: leading digits become the stem, the last digit the leaf.

Stems 5, 6, 7, 8, 9 (the tens digit). Leaf unit = 1.

Place every leaf in rank order beside its stem.

5 | 2 7 6 | 1 3 4 8 8 7 | 0 1 3 4 4 6 9 8 | 2 5 6 9 | 1 4 7 Leaf unit = 1, so "5 | 2" means 52

$n=20$ is even, so the median is the average of the 10th and 11th values — count leaves from the top.

10th = 73, 11th = 74 \[ M = \frac{73+74}{2} = \mathbf{73.5 \text{ min}} \]

The mode is the most frequent value — look for a repeated leaf on the same stem.

68 appears twice and 74 appears twice ⇒ the data is bimodal: modes 68 and 74.

Marker's note: the display's advantage over a histogram is the sentence to write: it shows both the shape and the actual data values slide 37. The shape here is roughly symmetric with a slight right tail.

A8Leaf unit

Q. Twelve reaction times (seconds): 3.4, 3.7, 3.9, 4.1, 4.2, 4.2, 4.5, 4.8, 5.0, 5.3, 5.5, 5.9. Build the stem-and-leaf display and state what "4 | 5" means. Find the median.

Why this method: the data has one decimal, so the leaf must carry that decimal. Declaring the leaf unit is what makes the display readable — without it the numbers are ambiguous.

Why we do itWhat to write

The units digit becomes the stem; the first decimal becomes the leaf.

3 | 4 7 9 4 | 1 2 2 5 8 5 | 0 3 5 9 Leaf unit = 0.1

Read the notation back using the declared unit.

"4 | 5" means $4 + 5(0.1) = \mathbf{4.5}$ seconds.

$n=12$ is even ⇒ average the 6th and 7th values.

6th = 4.2, 7th = 4.5 \[ M = \frac{4.2+4.5}{2} = \mathbf{4.35 \text{ s}} \]

The trap: if you forget to write "Leaf unit = 0.1", "4 | 5" reads as 45 seconds — an order of magnitude wrong. Where no leaf unit is shown, assume it is 1 slide 38.

A9Scatter diagram

Q. Monthly advertising spend (₹ lakh) and sales (₹ crore) for eight months: (2, 12), (3, 15), (3, 13), (4, 18), (5, 19), (5, 22), (6, 23), (7, 27). Describe the relationship and state what a manager may and may not conclude.

Why this method: a scatter question is graded on three sentences: the direction, the strength ("not perfect"), and the causation caveat. Write all three every time.

Why we do itWhat to write

Check whether sales rise or fall as spend rises — compare the two ends.

At spend 2 → sales 12; at spend 7 → sales 27. Sales rise with spend ⇒ positive relationship.

Look for scatter around the trend — two different sales values occur at the same spend.

At spend 3, sales are 15 and 13; at spend 5, 19 and 22. So "the relationship is not perfect; all plotted points are not on a straight line."

The caveat that separates a description from an interpretation.

The manager may conclude that higher spend is associated with higher sales — not that it causes it. Season, pricing and competitor activity are uncontrolled here.

Source: the middle sentence is lifted from the deck's Panthers example, which pairs "positive relationship" with "the relationship is not perfect" slide 45. Examiners recognise it.

A10Critique a chart

Q. A company's annual report shows a bar chart of quarterly revenue where the $y$ axis runs from ₹95 crore to ₹105 crore, the bars use unequal widths for "Q1–Q2" and "Q3", and 3-D perspective is applied. Revenue actually moved from ₹98 crore to ₹102 crore. What is misleading, and how would you redraw it?

Why this method: chart-critique questions have a fixed checklist. Walk it: axis origin, equal widths, area proportionality, decoration, and whether the base $n$ is stated.

Why we do itWhat to write

A truncated axis exaggerates small differences — the classic distortion.

The $y$ axis starts at 95, not 0. A 4% rise (98 → 102) fills most of the chart and looks like a doubling. Fix: start the axis at zero, or mark the break explicitly.

Combining two quarters into one wide bar breaks the area-proportionality rule.

"Q1–Q2" covers two periods but is drawn as one bar. Fix: equal widths, one bar per quarter — or use frequency density if the merge is unavoidable.

3-D adds a third dimension that carries no data and inflates the apparent size of the front bars.

Drop the 3-D perspective; a flat 2-D bar chart is read more accurately.

Say what the honest chart shows — that is the interpretation mark.

Redrawn from zero with equal widths, the chart shows revenue essentially flat — a rise of about 4% across the year, not the surge the original implies.

Section B — Solve a problem ten fresh questions · one or two lines of reasoning, then the answer

For the "which chart" items, write the justification clause too — that is where the mark is.

B1Choose

Q. Which graph and why: (a) proportion of a budget spent on five heads; (b) heights of 150 recruits; (c) number of accidents per year 2015–2025?

Solution

Why: shares of a whole → pie; one continuous variable → histogram; a quantity over time → a line graph (or a bar graph by year).

Parts of one budget.

(a) Pie chart from a percent frequency distribution.

Continuous measurement, grouped into intervals.

(b) Histogram, bars touching on class boundaries.

Chronological classification — the order of the years carries the message.

B2Pie angles

Q. Expenditure heads: Food 40%, Rent 25%, Transport 15%, Education 12%, Other 8%. Find the central angles.

Solution

Why: percentage ÷ 100 × 360, or simply percentage × 3.6.

Each percentage point is 3.6°.

\[ 144^\circ,\;90^\circ,\;54^\circ,\;43.2^\circ,\;28.8^\circ \qquad \Sigma = 360^\circ \;\checkmark \]

B3Pie read-back

Q. In a pie of 1,200 customers, the "dissatisfied" sector measures 54°. How many customers, and what percentage?

Solution

Why: angle ÷ 360 is the proportion; multiply by $n$.

Proportion then count.

\[ \frac{54}{360}=0.15 = \mathbf{15\%},\qquad 0.15\times1200 = \mathbf{180 \text{ customers}} \]

B4Histogram prep

Q. You must draw a histogram of the inclusive series 10–19, 20–29, 30–39, 40–49. What values go on the horizontal axis?

Solution

Why: a histogram needs a continuous axis, so the limits must first be converted to boundaries.

Shift each limit by $d = 0.5$.

Axis marks at 9.5, 19.5, 29.5, 39.5, 49.5. Drawn on the raw limits the bars would show false gaps.

B5Unequal widths

Q. Classes 0–20 (f = 40), 20–30 (f = 30), 30–50 (f = 40). Which bar is tallest on a correctly drawn histogram?

Solution

Why: plot $f/h$, not $f$; the widths are 20, 10, 20.

Densities.

\[ \tfrac{40}{20}=2.0,\quad \tfrac{30}{10}=\mathbf{3.0},\quad \tfrac{40}{20}=2.0 \] The 20–30 bar is tallest, despite having the fewest observations.

B6Ogive

Q. Classes 0–20, 20–40, 40–60, 60–80 with f = 10, 25, 40, 25. List the points you would plot for a "less than" ogive and estimate the median from it.

Solution

Why: cf plotted at upper boundaries; the median is where the curve crosses $n/2$.

Cumulate, then pair with upper boundaries.

(0,0), (20,10), (40,35), (60,75), (80,100). $n = 100$, so $n/2 = 50$.

Median class 40–60; interpolate.

\[ M = 40+\frac{50-35}{40}\times20 = 40+7.5 = \mathbf{47.5} \]

B7Stem-and-leaf

Q. Build a stem-and-leaf display for: 23, 25, 31, 34, 34, 38, 41, 42, 45, 49, 51, 56. What is the median?

Solution

Why: tens digit = stem, units digit = leaf; then count to the middle.

Leaves in rank order, leaf unit 1.

2 | 3 5 3 | 1 4 4 8 4 | 1 2 5 9 5 | 1 6

$n = 12$ ⇒ average the 6th and 7th.

6th = 38, 7th = 41 ⇒ $M = \mathbf{39.5}$

B8Leaf unit

Q. A display shows "7 | 3" with leaf unit 0.1, and another shows "7 | 3" with leaf unit 1. What value is each?

Solution

Why: the leaf unit is the place value of the leaf digit.

Multiply the leaf by its unit and add to the stem.

Leaf unit 0.1 ⇒ $7 + 3(0.1) = \mathbf{7.3}$. Leaf unit 1 ⇒ $70 + 3 = \mathbf{73}$.

B9Scatter

Q. A scatter of "hours of machine downtime" against "units produced" slopes downward with visible scatter. Write the two-sentence interpretation.

Solution

Why: name the direction, then qualify the strength — and never claim causation from a scatter alone.

Direction + strength + caveat.

"There is a negative relationship between downtime and output — as downtime rises, units produced fall. The relationship is not perfect: the points do not lie on a straight line, so downtime is associated with lower output rather than being its sole cause."

B10Critique

Q. A chart claims "sales doubled" but the $y$ axis runs 480 to 520 and sales moved 490 → 510. Comment.

Solution

Why: check the axis origin before believing any visual comparison.

Compute the real change, then name the distortion.

The rise is $ \frac{510-490}{490} = 4.1\% $, not 100%. The truncated axis (starting at 480, not 0) makes a 4% move fill half the chart. Redraw from zero, or mark an explicit axis break.

Textbook bank — every Levin exercise on this topic§2.5 graphing frequency distributions

12 exercisesLevin §2.5 pp. 38–57Nine of the twelve are ogive questions — learn that one construction well.

For an ogive: cumulate, plot at upper boundaries, join, then read across.

2-34Histogram + shapeLevin Ex. 2-34 · p. 53

Q. Phone-call lengths: 1–7 → 45, 8–14 → 32, 15–21 → 34, 22–28 → 22, 29–35 → 16, 36–42 → 12, 43–49 → 9, 50–56 → 5 (n = 175). Construct a histogram. (a) Describe the shape. (b) With 10 phones and each person knowing their time class, suggest an ordering that minimises waiting.

Solution

Inclusive series ⇒ convert to boundaries before drawing.

0.5–7.5, 7.5–14.5, 14.5–21.5, … so the bars touch. Heights are the frequencies, since all widths are equal (7).

(a) Read the shape off the heights 45, 32, 34, 22, 16, 12, 9, 5.

Strongly right-skewed (positively skewed). Most calls are short, with a long thin tail of very long calls. The pattern is clear: frequency falls steadily as call length rises, apart from a small bump at 15–21.

(b) The queuing question.

Shortest calls first. Sending the 45 people with 1–7 minute calls through first clears most of the queue quickly, so total waiting time across everyone is minimised.

This is the shortest-job-first rule, and it is provably optimal for minimising mean waiting time. Putting the 50-minute callers first would leave 174 people waiting behind them.

2-36Ogive readingLevin Ex. 2-36 · p. 54

Q. Catches on 20 fishing trips: 6500 6700 3400 3600 2000 7000 5600 4500 8000 5000 4600 8100 6500 9000 4200 4800 7000 7500 6000 5400. Break-even is 5,000 lb. Construct an ogive and use it to find (a) roughly what proportion of trips break even, (b) the approximate middle value.

Solution

Group into classes, cumulate, and plot at upper boundaries.

Using classes of width 1,000 from 2,000: $f = 1, 0, 4, 4, 3, 3, 3, 2$ for 2000–2999, …, 9000–9999 (Σ = 20 ✓)
"Less than" cf: 1, 1, 5, 9, 12, 15, 18, 20

(a) Break-even means at or above 5,000 lb — so read the cf at 5,000 and take the complement.

Below 5,000: 3400, 3600, 2000, 4500, 4600, 4200, 4800 ⇒ 7 trips.
\[ P(\text{break even})=\frac{13}{20}=\mathbf{0.65 = 65\%} \]

(b) The middle value is where the curve crosses $n/2 = 10$.

Sorted, the 10th and 11th values are 5,400 and 5,600. \[ \text{median}\approx\mathbf{5{,}500 \text{ lb}} \] Reading the ogive at cf = 10 gives the same figure.

For Homer: 65% of trips break even, and the typical trip lands about 500 lb above break-even. The business is viable but the margin is thin — a third of trips lose money.

2-37Ogive + quartileLevin Ex. 2-37 · p. 54

Q. Pollutants at 150 sites: 5.0–8.9 → 14, 9.0–12.9 → 16, 13.0–16.9 → 28, 17.0–20.9 → 36, 21.0–24.9 → 20, 25.0–28.9 → 16, 29.0–32.9 → 9, 33.0–36.9 → 7, 37.0–40.9 → 4. Construct an ogive to find (a) below what value the lowest quarter fall, (b) which sites to monitor most heavily.

Solution

Cumulate.

cf: 14, 30, 58, 94, 114, 130, 139, 146, 150 (N = 150 ✓)

(a) The lowest quarter is $N/4 = 37.5$ sites. Find where the curve crosses that height.

cf reaches 30 at 12.95 and 58 at 16.95, so 37.5 falls in the 13.0–16.9 class: \[ 12.95 + \frac{37.5-30}{28}\times4 \approx \mathbf{14.0 \text{ ppm}} \] The lowest 25% of sites are below about 14 ppm.

(b) The monitoring decision.

Monitor the upper tail — the 20 sites above 29 ppm (139 to 150 on the cf), which are the top 13%. Reading the ogive from the top down identifies them instantly.
The distribution is right-skewed: most sites cluster at 13–21 ppm with a thin tail of heavily polluted ones. Those few sites are where the environmental risk is concentrated.

2-38"More-than" ogiveLevin Ex. 2-38 · p. 55

Q. River flow (thousand gal/min): 1001–1050 → 7, 1051–1100 → 21, 1101–1150 → 32, 1151–1200 → 49, 1201–1250 → 58, 1251–1300 → 41, 1301–1350 → 27, 1351–1400 → 11. Construct a "more-than" cumulative frequency distribution and ogive.

Solution

Why: a "more-than" ogive accumulates upward and is plotted at lower boundaries — the mirror image of the usual one. For a dam, "how often does flow exceed X?" is the design question, so this is the right curve.

Total first, then accumulate from the top.

$N = 7+21+32+49+58+41+27+11 = \mathbf{246}$

"More than" cf, plotted at the lower boundary of each class.

≥1000.5 → 246 · ≥1050.5 → 239 · ≥1100.5 → 218 · ≥1150.5 → 186
≥1200.5 → 137 · ≥1250.5 → 79 · ≥1300.5 → 38 · ≥1350.5 → 11

Check and interpret.

The first value equals $N$ ✓ and the curve falls from left to right — the defining feature of a "more-than" ogive.

For the Corps of Engineers: flow exceeds 1,300 thousand gal/min on $38/246 = \mathbf{15\%}$ of readings, and exceeds 1,350 on just 4.5%. Those are the figures a spillway must be designed against.

2-40Full pipelineLevin Ex. 2-40 · p. 55

Q. Typesetting times for 50 days, ranging 19.0 to 25.3 minutes. (a) Array lowest to highest. (b) Construct a frequency distribution. (c) Construct an ogive and read off the requested values.

Solution

Why: the complete Unit 1 pipeline in one question — raw data → array → distribution → graph → reading.

(a) Range, then a sensible class width.

$S = 19.0$, $L = 25.3$, $R = 6.3$. Sturges gives $k = 1+3.322\log_{10}50 = 6.6 \approx 7$, so $h = 6.3/7 = 0.9 \to 1.0$.

(b) Classes of width 1.0 from 19.0.

19.0–19.9 · 20.0–20.9 · 21.0–21.9 · 22.0–22.9 · 23.0–23.9 · 24.0–24.9 · 25.0–25.9 — seven classes, tally the 50 values into them and check $\Sigma f = 50$.

Plot (19.95, cf₁), (20.95, cf₂), … up to (25.95, 50), starting the curve at (18.95, 0). Read the median where the curve crosses cf = 25, and any "what proportion under X minutes" straight off the height above X.

2-41Histogram + polygonLevin Ex. 2-41 · p. 56

Q. Monthly CD sales over 35 months: $10,000–12,499 → 2, 12,500–14,999 → 4, 15,000–17,499 → 7, 17,500–19,999 → 5, 20,000–22,499 → 6, 22,500–24,999 → 8, 25,000–27,499 → 2, 27,500–29,999 → 1. (a) Relative frequency distribution. (b) Relative frequency histogram and polygon on the same graph.

Solution

(a) Divide each frequency by 35.

0.057, 0.114, 0.200, 0.143, 0.171, 0.229, 0.057, 0.029 (Σ ≈ 1.000 ✓)

(b) The histogram — bars on class boundaries, touching, heights = relative frequencies.

Boundaries 9,999.5 · 12,499.5 · 14,999.5 · … all of equal width 2,500, so relative frequencies can be plotted directly as heights.

The polygon — plot at midpoints, not boundaries, and join with straight lines.

Midpoints $11{,}249.5,\ 13{,}749.5,\ 16{,}249.5,\ \dots$
Close the polygon by adding a midpoint with zero frequency at each end — otherwise it hangs in mid-air.

The shape: roughly bimodal, with peaks at 15,000–17,499 and 22,500–24,999. Worth flagging to Lee: two distinct kinds of month rather than one typical one.

2-42Ogive + unequal widthsLevin Ex. 2-42 · p. 56

Q. Commission earnings of 130 salespeople: ≤$5,000 → 5, 5,001–10,000 → 9, 10,001–15,000 → 11, 15,001–20,000 → 33, 20,001–30,000 → 37, 30,001–40,000 → 19, 40,001–50,000 → 9. Construct an ogive to find (a) roughly what proportion earn more than $25,000, (b) what the "middle" salesperson earns.

Solution

Note the widths are unequal — 5,000 for the first four classes, then 10,000. For an ogive this does not matter (cumulative frequency is plotted at boundaries), but it would for a histogram.

cf: 5, 14, 25, 58, 95, 114, 123 (N = 123 in the listed classes)

(a) $25,000 falls inside the 20,001–30,000 class, so interpolate.

\[ cf_{<25{,}000} \approx 58 + \tfrac{1}{2}(37) = 76.5 \] \[ P(>25{,}000)=\frac{123-76.5}{123}\approx\mathbf{38\%} \]

(b) The middle salesperson is at cf = $N/2 \approx 61.5$.

cf reaches 58 at $20,000 and 95 at $30,000, so \[ \text{median}\approx 20{,}000+\frac{61.5-58}{37}\times10{,}000\approx\mathbf{\$20{,}950} \]

Both answers required interpolation because the thresholds fall inside classes — which is exactly what an ogive does graphically. Reading a value off the curve is linear interpolation.

2-43Both ogivesLevin Ex. 2-43 · p. 56

Q. Minutes spent on parking-ticket appeals: 2–3 → 40, 4–5 → 40, 6–7 → 90, 8–9 → ?, 10–11 → 50, 12–13 → 50, 14–15 → 30, total 400. Construct a "less-than" cumulative distribution and ogive.

Solution

Recover the missing frequency from the stated total — a Topic 2 skill reappearing.

\[ f_{8\text{–}9} = 400-(40+40+90+50+50+30)=\mathbf{100} \]

Cumulate, and plot at the upper boundaries of the inclusive classes.

cf: 40, 80, 170, 270, 320, 370, 400
plotted at 3.5, 5.5, 7.5, 9.5, 11.5, 13.5, 15.5

Read the median off the curve at cf = 200.

cf reaches 170 at 7.5 min and 270 at 9.5, so \[ \text{median}\approx 7.5+\frac{200-170}{100}\times2=\mathbf{8.1 \text{ minutes}} \] Half of all appeals are dealt with in under about 8 minutes — the figure the administrative officer needs for staffing.

2-35 / 2-39Applied distributionsLevin Ex. 2-35, 2-39 · pp. 54–55

Q. (2-35) Ages of 25 trailer-park residents, to decide whether to waive a clubhouse fee for over-60s. (2-39) Indifference points at which 50 senior citizens would invest $1,000.

Solution

(2-35) Group the ages, then read off the proportion above 60.

Build classes of width 10 from 50, cumulate, and read the cf at 60. The decision figure is the proportion over 60 — if most residents qualify, waiving the fee removes most of the clubhouse income, which is the board member's real problem.
The ages given run from 51 to 96, so a large majority are over 60 — the proposal would gut the social budget.

(2-39) Group the indifference points and build an ogive.

The ogive answers Pamela's actual question directly: "at a return of $r$, what proportion of seniors will invest?" — read the height of the "more-than" curve above $r$.

That curve is a demand schedule: it tells her what return she must offer to attract any given share of the market, which is precisely how she should design the product.

SC 2-5Read a histogramLevin SC 2-5 · p. 57

Q. A histogram of weights (82.5 to 217.5 lb) for a ski-lift operator. (a) Describe the tails. (b) What does that let the operator do?

Solution

(a) Compare the two ends of the distribution.

The lower tail is fatter than the upper — more observations sit at low weights than at high ones.

(b) Turn the shape into an operating decision.

Because very few people weigh 180 lb or more, the operator can pair each heavy rider with a lighter one to balance the chair — and can do so without noticeably delaying anyone, since heavy riders are rare enough that a suitable partner is almost always in the queue.

Matches Levin's printed answer. The transferable point: the shape of a distribution, not its average, is what made the operating rule possible. A mean weight would have told the operator nothing useful here.

SC 2-6Ogive, two readingsLevin SC 2-6 · p. 53

Q. Birth weights of 200 premature babies: 0.5–0.9 → 10, 1.0–1.4 → 19, 1.5–1.9 → 24, 2.0–2.4 → 27, 2.5–2.9 → 29, 3.0–3.4 → 34, 3.5–3.9 → 40, 4.0–4.4 → 17. Construct an ogive to find (a) the approximate middle value, (b) roughly what percentage need an incubator (under 3.0 lb).

Solution

Cumulate; plot at upper boundaries 0.95, 1.45, 1.95, …

cf: 10, 29, 53, 80, 109, 143, 183, 200 (N = 200 ✓)

(a) Median at cf = 100 — that falls in the 2.5–2.9 class (cf runs 80 → 109).

\[ \text{median}\approx 2.45+\frac{100-80}{29}\times0.5=\mathbf{2.79 \text{ lb}} \]

(b) Under 3.0 lb is the cf at the boundary 2.95.

\[ \frac{109}{200}=\mathbf{54.5\%} \] More than half of Central Carolina's premature babies need an incubator — a direct capacity-planning number for the hospital.

2-1Class width formulaLevin Eq. 2-1 · p. 28

Q. State and explain Levin's formula for the width of class intervals.

Solution

The formula, and the one subtlety in it.

\[ \text{width of class intervals} = \frac{\text{next unit value after the largest} - \text{smallest value}}{\text{total number of class intervals}} \]
Decide the number of classes first — Levin advises 6 to 15 — then compute the width.

The subtlety: the numerator uses the next value of the same units, not simply $L-S$. This is because the width measures the interval from the first value of one class to the first value of the next, so the last class must have room for its own top value.

Topic 4 of 5 · computation questions · the syllabus's named measure for Unit 1

The Arithmetic Mean — Every Variant You Can Be Asked

One idea, seven disguises. Every version below is still $\Sigma fx / \Sigma f$; what changes is what you have to build before you can use it — midpoints, weights, a coded column, or a frequency recovered from a cumulative table. Recognise the disguise and the arithmetic is three lines.

How to know a question belongs to this topic

It asks you to

Compute an average and compare it against a threshold. Almost every one of Levin's exercises ends in a decision — do they qualify? should he build? should she recalibrate?

Give-away words

as long as the average stays below compute the sample mean weighted average average percentage increase growth rate does it qualify? using the coding method

Which mean?

Plain arithmetic mean — a list of equally important values.
Weighted mean — the values have different importance: credits, quantities, plant sizes.
Geometric mean — the values are rates of change over time: growth, inflation, percentage increases. Multiply and take the $n$th root, never average them.

The growth-rate tell

If the question says average percentage increase, growth rate or gives a series of yearly values, it wants the geometric mean. Averaging percentages arithmetically is the single most common error in this topic.

Don't confuse it with

If the question…	then…
asks for the middle value or the most common one	Topic 5
asks how spread out the data is	Unit 2
asks for an average against a threshold	Here

Master Method — Topic 4

Formula bank

Ungrouped

\[ \mu = \frac{\Sigma x}{N} \quad\text{(population)} \qquad\qquad \bar{x} = \frac{\Sigma x}{n} \quad\text{(sample)} \]

Grouped
(direct method)

\[ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \qquad x_i = \frac{\text{lower} + \text{upper}}{2} \ \text{(class midpoint)} \]

Step-deviation
(assumed mean)

\[ \bar{x} = A + h\cdot\frac{\Sigma f_i d_i}{\Sigma f_i}, \qquad d_i = \frac{x_i - A}{h} \]

Weighted mean

\[ \bar{x}_w = \frac{\Sigma w_i x_i}{\Sigma w_i} \]

Combined mean
(two groups)

\[ \bar{x}_{12} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2} \]

Combined mean
(k groups)

\[ \bar{x}_c = \frac{\Sigma n_i \bar{x}_i}{\Sigma n_i} \]

Corrected mean

\[ \Sigma x_{\text{correct}} = \Sigma x_{\text{wrong}} - x_{\text{wrong}} + x_{\text{right}} \] \[ \bar{x}_{\text{correct}} = \frac{\Sigma x_{\text{correct}}}{n} \qquad (n \text{ is unchanged}) \]

Effect of a
linear change

\[ \text{if } y = a + bx \ \text{ then } \ \bar{y} = a + b\bar{x} \]

$A$ is any convenient midpoint (pick the one with the largest frequency, or the middle class); $h$ is the common class width. Step-deviation and the direct method always give the same answer — it is a shortcut, not a different statistic.

The routine — grouped mean, every time

Look at the class limits. Is there a gap between them (inclusive) or do they touch (exclusive)? Write which it is. For the mean only, an inclusive series needs no conversion.
Build the midpoint column $x_i$. $(\text{lower}+\text{upper})/2$. Open-end class? Give it its neighbour's width and say so.
Build the $f_ix_i$ column. Row by row. Do not try to do it in your head.
Total both columns: $\Sigma f$ and $\Sigma fx$. Write both totals on a ruled "Total" row — that layout is what your class notes use.
Divide, then say it in words with the unit: "the average post-operative stay is 7.7 days".
Sanity-check: the mean must land inside the range of the midpoints, and near the classes with the biggest frequencies. If it does not, you have an arithmetic slip.

Pattern recognition — which variant is this?

Pattern	Wording cue	Formula to reach for
Plain list of numbers	"the marks of 12 students are…"	$\bar{x}=\Sigma x/n$. Check whether it is a sample or a population, and use the matching symbol.
Class intervals given	a table of classes and frequencies	$\bar{x}=\Sigma fx/\Sigma f$ with midpoints.
Large / awkward midpoints	midpoints in hundreds or thousands, equal widths	Step-deviation — it turns big numbers into −3 … +3.
Items have different importance	"credits", "weights", "quantities purchased"	Weighted mean $\Sigma wx/\Sigma w$. Never the simple mean.
Two or more groups	"40 men averaged … and 60 women averaged …"	Combined mean. Weight by the group sizes.
A value was misread	"it was later found that 25 was actually 52"	Rebuild $\Sigma x$ from the old mean, swap the value, re-divide.
One frequency missing	"…and the mean is 38. Find the missing frequency."	Set $\Sigma fx = \bar{x}\,\Sigma f$ and solve for the unknown.
Cumulative table given	"less than 10 → 8, less than 20 → 25 …"	Difference consecutive $cf$ to get $f$ first, then the ordinary grouped mean.

How each of your sources says it

Source	Notation and layout it uses
Class notes	Writes $\mu = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$ and lays the work out as four columns — CI · xᵢ · fᵢ · fᵢxᵢ — with $\Sigma f_i$ and $\Sigma f_ix_i$ on the bottom row. Copy this layout exactly notes p. 4
Class notes	The red rule: inclusive series need NOT be converted for the mean; conversion IS required for median and mode notes p. 3
Levin	§3.2 arithmetic mean, §3.3 weighted mean — and the caution that the mean is pulled by extreme values pp. 77–95
Exam	Never asks you to compute a grouped mean. It hands you a mean and asks what it means next to the median and mode — Topic 5.

Traps

Using class limits instead of midpoints. $x$ is the midpoint, never the lower limit.
Averaging the group means instead of weighting by group size. $(32{,}000+28{,}000)/2$ is wrong unless the groups are equal in size.
Using the simple mean where weights exist. If the question mentions credits, quantities or importance, it wants $\bar{x}_w$.
Converting an inclusive series before finding the mean. Harmless but wasted time — the midpoint is unchanged either way.
Reporting a bare number. "50.83" scores less than "50.83 marks, i.e. the average student scored just over 50".

Section A — Ten questions, fully explained every column shown, every total checked

A1Ungrouped

Q. The daily sales (₹ '000) of a kiosk over 12 randomly chosen days were: 46, 52, 38, 61, 55, 49, 43, 58, 50, 44, 57, 47. Find the mean and write it with the correct symbol.

Why this method: a plain list with no frequencies — so the mean is the total divided by the count. The only judgement is the symbol: "12 randomly chosen days" out of many possible days makes this a sample.

Why we do itWhat to write

Add the values. Group them in pairs that make round numbers to reduce slips.

\[ \Sigma x = (46+54\text{-ish pairs})\ldots = 600 \] 46+52+38+61+55+49+43+58+50+44+57+47 = 600

Count the values.

\[ n = 12 \]

Randomly chosen days ⇒ a sample ⇒ Roman symbol $\bar{x}$, not $\mu$.

\[ \bar{x} = \frac{\Sigma x}{n} = \frac{600}{12} \]

State it with the unit and in words.

\[ \bar{x} = \mathbf{₹50{,}000 \text{ per day}} \] Mean daily sales are ₹50 thousand.

Sanity check: the values run 38 to 61, so a mean of 50 sits sensibly in the middle. If your answer lands outside the data's range, you have added wrongly.

A2Grouped — exclusive

Q. Eighty invoices were grouped by value (₹). Compute the mean invoice value.

Invoice value (₹)	100–200	200–300	300–400	400–500	500–600
No. of invoices	7	15	28	20	10

Why this method: we no longer have the individual invoices, so each class is represented by its midpoint and counted $f$ times. That is exactly what $\Sigma fx/\Sigma f$ computes.

Why we do itWhat to write

The classes are exclusive (they touch), so the midpoint is simply the average of the two limits.

\[ x = \frac{100+200}{2} = 150,\;\; 250,\;\; 350,\;\; 450,\;\; 550 \]

Build the $fx$ column row by row — this is the whole calculation.

\[ fx = 1050,\; 3750,\; 9800,\; 9000,\; 5500 \]

Total both columns.

\[ \Sigma f = 80, \qquad \Sigma fx = 29{,}100 \]

Divide, then state with the unit.

\[ \bar{x} = \frac{29{,}100}{80} = \mathbf{₹363.75} \text{ per invoice} \]

Class (CI)	xᵢ	fᵢ	fᵢ·xᵢ
100–200	150	7	1,050
200–300	250	15	3,750
300–400	350	28	9,800
400–500	450	20	9,000
500–600	550	10	5,500
Total	—	Σf = 80	Σfx = 29,100

Layout note: this four-column table — CI · xᵢ · fᵢ · fᵢxᵢ with a ruled total row — is exactly the layout used in your class notes p. 4. Reproduce it and the method marks are automatic even if the arithmetic slips.

A3Grouped — inclusive

Q. Ninety employees were grouped by age. Find the mean age. Must the series be converted to exclusive form first?

Age (years)	20–29	30–39	40–49	50–59	60–69	70–79
No. of employees	8	12	20	25	15	10

Why this method: answer the conversion question first — it is a mark on its own, and getting it right saves you the conversion work.

Why we do itWhat to write

Deal with the conversion question explicitly, quoting the rule.

No. For the mean, an inclusive series need not be converted. (It would be required for the median, the mode, a histogram or an ogive.)

Show why: the boundaries would be 19.5–29.5, but the midpoint is unchanged either way.

\[ \frac{20+29}{2}=24.5 \qquad\text{and}\qquad \frac{19.5+29.5}{2}=24.5 \;\checkmark \]

Midpoints of the given limits.

\[ x = 24.5,\;34.5,\;44.5,\;54.5,\;64.5,\;74.5 \]

$fx$ column, then both totals.

\[ fx = 196,\;414,\;890,\;1362.5,\;967.5,\;745 \] \[ \Sigma f = 90,\qquad \Sigma fx = 4575 \]

Divide and state in words.

\[ \bar{x} = \frac{4575}{90} = \mathbf{50.83 \text{ years}} \] The average employee is about 51 years old.

A4Step-deviation

Q. Recompute the mean age of A3 using the step-deviation (assumed mean) method, and confirm it matches.

Why this method: multiplying 74.5 × 10 by hand is where marks get lost. The step-deviation method replaces the midpoints with small integers −3 … +2 by subtracting an assumed mean and dividing by the common width. Because the transformation is linear, the mean transforms back exactly.

Why we do itWhat to write

Pick $A$ = a convenient midpoint. The class with the largest frequency (50–59) is the standard choice, and $h$ is the common class width.

\[ A = 54.5, \qquad h = 10 \]

Code each midpoint. $d$ counts how many class-widths that class sits from $A$.

\[ d = \frac{x-A}{h} = -3,\,-2,\,-1,\,0,\,+1,\,+2 \]

Build $fd$ — now these are small numbers you can add in your head.

\[ fd = -24,\,-24,\,-20,\,0,\,+15,\,+20 \qquad \Sigma fd = -33 \]

Decode: the coded mean is $\Sigma fd/\Sigma f$; multiply back by $h$ and add $A$.

\[ \bar{x} = A + h\cdot\frac{\Sigma fd}{\Sigma f} = 54.5 + 10\left(\frac{-33}{90}\right) \]

Confirm it equals the direct answer — always state that it does.

\[ \bar{x} = 54.5 - 3.67 = \mathbf{50.83 \text{ years}} \quad\text{— identical to A3} \;\checkmark \]

Class	xᵢ	fᵢ	dᵢ = (xᵢ−54.5)/10	fᵢ·dᵢ
20–29	24.5	8	−3	−24
30–39	34.5	12	−2	−24
40–49	44.5	20	−1	−20
50–59	54.5	25	0	0
60–69	64.5	15	+1	+15
70–79	74.5	10	+2	+20
Total	—	Σf = 90	—	Σfd = −33

When to use it: only when the class widths are equal. Unequal widths break the single $h$, and you must fall back on the direct method.

A5Weighted mean

Q. A product's cost is made up of raw material at ₹120/unit, labour at ₹95/unit and overhead at ₹60/unit, consumed in the ratio 5 : 3 : 2. Find the average cost per unit. Why is the simple mean wrong here?

Why this method: the three costs are not consumed equally, so they must not count equally. The weight is the quantity each contributes.

Why we do itWhat to write

Show the wrong answer first, so the contrast is explicit.

\[ \text{Simple mean} = \frac{120+95+60}{3} = 91.67 \] — treats overhead as being as important as raw material, which it is not.

Assign the weights from the consumption ratio.

\[ w = 5,\;3,\;2 \qquad \Sigma w = 10 \]

Build $wx$ and total it.

\[ wx = 600,\;285,\;120 \qquad \Sigma wx = 1005 \]

Divide by the total weight, not by the number of items.

\[ \bar{x}_w = \frac{\Sigma wx}{\Sigma w} = \frac{1005}{10} = \mathbf{₹100.50 \text{ per unit}} \]

The mark-earning sentence: the weighted mean (₹100.50) is higher than the simple mean (₹91.67) because the most expensive component, raw material, is also the most heavily consumed. Ignoring the weights would understate cost by nearly ₹9 a unit.

A6Combined mean

Q. In a firm, 40 managers earn a mean of ₹32,000 a month and 60 executives a mean of ₹28,000. Find the mean salary of all 100 employees. Why is it not ₹30,000?

Why this method: a mean is a total divided by a count, so to combine two means you must first rebuild the two totals, add them, and divide by the combined count.

Why we do itWhat to write

Rebuild each group's total from $\Sigma x = n\bar{x}$.

\[ \Sigma x_1 = 40(32{,}000) = 12{,}80{,}000 \] \[ \Sigma x_2 = 60(28{,}000) = 16{,}80{,}000 \]

Add the totals and add the counts.

\[ \Sigma x = 29{,}60{,}000, \qquad n = 100 \]

Divide.

\[ \bar{x}_{12} = \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} = \frac{29{,}60{,}000}{100} = \mathbf{₹29{,}600} \]

Answer the "why not 30,000" part — this is the actual question.

₹30,000 is the average of the two means, which would only be right if the groups were the same size. There are more executives (60) than managers (40), so the combined mean is pulled toward the lower executive figure.

A7Combined mean, reversed

Q. A class of 100 students has a mean score of 66. The 50 students in Section A averaged 60. What did the 50 students in Section B average?

Why this method: the combined-mean formula has four quantities; give any three and the fourth is fixed. Work with totals, never with the means directly.

Why we do itWhat to write

Rebuild the grand total from the combined mean.

\[ \Sigma x_{\text{all}} = 100 \times 66 = 6600 \]

Rebuild the known group's total.

\[ \Sigma x_A = 50 \times 60 = 3000 \]

Subtract to isolate the unknown group's total.

\[ \Sigma x_B = 6600-3000 = 3600 \]

Divide by that group's size, then check.

\[ \bar{x}_B = \frac{3600}{50} = \mathbf{72} \] Check: $\dfrac{50(60)+50(72)}{100} = \dfrac{6600}{100} = 66$ ✓

A8Missing frequency

Q. The mean of the distribution below is 35. Find the missing frequency $f$.

Class	10–20	20–30	30–40	40–50	50–60
Frequency	10	f	25	20	10

Why this method: the mean is a known relationship between $\Sigma fx$ and $\Sigma f$. Both sides contain $f$, so writing $\Sigma fx = \bar{x}\,\Sigma f$ gives one equation in one unknown.

Why we do itWhat to write

Midpoints first — nothing can be done without them.

\[ x = 15,\;25,\;35,\;45,\;55 \]

Express both totals in terms of $f$.

\[ \Sigma f = 65+f \] \[ \Sigma fx = 150+25f+875+900+550 = 2475+25f \]

Set the mean relationship and clear the fraction.

\[ \frac{2475+25f}{65+f} = 35 \;\Rightarrow\; 2475+25f = 2275+35f \]

Solve.

\[ 200 = 10f \;\Rightarrow\; \mathbf{f = 20} \]

Always verify — it takes one line and catches a sign error.

$\Sigma f = 85$, $\Sigma fx = 150+500+875+900+550 = 2975$, $\;2975/85 = 35$ ✓

A9Corrected mean

Q. The mean of 50 observations was computed as 38.4. It was later found that one observation had been recorded as 25 instead of 52. Find the corrected mean.

Why this method: you cannot "fix a mean" directly — you fix the total. Rebuild $\Sigma x$ from the wrong mean, swap one value, then re-divide by the same $n$.

Why we do itWhat to write

Recover the incorrect total. This is the step everyone forgets.

\[ \Sigma x_{\text{wrong}} = n\bar{x} = 50 \times 38.4 = 1920 \]

Remove the wrong value and insert the right one.

\[ \Sigma x_{\text{correct}} = 1920 - 25 + 52 = 1947 \]

$n$ does not change — no observation was added or removed, only corrected.

\[ n = 50 \ \text{(unchanged)} \]

Divide again.

\[ \bar{x}_{\text{correct}} = \frac{1947}{50} = \mathbf{38.94} \]

Sanity check: the corrected value is larger than the one used, so the mean must rise — and it does, by $27/50 = 0.54$. If your corrected mean moved the wrong way, you subtracted in the wrong order.

A10Mean from a cumulative table

Q. Find the mean from this "less than" cumulative distribution of 75 delivery times (minutes).

Time	< 10	< 20	< 30	< 40	< 50
No. of deliveries	8	25	52	68	75

Why this method: the mean formula needs class frequencies, and a cumulative table does not contain them directly — it contains running totals. Difference consecutive entries to recover $f$ first.

Why we do itWhat to write

The first class frequency is the first cumulative value; each later one is the difference from the previous.

\[ f = 8,\; 25-8=17,\; 52-25=27,\; 68-52=16,\; 75-68=7 \]

Check the recovered frequencies against the final cumulative total.

\[ \Sigma f = 8+17+27+16+7 = 75 \;\checkmark \]

Now it is an ordinary grouped table — classes 0–10, 10–20, … with their midpoints.

\[ x = 5,\;15,\;25,\;35,\;45 \]

Build $fx$ and total.

\[ fx = 40,\;255,\;675,\;560,\;315 \qquad \Sigma fx = 1845 \]

Divide and state.

\[ \bar{x} = \frac{1845}{75} = \mathbf{24.6 \text{ minutes}} \]

Section B — Solve a problem ten fresh means · one or two lines of reasoning, then the answer

Write the four-column table each time — CI · x · f · fx — even for the quick ones.

B1Ungrouped

Q. Eight orders had values 15, 22, 18, 25, 30, 19, 21, 26 (₹'00). Find the mean.

Solution

Why: plain list ⇒ total ÷ count.

Sum, then divide by 8.

\[ \Sigma x = 176,\quad \bar{x} = \frac{176}{8} = \mathbf{22 \;(₹2{,}200)} \]

B2Grouped

Q. Classes 0–10, 10–20, 20–30, 30–40 with frequencies 5, 10, 15, 10. Find the mean.

Solution

Why: exclusive classes ⇒ midpoints straight away, then $\Sigma fx/\Sigma f$.

Midpoints 5, 15, 25, 35.

\[ \Sigma f = 40,\quad \Sigma fx = 25+150+375+350 = 900 \] \[ \bar{x} = \frac{900}{40} = \mathbf{22.5} \]

B3Step-deviation

Q. Classes 0–10 … 40–50 with frequencies 4, 8, 12, 10, 6. Use the step-deviation method with A = 25.

Solution

Why: equal widths (h = 10) make the coded column small integers.

Code the midpoints 5, 15, 25, 35, 45 around A = 25.

\[ d = -2,-1,0,1,2 \qquad fd = -8,-8,0,10,12 \;\Rightarrow\; \Sigma fd = 6 \]

Decode.

\[ \bar{x} = 25 + 10\left(\tfrac{6}{40}\right) = \mathbf{26.5} \] Direct check: Σfx = 1060, 1060/40 = 26.5 ✓

B4Weighted

Q. A student scores 80 in a 4-credit subject, 70 in a 3-credit subject and 60 in a 3-credit subject. Find the weighted average, and compare it with the simple average.

Solution

Why: credits are the weights — subjects do not count equally.

$\Sigma wx/\Sigma w$, then contrast.

\[ \Sigma wx = 320+210+180 = 710,\quad \Sigma w = 10 \] \[ \bar{x}_w = \mathbf{71.0} \quad\text{vs simple mean } \tfrac{210}{3} = 70.0 \] Higher, because the best score carries the most credits.

B5Combined

Q. 30 units averaged 45 kg and 20 units averaged 55 kg. Find the combined mean.

Solution

Why: rebuild both totals, then divide by 50 — never average the two means.

Totals then divide.

\[ \frac{30(45)+20(55)}{50} = \frac{1350+1100}{50} = \mathbf{49 \text{ kg}} \] Not 50 — the larger group has the lower mean.

B6Missing frequency

Q. Classes 0–10, 10–20, 20–30, 30–40 with frequencies 5, f, 15, 10. The mean is 21. Find f.

Solution

Why: one unknown, one fact ⇒ set $\Sigma fx = \bar{x}\,\Sigma f$.

Both totals in terms of f (midpoints 5, 15, 25, 35).

\[ \Sigma f = 30+f,\qquad \Sigma fx = 750+15f \] \[ 750+15f = 21(30+f) = 630+21f \]

Solve and verify.

\[ 120 = 6f \;\Rightarrow\; \mathbf{f = 20} \] Check: Σf = 50, Σfx = 1050, 1050/50 = 21 ✓

B7Corrected mean

Q. The mean of 25 readings was 48. One reading of 36 should have been 63. Find the corrected mean.

Solution

Why: fix the total, not the mean; $n$ is unchanged.

Rebuild, swap, re-divide.

\[ \Sigma x = 25(48) = 1200 \;\rightarrow\; 1200-36+63 = 1227 \] \[ \bar{x} = \frac{1227}{25} = \mathbf{49.08} \]

B8Linear change

Q. The mean of a dataset is 40. What is the mean if (i) every value rises by 5; (ii) every value doubles; (iii) every value is transformed as $(x-10)/2$?

Solution

Why: the mean is linear — $\bar{y} = a + b\bar{x}$ for $y = a+bx$. No recomputation needed.

Apply the same transformation to the mean.

(i) $40+5 = \mathbf{45}$ · (ii) $2(40) = \mathbf{80}$ · (iii) $\dfrac{40-10}{2} = \mathbf{15}$

B9From cumulative

Q. "Less than" cf: <20 → 5, <40 → 18, <60 → 40, <80 → 50. Find the mean.

Solution

Why: difference the cf column to recover $f$, then the ordinary grouped mean.

Successive differences.

\[ f = 5,\;13,\;22,\;10 \qquad \Sigma f = 50 \;\checkmark \]

Midpoints 10, 30, 50, 70.

\[ \Sigma fx = 50+390+1100+700 = 2240 \] \[ \bar{x} = \frac{2240}{50} = \mathbf{44.8} \]

B10Inclusive series

Q. Classes 5–9, 10–14, 15–19, 20–24 with frequencies 6, 12, 14, 8. Find the mean. Is conversion needed?

Solution

Why: inclusive series — but for the mean no conversion is required, because the midpoint is the same either way.

Midpoints from the given limits.

\[ x = 7,\;12,\;17,\;22 \]

Grouped mean.

\[ \Sigma f = 40,\quad \Sigma fx = 42+144+238+176 = 600 \] \[ \bar{x} = \frac{600}{40} = \mathbf{15.0} \] No conversion needed for the mean; it would be needed for the median and mode.

Textbook bank — every Levin exercise on this topic§3.1 summary statistics · §3.2 the mean · §3.3 weighted mean · §3.4 geometric mean

35 exercisesLevin §3.1–§3.4 pp. 74–96Section A above already works the grouped, step-deviation, weighted and combined shapes in full.

Every one of these ends in a decision — give it.

3-6Mean vs thresholdLevin Ex. 3-6 · p. 84

Q. Child-Care keeps its grant only while the average age stays below 9. Ages: 8 5 9 10 9 12 7 12 13 7 8. Do they qualify?

Solution

Sum, divide, compare.

\[ \bar{x}=\frac{100}{11}=\mathbf{9.09 \text{ years}} \] $9.09 > 9$, so NO — they do not qualify. They miss by only 0.09 of a year; enrolling one 7-year-old would bring the mean to 8.92 and restore eligibility.

3-7Mean + marginLevin Ex. 3-7 · p. 84

Q. Support continues while average family income is below $12,500. Incomes: 14,500 15,600 12,500 8,600 7,800 6,500 5,900 10,200 8,800 14,300 13,900. (a) Do they qualify? (b) If not, by how much must income fall? (c) If yes, by how much can it rise?

Solution

Mean of 11 incomes.

\[ \bar{x}=\frac{118{,}600}{11}=\mathbf{\$10{,}781.82} \]

(a) yes ⇒ answer (c), the headroom.

(a) Yes, Child-Care qualifies — $10,781.82 is comfortably below $12,500.
(c) Average income can rise by $12{,}500-10{,}781.82=\mathbf{\$1{,}718.18}$ before eligibility is lost — a margin of about 16%.

3-8Grouped vs rawLevin Ex. 3-8 · p. 84

Q. Ages of 20 patients: 85 75 66 43 40 88 80 56 56 67 89 83 65 53 75 87 83 52 44 48. (a) Build a frequency distribution with classes 40–49, 50–59, … (b) Compute the mean from it. (c) Compute the mean from the raw data. (d) Compare.

Solution

(a) Tally into five classes.

40–49 → 4 · 50–59 → 4 · 60–69 → 3 · 70–79 → 2 · 80–89 → 7 (Σ = 20 ✓)

(b) Grouped mean using midpoints 44.5, 54.5, 64.5, 74.5, 84.5.

\[ \bar{x}=\frac{1330}{20}=\mathbf{66.50 \text{ years}} \]

\[ \bar{x}=\frac{1335}{20}=\mathbf{66.75 \text{ years}} \]

(d) The comparison — this is the mark.

The grouped mean is 0.25 years lower. The difference is grouping error: replacing every value by its class midpoint discards the actual positions within each class.

Here the error is tiny (0.4%) because the values happen to sit fairly evenly within their classes. The raw mean is exact and always preferable when the raw data is available — grouping is a convenience for large datasets, not an improvement.

3-9Grouped + codingLevin Ex. 3-9 · p. 85

Q. BullsEye service times (seconds): 20–29 → 6, 30–39 → 16, 40–49 → 21, 50–59 → 29, 60–69 → 25, 70–79 → 22, 80–89 → 11, 90–99 → 7, 100–109 → 4, 110–119 → 0, 120–129 → 2. (a) Compute the mean directly. (b) Compute it by the coding method with 0 assigned to the 70–79 class.

Solution

(a) Midpoints 24.5, 34.5, …, 124.5; total both columns.

\[ \Sigma f = 143, \qquad \Sigma fx = 8723.5 \] \[ \bar{x}=\frac{8723.5}{143}=\mathbf{61.00 \text{ seconds}} \]

(b) Coding with $A = 74.5$ (the 70–79 midpoint) and $h = 10$.

\[ d = -5,-4,-3,-2,-1,0,1,2,3,4,5 \] \[ \bar{x}=A+h\cdot\frac{\Sigma fd}{\Sigma f}=74.5+10\left(\frac{-193}{143}\right) \] \[ =74.5-13.50=\mathbf{61.00 \text{ seconds}} \] Identical, as it must be — coding is a shortcut, not a different statistic.

Note the empty class (110–119, frequency 0). It contributes nothing to either total but must be kept — deleting it would break the equal spacing that the coding method depends on.

3-10Two thresholdsLevin Ex. 3-10 · p. 85

Q. Pets 'R Us builds only if the average for the first 6 months is at least 300 and the yearly monthly average is at least 285. Data: 234 216 195 400 315 274 302 291 275 300 375 450. What is the decision?

Solution

Why: two conditions joined by and — both must hold, so compute both and check each.

First six months.

\[ \frac{234+216+195+400+315+274}{6}=\frac{1634}{6}=\mathbf{272.33} \] $272.33 < 300$ ✗

Whole year.

\[ \frac{3627}{12}=\mathbf{302.25} \] $302.25 \ge 285$ ✓

Both are required.

The owner does NOT build. The yearly condition passes comfortably but the first-half condition fails by 27.7 animals a month.

Worth adding: sales are clearly growing — the second half averages 332 against the first half's 272. The owner's first-half test may be the wrong criterion for a business on an upward trend.

3-11ToleranceLevin Ex. 3-11 · p. 85

Q. A machine fills 3-ounce bottles; recalibration is unnecessary if the average volume is within 0.04 of 3.00 oz. Trial volumes: 3.02 2.89 2.92 2.84 2.90 2.97 2.95 2.94 2.93 3.01 2.97 2.95 2.90 2.94 2.96 2.99 2.99 2.97. Should it recalibrate?

Solution

Mean of 18 trials.

\[ \bar{x}=\frac{53.04}{18}=\mathbf{2.9467 \text{ oz}} \]

Compare the deviation with the tolerance.

\[ |3.00-2.9467|=\mathbf{0.0533} \] $0.0533 > 0.04$, so YES — the machine should be recalibrated.

It is systematically underfilling by more than the allowed tolerance. Note that underfilling is the worse direction commercially: customers receive less than the labelled 3 ounces, which is a labelling compliance problem as well as a quality one.

3-12Mean vs thresholdLevin Ex. 3-12 · p. 85

Q. Plate-photographing times (seconds): 20.4 20.0 22.2 23.8 21.3 25.1 21.2 22.9 28.2 24.3 22.0 24.7 25.7 24.9 22.7 24.4 24.3 23.6 23.2 21.0. Productivity is satisfactory below 23.0 seconds. Should the manager be concerned?

Solution

Mean of 20 observations, then compare.

\[ \bar{x}=\frac{465.9}{20}=\mathbf{23.295 \text{ seconds}} \] $23.295 > 23.0$, so yes — he should be concerned, though only marginally (1.3% over).

Worth adding: the single value of 28.2 is well above the rest. Removing it drops the mean to 23.03 — still just over. The problem is a general drift, not one bad plate.

3-13Several means + a targetLevin Ex. 3-13 · p. 86

Q. Daily balances ($ millions). Week 1: 1.973 1.970 1.972 1.975 1.976. Week 2: 1.969 1.892 1.893 1.887 1.895. Find the mean for (a) week 1, (b) week 2, (c) both weeks. (d) Does an average above $1.970 million qualify? (e) If not, how much would the last day need to rise?

Solution

(a), (b), (c) — three means.

Week 1: $\dfrac{9.866}{5}=\mathbf{\$1.9732\text{m}}$
Week 2: $\dfrac{9.536}{5}=\mathbf{\$1.9072\text{m}}$
Both: $\dfrac{19.402}{10}=\mathbf{\$1.9402\text{m}}$

(d) Compare with the threshold.

$1.9402 < 1.970$ ⇒ No, National does not qualify for the higher interest rate.

(e) Solve for the last day's balance that would just qualify.

Total needed: $10 \times 1.970 = 19.700$. Total of the first nine days: $19.402-1.895=17.507$. \[ \text{last day} = 19.700-17.507=\mathbf{\$2.1930\text{m}} \] It was $1.895m, so it would have to rise by $0.298m — a 16% jump in a single day, which is not realistic. The shortfall is structural: week 2 was simply much lower than week 1.

3-14Means of meansLevin Ex. 3-14 · p. 86

Q. M. T. Smith's quarterly earnings over 3 years. (a) Average for each of the four quarters. (b) Average for each of the three years. (c) Show that the mean of the four figures in (a) equals the mean of the three in (b), and that both equal the mean of all 12.

Solution

Why: the question proves a real property — averaging averages is valid only when the groups are equal in size, which is exactly the case here.

(a) Quarter means, each over three years.

Q1: $\dfrac{10{,}000+20{,}000+30{,}000}{3}=\$20{,}000$ · Q2: $\$10{,}000$
Q3: $\$30{,}000$ · Q4: $\$25{,}000$

(b) Year means, each over four quarters.

Year 1: $\$13{,}750$ · Year 2: $\$15{,}000$ · Year 3: $\$35{,}000$

Mean of the four quarter means: $\dfrac{85{,}000}{4}=\mathbf{\$21{,}250}$
Mean of the three year means: $\dfrac{63{,}750}{3}=\mathbf{\$21{,}250}$
Mean of all 12 values: $\dfrac{255{,}000}{12}=\mathbf{\$21{,}250}$ ✓

Why they agree: every group contains the same number of observations, so each value carries the same weight whichever way you group it. With unequal groups this would fail — and that is precisely why the combined mean must be weighted by $n$ (A6 above).

3-16 / 3-17 / 3-18Weighted meanLevin Ex. 3-16 to 3-18 · pp. 90–91

Q. (3-16) Final grades weighted homework 20%, midterm 25%, final 35%, term paper 10%, quizzes 10%. (3-17) Jim orders 6 cases at $28, 4 at $36, 8 at $16, 3 at $18, 1 at $12 — find the average cost per case. (3-18) Average number of times a subscriber saw an advertisement, from a frequency distribution.

Solution

Why: all three are $\Sigma wx/\Sigma w$. The only skill is spotting what plays the role of the weight — a percentage, a quantity, or a frequency.

(3-16) The weights are the stated percentages; they already sum to 100.

\[ \bar{x}_w = 0.20H + 0.25M + 0.35F + 0.10T + 0.10Q \] Apply to each student's five scores. Do not average the five scores — the final is worth three and a half times the term paper.

(3-17) The weights are the numbers of cases.

\[ \bar{x}_w=\frac{6(28)+4(36)+8(16)+3(18)+1(12)}{6+4+8+3+1} \] \[ =\frac{168+144+128+54+12}{22}=\frac{506}{22}=\mathbf{\$23.00 \text{ per case}} \] The unweighted mean of the five prices is $22.00 — different, because Jim buys most of the cheap Standard cases and only one Low Grade.

(3-18) The weights are the frequencies of subscribers.

\[ \bar{x}=\frac{\Sigma f x}{\Sigma f} \] where $x$ is the number of times seen and $f$ the number of subscribers seeing it that often. This is the grouped mean under another name — a frequency is a weight.

3-19 / 3-20 / 3-21Weighted mean, appliedLevin Ex. 3-19 to 3-21 · pp. 91–92

Q. (3-19) Nelson Window's five plants forecast different growth rates on different production volumes — find the company-wide growth rate. (3-20) Postal volumes and prices per ounce for seven mail types — find the average price per ounce. (3-21) Four staff grades billed at $75, $40, $30 and $15 an hour — find the average rate charged.

Solution

Why: in all three, an unweighted average would be badly wrong, and the question is designed so you can see it.

(3-19) Weight each plant's growth rate by its production volume.

\[ \text{growth}_{\text{company}}=\frac{\Sigma (\text{volume}\times\text{rate})}{\Sigma \text{volume}} \] Orlando's 11.5% on 72 million windows counts far more than a small plant's rate. Averaging the five percentages would treat a 5-million plant as equal to a 72-million one.

(3-20) Weight each price by ounces delivered.

\[ \bar{p}=\frac{\Sigma(\text{ounces}\times\text{price})}{\Sigma\text{ounces}} \] First class dominates at 77,600 million ounces, so the weighted average sits close to its $0.13 — nothing like the plain average of the seven prices.

(3-21) Weight each rate by hours billed.

\[ \bar{r}=\frac{75h_1+40h_2+30h_3+15h_4}{h_1+h_2+h_3+h_4} \] The average realised rate depends entirely on the mix of hours. A firm billing mostly office staff earns nowhere near the $40 midpoint of its rate card — which is the practical point of the exercise.

3-22Geometric meanLevin Ex. 3-22 · p. 95

Q. Hayes Textiles' net worth grew 5%, 10.5%, 9.0%, 6.0%, 7.5% over five years. What is the average percentage increase?

Solution

Why: growth compounds, so the rates multiply rather than add. Averaging them arithmetically would overstate the true growth.

Convert each rate to a growth factor.

\[ 1.050,\ 1.105,\ 1.090,\ 1.060,\ 1.075 \]

Multiply and take the 5th root.

\[ G.M.=\sqrt[5]{1.050\times1.105\times1.090\times1.060\times1.075}=1.075817 \]

Convert back to a percentage.

\[ \text{average increase}=\mathbf{7.582\%\text{ per year}} \] The arithmetic mean of the five percentages is 7.600% — close here, but it is always at least as large as the geometric mean, and the gap widens as the rates become more variable.

3-24GM + forecastLevin Ex. 3-24 · p. 95

Q. Birch Company units produced: 1992 → 12,500, 1993 → 13,250, 1994 → 14,310, 1995 → 15,741, 1996 → 17,630. Find the average percentage increase and estimate 1999 production.

Solution

Build the year-on-year growth factors from the levels.

\[ \frac{13250}{12500}=1.060,\ \frac{14310}{13250}=1.080,\ \frac{15741}{14310}=1.100,\ \frac{17630}{15741}=1.120 \]

Geometric mean of the four factors.

\[ G.M.=\sqrt[4]{1.060(1.080)(1.100)(1.120)}=1.08977 \] \[ \Rightarrow \mathbf{8.977\% \text{ per year}} \]

Project three years forward from 1996.

\[ 1999 = 17{,}630\times(1.08977)^3=\mathbf{22{,}817 \text{ units}} \] State the assumption: this holds only if growth continues at the same average rate. Note the factors are rising (1.06 → 1.12), so growth is accelerating and 22,817 may be conservative.

3-25Work backwardsLevin Ex. 3-25 · p. 95

Q. Bob's geometric mean growth factor over 6 years was 1.24. The first five factors were 1.19, 1.35, 1.23, 1.19, 1.30. What was the sixth?

Solution

Why: reverse the definition. If the 6th root of the product is 1.24, the product itself is $1.24^6$ — divide out the five known factors.

Recover the full product.

\[ \prod_{i=1}^{6} g_i = (1.24)^6 = 5.6383 \]

Product of the five known factors.

\[ 1.19(1.35)(1.23)(1.19)(1.30) = 4.7413 \]

Divide.

\[ g_6 = \frac{5.6383}{4.7413}=\mathbf{1.1892} \] So the sixth year grew by about 18.9% — the weakest of the six, which is why it dragged the mean below the earlier factors.

3-28GM + projectionLevin Ex. 3-28 · p. 96

Q. Purchase-order processing costs over five years: $55.00, $58.00, $61.00, $65.00, $66.00. What has the average percentage increase been, and what will it cost in 3 more years at that rate?

Solution

Four growth factors from five levels.

\[ 1.0545,\ 1.0517,\ 1.0656,\ 1.0154 \]

Geometric mean.

\[ G.M.=\sqrt[4]{\text{product}}=1.046635 \;\Rightarrow\; \mathbf{4.664\% \text{ per year}} \]

Project three years on from $66.00.

\[ 66.00\times(1.046635)^3=\mathbf{\$75.67} \] Note the last factor (1.0154) is much smaller than the others — cost growth is slowing, so the projection may overstate. Say so.

3-26 / 3-27 / 3-29 / 3-23Geometric mean drillsLevin Ex. 3-23, 3-26, 3-27, 3-29 · pp. 95–96

Q. (3-23) MacroSwift's net worth over 9 years — find the average change and project 1998 from 1996. (3-26) Acrylic sheeting bought at $1.00, $1.10, $1.15 — find the average weekly rate of increase. (3-27) Lisa's milk prices over 8 weeks. (3-29) Yearly percentage changes in prisoner numbers, some negative.

Solution

ExerciseWhat to do

3-23 MacroSwift

Eight growth factors from nine values; take the 8th root. Then $1998 = 1996\text{ value}\times G.M.^2$. Two of the nine years show a decrease, giving factors below 1 — they still multiply in normally.

3-26 acrylic sheeting

Two growth factors: $1.10/1.00=1.100$ and $1.15/1.10=1.04545$. \[ G.M.=\sqrt{1.100\times1.04545}=1.0724 \] ⇒ about 7.24% per week. The three equal dollar purchases are a distractor — the question asks about the price series, not the quantities.

3-27 Lisa's milk

Seven factors from eight weekly prices; take the 7th root. Because the price both rises and falls, several factors are below 1, and the G.M. will be close to 1 — a near-flat average rate of change.

3-29 prisoner numbers with negatives

The trap: a −4% change is a factor of 0.96, not −0.04. Convert every percentage to a factor $1+r/100$ first — you can never take a root of a negative number, and the geometric mean is undefined if you try.
\[ G.M.=\sqrt[n]{0.96\times1.05\times1.10\times1.03\times\cdots} \]

3-15 / 3-1 to 3-5Remaining drillsLevin Ex. 3-1 to 3-5, 3-15 · pp. 76–86

Q. §3.1's five conceptual questions on summary statistics, plus 3-15 (Lillian Tyson's claim that she managed the bookmobile repair budget better than her predecessor, from 15 years of data).

Solution

§3.1 (3-1 to 3-5) — what summary statistics are for.

The examinable content: a summary statistic replaces a whole dataset with one number, which is both its value (comprehensible, comparable) and its danger (detail is destroyed). Measures of central tendency locate the data; measures of dispersion say how well that location represents it. Neither is adequate alone — see Unit 2, Topic 1.

3-15 — testing Lillian's claim.

Split the 15 years into her tenure and her predecessor's, and compute the mean repair budget for each.

But the comparison is not fair as it stands, and saying so is the mark: the two periods differ in length, the earlier figures are in older dollars (inflation makes any later budget look larger), and the vehicles were older in later years, which raises repair costs regardless of management. A defensible comparison would adjust for inflation and for fleet age.

SC 3-1 to SC 3-6Self-check setLevin SC 3-1 to SC 3-6 · pp. 87–96

Q. SC 3-1 grouped mean by two coding origins. SC 3-2 Davis Furniture's average balance against $65,000. SC 3-3 unweighted vs weighted averages at two stores. SC 3-4 a weighted percentage. SC 3-5 and SC 3-6 geometric means.

Solution

SC 3-1 — the coding method proved origin-independent.

The same distribution is coded twice, with 0 assigned to the fourth class and then the sixth. Both give the same mean, because $A + h(\Sigma fu/\Sigma f)$ shifts the origin and shifts it back. That is the point of the exercise — the choice of assumed mean is free.

SC 3-2 — Davis Furniture.

\[ \bar{x}=\frac{\Sigma x}{12}=\mathbf{\$68{,}967} \] Since $68{,}967 > 65{,}000$, they do qualify for the reduced interest rate.

SC 3-3 — where weighting flips the answer.

Unweighted: competition $5.20, Dave's $5.25 — Dave's looks dearer.
Weighted by quantities: competition $4.344, Dave's $4.399.
Both methods favour the competition here, but the gap changes, and the weighted figures are the ones a shopper actually pays — because they reflect what is actually bought.

SC 3-4 / SC 3-5 / SC 3-6.

SC 3-4: $\bar{x}_w = 2466.435/330.6 = \mathbf{7.46\%}$ — weights are the three component sizes.
SC 3-5: bad-debt factors over six years give $G.M.=1.0146$ ⇒ 1.461% per year; 1997 relative to 1995 is $0.120\times G.M.^2 = 0.1235$, a 2.94% increase.
SC 3-6: monthly factors 0.9167, 0.9091, 0.9600, 0.9375 give $G.M.=\sqrt[4]{0.75}=0.9306$ ⇒ the price is falling 6.94% per month. A factor below 1 means decline — read the sign, not just the number.

Topic 5 of 5 · this is the 5-mark exam question

Median, Mode, and Reading the Shape of a Distribution

Both 2025-26 papers put the same question in slot Q3 (a): a table of mean, median, mode, SD, skewness and kurtosis for two service channels, and three sub-parts asking what it all means. No arithmetic. Everything below exists to make that table readable — and the last two questions in Section A are that exam question, in both of its versions.

How to know a question belongs to this topic

It asks you to

Find the middle value or the most common one — or, in the exam, to read what three averages together say about shape.

Give-away words

calculate the median compute the mode modal value which is the better measure of central tendency what does the difference signify estimate the modal class

The "which is better" question

Levin asks this again and again, and the answer is always the same shape: if the data is skewed or has outliers, the median; if it is categorical or you want the most typical value, the mode; if it is symmetric or you need further arithmetic, the mean.

Boundaries, not limits

Both the median and mode formulas need $L$ = the lower boundary. On an inclusive series you must convert first — the mean does not care, but these two do.

This is where Q3 (a) lives

The exam never asks you to compute these. It hands you all three plus an SD, a skewness and a kurtosis, and asks what they mean together. A9 and A10 above are that question in both its versions.

Master Method — Topic 5

Formula bank

Median — ungrouped

\[ M = \text{value of the } \left(\frac{n+1}{2}\right)^{\text{th}} \text{ item, after arranging in order} \]

Median — grouped

\[ M = L + \frac{\dfrac{n}{2} - cf_{\text{prev}}}{f_m}\times h \]

Mode — grouped

\[ Z = L + \frac{f_1-f_0}{2f_1-f_0-f_2}\times h \]

Empirical relation

\[ Z = 3M - 2\bar{x} \qquad\Longleftrightarrow\qquad M = \frac{Z+2\bar{x}}{3} \]

Pearson's coefficient
of skewness

\[ S_k = \frac{\bar{x}-Z}{s} \qquad\text{or}\qquad S_k = \frac{3(\bar{x}-M)}{s} \]

$L$ lower boundary of the median / modal class

$cf_{\text{prev}}$ cumulative frequency of the class before it

$f_m$ frequency of the median class

$f_1$ frequency of the modal class

$f_0,\,f_2$ frequencies of the classes just before and just after it

$h$ class width

The one thing that breaks these formulas: $L$ must be a class boundary. On an inclusive series you must convert first — 30–39 becomes 29.5–39.5 and $L=29.5$, not 30. Skipping this shifts every answer by exactly $d$.

The routine — median or mode from a grouped table

Convert if inclusive. Write the boundaries out. Both formulas use $L$ = lower boundary.
Build the cf column (median only) and confirm the last entry is $n$.
Locate the class. Median: the first class whose $cf \ge n/2$. Mode: the class with the highest $f$.
Write down every symbol before substituting — $L$, $n/2$, $cf_{\text{prev}}$, $f_m$, $h$ (or $L, f_1, f_0, f_2, h$). This line alone usually carries a mark.
Substitute and compute. The answer must fall inside the located class — if it does not, you picked the wrong class or the wrong $L$.
Cross-check with the empirical relation $Z \approx 3M-2\bar{x}$. It will not be exact, but it should be close; a wild mismatch means an error.

The reading rules — this is what Q3 (a) marks

Averages & Skewness Rules

\[ \text{Mean} = \text{Median} = \text{Mode} \implies \text{Symmetric} \] \[ \text{Mean} < \text{Median} \le \text{Mode} \implies \text{Negatively Skewed (Left tail)} \] \[ \text{Mean} > \text{Median} \ge \text{Mode} \implies \text{Positively Skewed (Right tail)} \]

Consistency Rule

Low SD + HIGH kurtosis (leptokurtic): CONSISTENT, predictable service
High SD + LOW kurtosis (platykurtic): INCONSISTENT, variable service

Skewness and kurtosis are formally Unit 2 (Unit 2, Concepts 3–4), but Q3 (a) mixes them with the Unit 1 averages in a single table, so they are read here.

Pattern recognition

Pattern	Wording cue	What to do
Compute the median	"find the median of the following distribution"	cf column → locate → $M = L+\frac{n/2-cf}{f_m}h$.
Compute the mode	"find the modal wage / most typical value"	Highest $f$ → $Z = L+\frac{f_1-f_0}{2f_1-f_0-f_2}h$.
One average missing	"mean is 48 and mode is 54; find the median"	Empirical relation $Z = 3M-2\bar{x}$, rearranged.
Name the skew	"what does the difference between mean, median and mode signify?"	Order the three, name the direction, describe the tail in the story's own language.
Consistency comparison	"what do skewness and kurtosis imply about consistency?"	SD first (spread), then kurtosis (peakedness). Low SD + high K = consistent.
Recommend actions	"what strategies should the firm implement?"	Each recommendation must name the statistic it attacks. That link is the mark.
Which average?	"income data with a few very high earners"	Median — the mean is dragged by extremes; the mode for categorical / most-common questions.

How the exam actually asks it

Paper	The question
Final 2025-26	BrightWave Financial: mean/median/mode/SD/skewness/kurtosis for Call Centre vs Branch Counter. Three parts — (a) what the mean/median/mode difference signifies, (b) what skewness and kurtosis imply for consistency, (c) what strategies to implement QP p. 2 · worked in full
Re-Exam 2025-26	UrbanStyles Retail: the same six statistics for Online Chat vs Walk-In Desk. Structurally identical — only the story and the numbers change QP p. 2 · worked in full
Official key	Structured under exactly those three headings, and awards the strategy marks only where a recommendation is tied to a statistic Key p. 3

Traps

Using $L$ = the class limit on an inclusive series. Every median and mode is then wrong by $d = 0.5$.
Getting the skew direction backwards. Anchor on the tail: the mean is dragged toward the tail, so mean below median means the tail is on the left = negative.
Reading kurtosis on the wrong scale. If the table gives excess kurtosis the neutral value is 0, not 3. If a value like 2.4 or 4.8 is given, it is the "normal = 3" scale.
Missing a bimodal distribution. The Final's Call Centre had mode "4 and 5" — say so: opinion is split between two clusters, not settled on one experience.
Listing strategies with no statistic attached. "Improve training" earns nothing; "reduce the SD of 0.97 by standardising call scripts" earns the mark.

Section A — Ten questions, fully explained A9 and A10 are the actual exam question, both versions

A1Median — ungrouped

Q. Find the median of (i) 27, 19, 34, 22, 41, 30, 25 and (ii) the same seven values with 38 added.

Why this method: the median is a position, not a calculation — so the data must be arranged in order first, every time, before anything else happens.

Why we do itWhat to write

Arrange in ascending order. Skipping this is the single most common error.

19, 22, 25, 27, 30, 34, 41

$n=7$ is odd, so the middle item exists.

\[ \text{position} = \frac{n+1}{2} = \frac{8}{2} = 4^{\text{th}} \;\Rightarrow\; M = \mathbf{27} \]

Add 38 and re-order. $n=8$ is even, so the "middle" falls between two items.

19, 22, 25, 27, 30, 34, 38, 41 \[ \text{position} = \frac{9}{2} = 4.5^{\text{th}} \]

Average the two items either side of that position.

\[ M = \frac{27+30}{2} = \mathbf{28.5} \]

Note: $\frac{n+1}{2}$ gives the position, not the value. Writing "M = 4" instead of "the 4th item, which is 27" is a standard lost mark.

A2Median — grouped

Q. Find the median call duration.

Duration (min)	0–10	10–20	20–30	30–40	40–50
No. of calls	5	12	20	9	4

Why this method: we cannot see individual calls, so we find the class that contains the half-way call and then interpolate inside it — assuming the calls in that class are spread evenly.

Why we do itWhat to write

Build the cumulative frequency column; the last value must equal $n$.

\[ cf: \; 5,\; 17,\; 37,\; 46,\; 50 \qquad n = 50 \;\checkmark \]

Find the half-way position and the first class that reaches it.

\[ \frac{n}{2} = 25 \] $cf$ jumps 17 → 37 at 20–30, so that is the median class.

Write every symbol down before substituting — the classes are exclusive, so $L$ is 20 as given.

\[ L = 20,\quad cf_{\text{prev}} = 17,\quad f_m = 20,\quad h = 10 \]

Substitute into the median formula.

\[ M = L + \frac{\frac{n}{2}-cf_{\text{prev}}}{f_m}\times h = 20 + \frac{25-17}{20}\times 10 \]

Compute and check the answer lies inside the median class.

\[ M = 20 + 4 = \mathbf{24 \text{ minutes}} \] 24 lies inside 20–30 ✓ Half of all calls last under 24 minutes.

Cross-check: this is the same table as the ogive in Topic 3 A6 — the graphical reading gave 24 as well. The formula is the ogive, done arithmetically.

A3Median — inclusive series

Q. Find the median mark, and show how much you would be wrong by if you skipped the conversion.

Marks	10–19	20–29	30–39	40–49	50–59
No. of students	6	14	22	12	6

Why this method: the median formula measures a distance from the start of the class, so the "start" must be the real boundary where one class ends and the next begins — not the printed limit with a gap sitting under it.

Why we do itWhat to write

Convert first — this is the step the whole question is testing.

\[ d = \frac{20-19}{2}=0.5 \] Boundaries: 9.5–19.5, 19.5–29.5, 29.5–39.5, 39.5–49.5, 49.5–59.5

cf column and the half-way position.

\[ cf: \; 6,\;20,\;42,\;54,\;60 \qquad \frac{n}{2} = \frac{60}{2}=30 \] $cf$ jumps 20 → 42 ⇒ median class 29.5–39.5.

List the symbols, taking $L$ from the boundary.

\[ L = 29.5,\quad cf_{\text{prev}} = 20,\quad f_m = 22,\quad h = 10 \]

Substitute.

\[ M = 29.5 + \frac{30-20}{22}\times10 = 29.5 + 4.55 = \mathbf{34.05 \text{ marks}} \]

Show the cost of the shortcut, since the question asks for it.

Using $L=30$ instead gives $30+4.55 = 34.55$ — wrong by exactly $d = 0.5$, the width of the gap you failed to close.

A4Mode — ungrouped

Q. Find the mode of (i) 4, 5, 3, 5, 4, 5, 2, 4, 5, 3, 5, 4 and (ii) 7, 8, 9, 8, 7, 10, 7, 8, 11. What do you call case (ii)?

Why this method: the mode is a straight frequency count — build the tally, then read the peak. The only subtlety is what to do when two values tie.

Why we do itWhat to write

Tally the first set.

2 → 1, 3 → 2, 4 → 4, 5 → 5

The peak count identifies the mode.

\[ Z = \mathbf{5} \quad\text{(occurs 5 times — unimodal)} \]

Tally the second set.

7 → 3, 8 → 3, 9 → 1, 10 → 1, 11 → 1

Two values tie for the peak — report both and name the situation.

\[ Z = \mathbf{7 \text{ and } 8} \] The distribution is bimodal: there is no single typical value — the data splits into two clusters.

Why this matters for the exam: the Final paper's Call Centre channel is reported with mode "4 and 5" QP p. 2. The interpretation mark is for saying that customer opinion is split between two clusters rather than settled on one typical experience.

A5Mode — grouped

Q. Find the modal call duration for the A2 data: 0–10 (5), 10–20 (12), 20–30 (20), 30–40 (9), 40–50 (4).

Why this method: the modal class is obvious (the tallest bar), but the mode sits somewhere inside it. The formula leans the answer toward whichever neighbour is smaller — that is what $f_1-f_0$ and $f_1-f_2$ measure.

Why we do itWhat to write

Identify the class with the highest frequency.

Highest $f = 20$ ⇒ modal class 20–30.

Name the three frequencies the formula needs — the modal class and its two neighbours.

\[ L = 20,\quad f_1 = 20,\quad f_0 = 12,\quad f_2 = 9,\quad h = 10 \]

Substitute. Compute the denominator separately — it is where sign slips happen.

\[ 2f_1-f_0-f_2 = 40-12-9 = 19 \] \[ Z = 20 + \frac{20-12}{19}\times 10 = 20 + \frac{80}{19} \]

Compute and check it lies inside the modal class.

\[ Z = 20 + 4.21 = \mathbf{24.21 \text{ minutes}} \] 24.21 lies inside 20–30 ✓ — and it sits past the middle because the class below (12) is heavier than the class above (9).

A6All three, then the shape

Q. For the distribution below, find the mean, median and mode, and state what their ordering says about the shape.

Class	0–10	10–20	20–30	30–40	40–50
Frequency	4	10	24	14	8

Why this method: this is the exam's Q3 (a) in miniature — compute the three averages, then order them, because the order is what names the shape.

Why we do itWhat to write

Mean: midpoints × frequencies, totalled.

\[ x = 5,15,25,35,45 \quad fx = 20,150,600,490,360 \] \[ \bar{x} = \frac{1620}{60} = \mathbf{27.0} \]

Median: cf column, locate $n/2 = 30$.

\[ cf: 4,\;14,\;38,\;52,\;60 \;\Rightarrow\; \text{median class } 20\text{–}30 \] \[ M = 20+\frac{30-14}{24}\times10 = \mathbf{26.67} \]

Mode: highest $f = 24$, same class.

\[ Z = 20+\frac{24-10}{48-10-14}\times10 = 20+\frac{140}{24} = \mathbf{25.83} \]

Order them — this single line is the interpretation.

\[ \bar{x}\,(27.0) \;>\; M\,(26.67) \;>\; Z\,(25.83) \]

Name the shape and describe the tail in plain language.

Positively (right) skewed — a minority of high values in the 30–50 range drags the mean above the median and mode. The typical observation is about 26; the average is higher only because of that right tail.

Cross-check with the empirical relation: $3M-2\bar{x} = 3(26.67)-2(27) = 26.0$, against the computed mode of 25.83. Close, as it should be for a mildly skewed distribution — a large mismatch would mean an arithmetic error somewhere.

A7Empirical relation

Q. (i) A distribution has mean 25 and median 26. Estimate the mode and name the skew. (ii) Another has mean 48 and mode 54. Find the median. (iii) A third has mean 46, median 48, mode 52 and SD 8 — compute Pearson's coefficient of skewness.

Why this method: for a moderately skewed distribution the three averages sit in a fixed proportion — the mode is roughly twice as far from the mean as the median is. That is all $Z = 3M-2\bar{x}$ says.

Why we do itWhat to write

(i) Substitute directly into the empirical relation.

\[ Z = 3M-2\bar{x} = 3(26)-2(25) = 78-50 = \mathbf{28} \]

Order the three to name the skew.

$\bar{x}(25) < M(26) < Z(28)$ ⇒ negatively skewed — a left tail of low values.

(ii) Rearrange the same relation for $M$.

\[ M = \frac{Z+2\bar{x}}{3} = \frac{54+96}{3} = \mathbf{50} \]

(iii) Pearson's coefficient standardises the mean–mode gap by the SD, so it can be compared across datasets.

\[ S_k = \frac{\bar{x}-Z}{s} = \frac{46-52}{8} = \mathbf{-0.75} \]

Read the sign and the size.

Negative ⇒ left tail. $|S_k| = 0.75 < 1$ ⇒ moderate, not extreme, skew. Consistent with \(\bar{x}

A8Which average?

Q. A town's annual household income has mean ₹8.4 lakh, median ₹4.6 lakh and mode ₹3.8 lakh. A politician quotes the mean; a journalist quotes the median. Which is the fairer summary, and why? Which would you use to decide the most common shirt size to stock?

Why this method: choosing an average is a question about what the extremes do. The mean uses every value, so a handful of very large ones move it; the median only cares about position, so it does not.

Why we do itWhat to write

Read the ordering first — it diagnoses the problem before you argue about it.

$\bar{x}(8.4) > M(4.6) > Z(3.8)$ ⇒ strongly positively skewed: a small number of very high incomes.

Explain what that does to the mean.

The mean is pulled up by the high tail. ₹8.4 lakh is nearly double the median — so more than half of households earn far less than "the average".

Answer the question asked, with the reason attached.

The journalist is fairer: the median (₹4.6 lakh) is resistant to extreme values and genuinely describes the typical household.

The second part changes the goal, so it changes the average.

For shirt sizes, use the mode — you need the most frequently occurring size, and a "mean size" of 39.6 corresponds to no shirt that exists.

The one-line rule: mean for symmetric numerical data and further calculation; median for skewed or open-ended data; mode for categorical data and "most common" questions.

A9The exam question — Version 1

Q. A bank measured customer satisfaction (1–5) across two service channels:

Statistic	Phone Support	Relationship Desk
Mean	3.6	4.5
Median	4	5
Mode	4	5
Standard deviation	1.15	0.62
Skewness	−0.62	−1.45
Kurtosis	2.1	4.2

(a) What does the difference between mean, median and mode signify? (b) What do skewness and kurtosis imply about consistency? (c) What should the bank do about Phone Support? [5 marks]

Why this method: answer the three sub-parts in the question's own order and under its own headings. No arithmetic is required or wanted — every mark is for a reading.

Why we do itWhat to write

(a) Order the three averages for each channel — the ordering names the shape.

Phone: $3.6 < 4 = 4$. Desk: $4.5 < 5 = 5$. In both, Mean < Median = Mode ⇒ negatively skewed — a left tail of low ratings pulling each mean down.

Size the gap — a bigger gap is a heavier tail.

Phone's gap is 0.4 with skew −0.62; the Desk's is 0.5 with skew −1.45. The Desk is more strongly skewed, but from a much higher base: most of its ratings sit at 5, with only a very thin tail of unhappy customers.

(b) SD first — it measures spread, which is what "consistency" means.

SD 1.15 vs 0.62: the Desk's ratings vary about half as much.

Then kurtosis on the "normal = 3" scale.

Phone 2.1 < 3 ⇒ platykurtic — flat and spread out. Desk 4.2 > 3 ⇒ leptokurtic — tall and narrow, tightly clustered.

Combine them into the sentence the marker is looking for.

Low SD + high kurtosis ⇒ the Relationship Desk is consistent and predictable. High SD + low kurtosis ⇒ Phone Support is inconsistent: two callers can have completely different experiences.

• Cut the SD of 1.15 — standardise call scripts and resolution procedures so every agent handles a call the same way.
• Attack the −0.62 left tail — diagnose and fix the specific causes of the low ratings (wait times, unresolved first contacts).
• Raise kurtosis from 2.1 toward the Desk's shape — continuous training, post-call feedback loops, and rewards for consistent performers rather than peak ones.
• Track the metrics monthly with explicit targets: mean up, SD down, skewness toward 0, kurtosis up.

This is the Final 2025-26 Q3 (a) with the numbers changed. The official key is structured under exactly these three headings and awards the (c) marks only where a recommendation is tied to a statistic Key p. 3. The real version is worked at Solved Papers → Final Q3 (a).

A10The exam question — Version 2 (positive skew)

Q. A logistics firm reports delivery times (minutes) for two hubs:

Statistic	Hub North	Hub South
Mean	18.4	16.2
Median	15.0	16.0
Mode	12.0	16.0
Standard deviation	9.8	3.1
Skewness	+1.35	+0.10
Kurtosis	5.1	2.9

(a) Interpret the three averages for each hub. (b) Which hub is more reliable, and what does the kurtosis add to that judgement? (c) Recommend two actions for the weaker hub. [5 marks]

Why this method: same routine as A9, but the skew has flipped sign — and the kurtosis reading is subtler here, because Hub North has both a high SD and a high kurtosis. Do not read kurtosis alone.

Why we do itWhat to write

(a) Order the averages for each hub.

North: $18.4 > 15.0 > 12.0$ ⇒ positively skewed. South: $16.2 \approx 16.0 = 16.0$ ⇒ essentially symmetric.

Turn the ordering into a business statement.

At North the typical delivery takes 12–15 minutes, but the mean is 18.4 because a minority of deliveries run very long — a right tail. Quoting "18 minutes average" flatters neither the typical customer nor the delayed one. At South, mean = median = mode, so the average is the typical experience.

(b) SD is the primary reliability measure.

SD 9.8 vs 3.1 — North's delivery times vary more than three times as much. Hub South is far more reliable.

Now read kurtosis in the presence of the skew, which is the discriminating part.

North's 5.1 (>3, leptokurtic) here means a sharp peak plus heavy tails — most deliveries cluster near 12–15 minutes, but extreme delays occur far more often than a normal curve would predict. South's 2.9 ≈ 3 is mesokurtic: near-normal, no fat tail of disasters.

• Attack the right tail (skew +1.35, kurtosis 5.1) — investigate the specific long-delivery cases: route failures, address errors, single-point congestion. Fixing the tail cuts the mean without touching the typical delivery.
• Cut the SD of 9.8 — standardise dispatch scheduling and route allocation so the spread narrows toward South's 3.1. Track the median and the 90th percentile, not the mean, since the mean hides the tail.

The subtlety worth a mark: "leptokurtic = consistent" is only safe when the SD is also low. High kurtosis with a high SD means a sharp peak and fat tails — mostly fine, occasionally disastrous. Say which of the two you mean.

Section B — Solve a problem ten fresh questions · one or two lines of reasoning, then the answer

B9 and B10 are interpretation questions — write full sentences, not labels.

B1Median — ungrouped

Q. Find the median of 12, 7, 19, 15, 9, 22, 11, 14.

Solution

Why: order first; $n = 8$ is even, so average the 4th and 5th values.

Sorted, then the middle pair.

7, 9, 11, 12, 14, 15, 19, 22 \[ M = \frac{12+14}{2} = \mathbf{13} \]

B2Median — grouped

Q. Classes 0–20, 20–40, 40–60, 60–80 with frequencies 10, 25, 40, 25. Find the median.

Solution

Why: cf column, locate $n/2$, interpolate inside that class.

cf and position.

\[ cf: 10,\;35,\;75,\;100 \qquad \frac{n}{2}=50 \;\Rightarrow\; \text{class } 40\text{–}60 \]

Substitute $L=40,\,cf=35,\,f_m=40,\,h=20$.

\[ M = 40+\frac{50-35}{40}\times20 = \mathbf{47.5} \]

B3Median — inclusive

Q. Classes 1–10, 11–20, 21–30, 31–40 with frequencies 8, 16, 20, 6. Find the median.

Solution

Why: inclusive series ⇒ convert to boundaries before using $L$.

$d = 0.5$; cf column; $n/2 = 25$.

Boundaries 0.5–10.5, 10.5–20.5, 20.5–30.5, 30.5–40.5. $cf: 8,\;24,\;44,\;50$.

$L = 20.5,\, cf_{\text{prev}} = 24,\, f_m = 20,\, h = 10$.

\[ M = 20.5+\frac{25-24}{20}\times10 = 20.5+0.5 = \mathbf{21.0} \]

B4Mode — ungrouped

Q. Find the mode of 3, 6, 3, 9, 6, 3, 12, 6, 15. What is unusual about it?

Solution

Why: tally and look for a tie at the top.

3 appears 3 times, 6 appears 3 times.

\[ Z = \mathbf{3 \text{ and } 6} \] Bimodal — there is no single typical value; the data has two clusters.

B5Mode — grouped

Q. Classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 6, 14, 30, 18, 12. Find the mode.

Solution

Why: modal class is the tallest; the formula leans toward the lighter neighbour.

Highest $f=30$ ⇒ class 20–30; $f_0 = 14,\, f_2 = 18,\, h = 10$.

\[ 2f_1-f_0-f_2 = 60-14-18 = 28 \]

Substitute.

\[ Z = 20+\frac{30-14}{28}\times10 = 20+5.71 = \mathbf{25.71} \]

B6Empirical relation

Q. A distribution has median 42 and mode 45. Estimate the mean and name the skew.

Solution

Why: rearrange $Z = 3M-2\bar{x}$ for $\bar{x}$.

Solve, then order the three.

\[ \bar{x} = \frac{3M-Z}{2} = \frac{126-45}{2} = \mathbf{40.5} \] $40.5 < 42 < 45$ ⇒ negatively skewed.

B7Pearson's Sk

Q. Mean 58, mode 50, SD 10. Compute Pearson's coefficient of skewness and interpret it.

Solution

Why: $(\bar{x}-Z)/s$ standardises the gap so it can be compared across datasets.

Substitute, then read the sign and size.

\[ S_k = \frac{58-50}{10} = \mathbf{+0.8} \] Positive ⇒ right tail of high values. $|S_k| < 1$ ⇒ moderate, not extreme.

B8Which average?

Q. Which average would you use for: (i) house prices in a city; (ii) the most popular smartphone colour; (iii) average marks to compute a class total?

Solution

Why: skewed data → median; categorical data → mode; further arithmetic → mean.

Match the goal to the average.

(i) Median — house prices are strongly right-skewed by a few luxury sales.
(ii) Mode — colour is nominal; only "most frequent" is meaningful.
(iii) Mean — it is the only average from which the total can be recovered ($\Sigma x = n\bar{x}$).

B9Read the shape

Q. Branch A: mean 4.2, median 4, mode 4, SD 0.55, skewness +0.20, kurtosis 3.8. Branch B: mean 3.5, median 4, mode 4 and 5, SD 1.30, skewness −0.85, kurtosis 2.2. Compare the two in three sentences.

Solution

Why: ordering → shape; SD + kurtosis → consistency; then one business sentence.

Order the averages for each.

A: $4.2 > 4 = 4$ ⇒ mildly positive skew (+0.20), close to symmetric. B: $3.5 < 4$ ⇒ negative skew (−0.85), with a bimodal mode of 4 and 5.

SD then kurtosis.

A: SD 0.55, kurtosis 3.8 ⇒ leptokurtic and tightly clustered — consistent. B: SD 1.30, kurtosis 2.2 ⇒ platykurtic and spread out — inconsistent.

The business sentence.

Branch A delivers a uniform experience around a rating of 4. Branch B is polarised: a cluster at 4–5 and a left tail of dissatisfied customers dragging the mean to 3.5, so two customers of B can have completely different experiences.

B10Recommend actions

Q. For Branch B in B9, give three improvement actions, each tied to the statistic it targets.

Solution

Why: the marks are for the link — a recommendation with no statistic attached scores nothing.

One action per statistic.

• SD 1.30 → standardise service procedures and scripts so every customer gets the same process, narrowing the spread toward Branch A's 0.55.
• Skewness −0.85 → root-cause the specific low ratings in the left tail (wait time, unresolved complaints) rather than trying to lift the average generally.
• Kurtosis 2.2 → raise it toward 3–4 through continuous training and post-interaction feedback, so outcomes cluster instead of spreading; and treat the bimodality as evidence that two different service experiences are being delivered — identify which staff or shifts produce which.

Textbook bank — every Levin exercise on this topic§3.5 the median · §3.6 the mode

19 exercisesLevin §3.5 pp. 96–103 · §3.6 pp. 104–111Most end in "which is the better measure?" — answer that part.

Sort first for the median; tally first for the mode.

3-30Median vs meanLevin Ex. 3-30 · p. 102

Q. Weekly truck mileages: 810 450 756 789 210 657 589 488 876 689 1450 560 469 890 987 559 788 943 447 775. (a) Median. (b) Mean. (c) Which is the better measure?

Solution

(a) Sort; $n=20$ is even, so average the 10th and 11th.

210 447 450 469 488 559 560 589 657 689 | 756 775 788 789 810 876 890 943 987 1450 \[ M=\frac{689+756}{2}=\mathbf{722.5 \text{ miles}} \]

(b) Mean.

\[ \bar{x}=\frac{14{,}182}{20}=\mathbf{709.1 \text{ miles}} \]

$\bar{x}(709.1) < M(722.5)$ — a mild negative skew overall, but look at the extremes: 210 at the bottom and 1,450 at the top are both far from the rest.

The median is better here. Both outliers distort the mean; the median ignores them and describes the typical truck. If Meridian wants total fleet mileage for fuel budgeting, however, the mean is required — only it recovers the total via $\Sigma x=n\bar{x}$.

3-31Median vs meanLevin Ex. 3-31 · p. 102

Q. Cable channels offered: 32 28 31 15 25 14 12 29 22 28 29 32 33 24 26 8 35. (a) Median. (b) Mean. (c) Which is better?

Solution

(a) $n=17$ is odd, so the median is the 9th value.

8 12 14 15 22 24 25 26 [28] 28 29 29 31 32 32 33 35 \[ M=\mathbf{28 \text{ channels}} \]

(b) Mean.

\[ \bar{x}=\frac{423}{17}=\mathbf{24.88 \text{ channels}} \]

$\bar{x}(24.88) < M(28)$ ⇒ negatively skewed. The low values 8, 12, 14, 15 drag the mean down while the bulk of providers sit in the high 20s and low 30s.

The median is the better measure — 28 channels genuinely describes a typical provider, whereas the mean of 24.88 is below what most providers actually offer.

3-32Median step by stepLevin Ex. 3-32 · p. 102

Q. For a grouped distribution: (a) which item number is the median? (b) which class contains it? (c) what is the width of the equal steps in the median class? (d) estimate the median by that step method; (e) estimate it by Equation 3-8. Are the two close?

Solution

Why: the question walks you through the reasoning behind the median formula, then asks you to confirm the formula reproduces it. Worth doing once — after this the formula is no longer something to memorise.

(a) The median item.

The $\left(\dfrac{n+1}{2}\right)$th item for the step method, or the $\dfrac{n}{2}$th for the formula — state which convention you use.

(b) The median class.

Build the cumulative frequency column; the median class is the first one whose cf reaches or passes that item number.

\[ \text{step} = \frac{\text{class width}}{\text{frequency of the median class}} \] i.e. spread that class's observations evenly across its width, and each one occupies one step.

(d) and (e) — the two routes.

Step method: start at the class's lower boundary and take as many steps as there are items between $cf_{\text{prev}}$ and the median item.
Equation 3-8: \[ M = L + \frac{\frac{n}{2}-cf_{\text{prev}}}{f_m}\times h \]
They are the same calculation written two ways — the fraction $\frac{n/2-cf}{f_m}$ counts the steps and $h$ supplies the width. Any small difference comes only from using $(n+1)/2$ versus $n/2$.

3-33Grouped median + meanLevin Ex. 3-33 · p. 102

Q. Gamefish weights: 0–24.9 → 5, 25–49.9 → 13, 50–74.9 → 16, 75–99.9 → 8, 100–124.9 → 6. (a) Estimate the median. (b) Compute the mean. (c) Which is better?

Solution

(a) cf column, locate $n/2$.

$N=48$, $n/2=24$; cf: 5, 18, 34, 42, 48 ⇒ median class 50–74.9 \[ M=50+\frac{24-18}{16}\times25=50+9.375=\mathbf{59.38 \text{ lb}} \]

(b) Grouped mean, midpoints 12.5, 37.5, 62.5, 87.5, 112.5.

\[ \bar{x}=\frac{2925}{48}=\mathbf{60.94 \text{ lb}} \]

$\bar{x}(60.94) > M(59.38)$ ⇒ mildly positively skewed — the 6 fish over 100 lb pull the mean up.

The median is slightly better as a description of a typical catch, but the gap is small (1.6 lb on a 60 lb average, under 3%), so either is defensible here — and saying that the difference is immaterial is itself a good answer.

3-34Median vs a targetLevin Ex. 3-34 · p. 103

Q. The Transit Authority thinks a reasonable median trip time is 30 minutes and suspects excessive speed raises maintenance costs. From the sample times, can you tell whether buses are driven too fast? If so, what explanation might the drivers give?

Solution

Compute the sample median and compare with 30.

Sort the times and take the middle value. If the median is materially below 30 minutes, buses are completing the route faster than the Authority considers reasonable — evidence consistent with excessive speed.

The explanation the drivers would give — this is the real question.

Speed is not the only way to finish early. Drivers could reasonably reply that:
• traffic was light during the sampled period,
• fewer passengers boarded, so fewer stops and shorter dwell times,
• the sample was taken off-peak, or on days without roadworks.

The statistical point: a low median is consistent with speeding but does not establish it. The Authority has an association, not a cause — it would need to measure speed directly, or control for traffic and passenger loads, before disciplining anyone.

3-35FractionsLevin Ex. 3-35 · p. 103

Q. Upholstery yardages recorded as mixed fractions (5¼, 6¼, 6, 7⅜, 9¼, 9½, 10½, …). Calculate the median yardage.

Solution

Why: the only difficulty is the fractions. Convert everything to decimals before sorting — comparing ⅜ with ½ by eye is where the errors come from.

Convert to decimals.

$5\tfrac14 = 5.25$ · $7\tfrac38 = 7.375$ · $9\tfrac12 = 9.5$ · $10\tfrac12 = 10.5$, and so on.

Sort, then take the middle position.

With $n$ values, the median is the $\left(\frac{n+1}{2}\right)$th item — or the average of the two middle items if $n$ is even.

Report the answer back in the original units — "the median job used about 8¼ yards" is more useful to Mark than a decimal, because that is how fabric is ordered.

3-36Open-ended classesLevin Ex. 3-36 · p. 103

Q. Insurance claims: less than $250 → 52, 250–499.99 → 337, 500–749.99 → 1,066, 750–999.99 → 1,776, $1,000 and above → 1,492. Find the median by the step method and verify with Equation 3-8.

Solution

Why: note that both end classes are open — and note that this does not matter, because the median class is in the middle. That is a real advantage of the median over the mean.

Total and locate the median item.

$N = 52+337+1066+1776+1492 = \mathbf{4723}$; $N/2 = 2361.5$
cf: 52, 389, 1455, 3231, 4723 ⇒ median class 750–999.99

Apply Equation 3-8 with $L=750$, $cf_{\text{prev}}=1455$, $f_m=1776$, $h=250$.

\[ M=750+\frac{2361.5-1455}{1776}\times250 \] \[ =750+\frac{906.5}{1776}\times250=750+127.6 \] \[ M=\mathbf{\$877.60} \]

The point about the open classes.

The open-ended classes never entered the calculation. The median needed only $N$, the cumulative frequency below the median class, and that class's own boundaries and frequency.

A mean could not have been computed at all without inventing midpoints for "less than 250" and "$1,000 and above". This is the single strongest argument for the median on open-ended data.

3-37Ordinal medianLevin Ex. 3-37 · p. 103

Q. Survey answers were: strongly disagree, disagree, mildly disagree, agree somewhat, agree, strongly agree. Of the six, which is the median?

Solution

Why: a trick question with a real point. The median is a position, so it works on ordered categories even though no arithmetic is possible on them.

Six ordered categories, so the middle falls between the 3rd and 4th.

Position $\dfrac{6+1}{2}=3.5$ ⇒ between mildly disagree and agree somewhat.

State the answer honestly.

There is no single median category. With an even number of ordered categories the median lies between mildly disagree and agree somewhat — and unlike numbers, these cannot be averaged to produce a value in between.

The lesson: the median is defined for ordinal data (the mean is not), but with an even count it may fall between two categories and simply has to be reported that way.

3-38Mode vs meanLevin Ex. 3-38 · p. 109

Q. Books checked out by 20 people: 1 0 2 2 3 4 2 1 2 0 2 2 3 1 0 7 3 5 4 2. (a) Mode. (b) Mean. (c) Graph the data — is the mean or the mode better?

Solution

(a) Tally.

0 → 3 · 1 → 3 · 2 → 7 · 3 → 3 · 4 → 2 · 5 → 1 · 7 → 1 \[ \text{Mode}=\mathbf{2 \text{ books}} \]

(b) Mean.

\[ \bar{x}=\frac{46}{20}=\mathbf{2.30 \text{ books}} \]

The plot rises to a clear peak at 2 and tails off to the right, with one person taking 7 — right-skewed.

The mode is better here, for two reasons: books are discrete, so "2.30 books" describes nobody, and the mode is the value the librarian would actually plan around. The single reader taking 7 pulls the mean up without being typical of anyone.
Median = 2 as well, agreeing with the mode and confirming that 2 is the typical value.

3-39Grouped modeLevin Ex. 3-39 · p. 109

Q. Ages at Twin Lakes: 47–51.9 → 4, 52–56.9 → 9, 57–61.9 → 13, 62–66.9 → 42, 67–71.9 → 39, 72–76.9 → 20, 77–81.9 → 9. Estimate the modal value.

Solution

Identify the modal class and its neighbours.

Highest $f = 42$ ⇒ modal class 62–66.9, with $f_0 = 13$, $f_2 = 39$, $L = 62$, $h = 5$.

Substitute — note the denominator, where sign errors happen.

\[ 2f_1-f_0-f_2 = 84-13-39 = 32 \] \[ Z = 62+\frac{42-13}{32}\times5 = 62+4.53 \]

Answer, and check it lies inside the modal class.

\[ Z=\mathbf{66.53 \text{ years}} \] 66.53 lies in 62–66.9 ✓, and it sits near the top of that class because the class above (39) is far heavier than the class below (13) — the formula leans toward the heavier neighbour, exactly as it should.
Mean = 66.82, median = 67.00 — all three nearly equal, so the distribution is close to symmetric.

3-40Mode on categoriesLevin Ex. 3-40 · p. 109

Q. What are the modal values for distributions of (a) hair colour, (b) blood type, (c) day of birth?

Solution

AnswerIn each case the mode is simply the category with the highest frequency — the most common hair colour, the most common blood type, the most common day of birth.

And that is the whole point of the question: these are nominal data. The mean and the median are undefined for them — there is no arithmetic on "brunette", and no meaningful order in which to find a middle value.

The mode is the only measure of central tendency available for nominal data, which is exactly why it exists as a separate measure rather than being displaced by the mean.

3-41Build + mode + meanLevin Ex. 3-41 · p. 109

Q. Apartments in 27 complexes: 91 79 66 98 127 139 154 147 192 88 97 92 87 142 127 184 145 162 95 89 86 98 145 129 149 158 241. (a) Build a distribution with intervals 66–87, 88–109, … (b) Estimate the mode. (c) Compute the mean of the raw data. (d) Compare.

Solution

(a) Classes of width 22.

66–87 → 4 · 88–109 → 8 · 110–131 → 3 · 132–153 → 6 · 154–175 → 3 · 176–197 → 2 · 198–219 → 0 · 220–241 → 1 (Σ = 27 ✓)

(b) Modal class 88–109; $f_0=4$, $f_1=8$, $f_2=3$, $L=88$, $h=22$.

\[ Z=88+\frac{8-4}{16-4-3}\times22=88+\frac{4}{9}\times22 \] \[ Z=\mathbf{97.78 \text{ apartments}} \]

\[ \bar{x}=\frac{3407}{27}=\mathbf{126.19 \text{ apartments}} \]

(d) The comparison — a gap of nearly 30.

$\bar{x}(126.19) \gg Z(97.78)$ ⇒ strongly positively skewed. Most complexes are small (88–109 units) but a few large ones — 184, 192 and especially 241 — drag the mean far above the typical value.

The mode describes a typical complex; the mean does not. For a question like "how big is a typical complex in Cary?", the mode or median is the honest answer.

3-43Grouped modeLevin Ex. 3-43 · p. 110

Q. Heat-storage capacity of 28 solar systems (days): 0–0.99 → 2, 1–1.99 → 4, 2–2.99 → 6, 3–3.99 → 7, 4–4.99 → 5, 5–5.99 → 3, 6–6.99 → 1. Find the modal value.

Solution

Modal class and neighbours.

Highest $f=7$ ⇒ class 3–3.99; $f_0=6$, $f_2=5$, $L=3$, $h=1$.

Substitute.

\[ Z=3+\frac{7-6}{14-6-5}\times1=3+\frac{1}{3}=\mathbf{3.33 \text{ days}} \] Mean = 3.286, median = 3.286 — all three within 0.05 of each other, so this distribution is almost perfectly symmetric and any of the three would serve. Worth saying in the report University Laboratories is writing.

3-44Modal class + open endLevin Ex. 3-44 · p. 110

Q. Summer earnings: $0–499 → 231, 500–999 → 304, 1,000–1,499 → 400, 1,500–1,999 → 296, 2,000–2,499 → 123, 2,500–2,999 → 68, $3,000 or more → 23. (a) Find the modal class. (b) Estimate the mode.

Solution

(a) Highest frequency.

$f = 400$ ⇒ modal class $1,000–1,499.

(b) Substitute, with $f_0=304$, $f_2=296$, $L=1000$, $h=500$.

\[ 2f_1-f_0-f_2 = 800-304-296=200 \] \[ Z=1000+\frac{400-304}{200}\times500=1000+240 \] \[ Z=\mathbf{\$1{,}240} \]

Why the mode is the right measure for Ed Grant.

The final class is open-ended ("$3,000 or more"), so a mean cannot be computed without inventing a midpoint. The mode needs only the modal class and its two neighbours, all of which are closed — so it is computable, and it answers Ed's actual question: what does a typical applicant earn over the summer?
Median = $1,234 — almost identical, confirming $1,240 as a solid figure to plan aid packages around.

3-42Mode of 3-36Levin Ex. 3-42 · p. 109

Q. Estimate the mode for the insurance-claim distribution of Exercise 3-36.

Solution

Modal class from the frequencies 52, 337, 1066, 1776, 1492.

Highest $f=1776$ ⇒ class 750–999.99; $f_0=1066$, $f_2=1492$, $L=750$, $h=250$.

Substitute.

\[ 2f_1-f_0-f_2=3552-1066-1492=994 \] \[ Z=750+\frac{1776-1066}{994}\times250=750+178.6=\mathbf{\$928.60} \]

Compare with the median of $877.60 from 3-36. $Z > M$, and with an open top class holding 1,492 claims the mean would be higher still — so the ordering is $M < Z < \bar{x}$, indicating a right-skewed claims distribution. For pricing, the insurer cares about the mean (total payout); for describing a typical claim, the median or mode.

SC 3-7Median vs meanLevin SC 3-7 · p. 103

Q. Prices: 0.98 1.05 1.08 1.08 1.09 1.14 1.22 1.24 1.33 1.55. (a) Median. (b) Mean. (c) Which is better?

Solution

(a) $n=10$ even ⇒ average items 5 and 6.

\[ M=\frac{1.09+1.14}{2}=\mathbf{\$1.115} \]

(b) Mean.

\[ \bar{x}=\frac{11.76}{10}=\mathbf{\$1.176} \]

$\bar{x} > M$ ⇒ mildly right-skewed (the $1.55 pulls the mean up). The median is a little better — but as Levin himself says, "there really isn't very much difference": 6 cents on a $1.15 average.
Matches Levin's printed answers exactly, including his conclusion that the choice hardly matters here.

SC 3-8Median, both methodsLevin SC 3-8 · p. 104

Q. Distribution: 100–149.5 → 12, 150–199.5 → 14, 200–249.5 → 27, 250–299.5 → 58, 300–349.5 → 72, 350–399.5 → 63, 400–449.5 → 36, 450–499.5 → 18 (N = 300). (a) Median class. (b) Which items? (c) Step width. (d) The median.

Solution

cf column and the median class.

cf: 12, 26, 53, 111, 183, 246, 282, 300 ⇒ $N/2=150$ falls in 300–349.5.

(b) With $N=300$, the median is the average of the 150th and 151st items.

Both lie in the 300–349.5 class, which starts after cf = 111.

\[ \text{step}=\frac{50}{72}=\mathbf{0.6944} \]

(d) Step to each item and average.

150th item: $300+38(0.6944)=326.3872$
151st item: $300+39(0.6944)=327.0816$ \[ M=\frac{326.3872+327.0816}{2}=\mathbf{326.7344} \]

Matches Levin's printed answer (326.7344) exactly. Equation 3-8 gives the same thing more quickly: $300+\frac{150-111}{72}\times50 = 327.08$ — the tiny difference is only the $(n+1)/2$ versus $n/2$ convention.

SC 3-9Mode vs meanLevin SC 3-9 · p. 108

Q. Ages of cars serviced: 5 6 3 6 11 7 9 10 2 4 10 6 2 1 5. (a) Mode. (b) Mean. (c) Which is better?

Solution

(a) Tally.

1→1 · 2→2 · 3→1 · 4→1 · 5→2 · 6→3 · 7→1 · 9→1 · 10→2 · 11→1 \[ \text{Mode}=\mathbf{6 \text{ years}} \]

(b) Mean.

\[ \bar{x}=\frac{87}{15}=\mathbf{5.80 \text{ years}} \]

Mode 6, mean 5.80, median 6 — all three essentially agree, so the distribution is roughly symmetric and either measure serves.

For the Autohaus's practical purpose — which parts to stock — the mode is marginally more useful, since it names an actual car age rather than a fractional one.

SC 3-10Mode on a skewed setLevin SC 3-10 · p. 111

Q. Classes 15–19, 20–24, 25–29, 30–34, 35+ with the modal class 15–19. (b) Compute the mode. (c) Compute the mean (Σx = 760, n = 30). (d) Which is the better measure?

Solution

(b) The mode by Levin's $d_1/(d_1+d_2)$ form, with $d_1 = 10$, $d_2 = 1$, $L = 15$, $w = 5$.

\[ Z = 15+\left(\frac{10}{10+1}\right)5 = 15+4.55=\mathbf{19.55} \]

\[ \bar{x}=\frac{760}{30}=\mathbf{25.33} \]

(d) The judgement — a gap of nearly 6.

$\bar{x}(25.33) \gg Z(19.55)$ ⇒ strongly positively skewed. The mode is the better measure, exactly as Levin concludes: the bulk of the data sits in the lowest class, and the mean is dragged upward by a thin tail of high values that describe almost nobody.

Matches Levin's printed answers (19.55, 25.33) exactly, including his conclusion. Note the $d_1/(d_1+d_2)$ form is algebraically identical to the $\frac{f_1-f_0}{2f_1-f_0-f_2}$ version used elsewhere on this page — $d_1 = f_1-f_0$ and $d_2 = f_1-f_2$.

THE 90-SECOND VERSION — WHAT TO CARRY INTO THE HALL

Summary Cheat Sheet

\[ \text{Sturges Rule: } k = 1 + 3.322 \log_{10} n, \qquad h = \frac{R}{k} \] \[ \text{Mean: } \bar{x} = \frac{\sum f x}{\sum f}, \qquad \text{Median: } M = L + \frac{\frac{n}{2} - cf_{\text{prev}}}{f_m} \cdot h, \qquad \text{Mode: } Z = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \cdot h \] \[ \text{Empirical Relation: } Z \approx 3M - 2\bar{x} \]

Consistency & Recommendations

Low SD + Leptokurtic $(K > 3)$ $\to$ Consistent
High SD + Platykurtic $(K < 3)$ $\to$ Inconsistent
Note: Every recommendation must explicitly name the statistic it targets.

◆ Where this sits in the rest of the site

This page is the drilling layer. The concepts behind it are in Unit 01 notes; the real exam questions in this unit's slot are worked in Solved Papers → Q3 (a), both versions; and the cross-unit problem set is at Practice Problems → Group A.

Skewness and kurtosis are read here because Q3 (a) mixes them into the Unit 1 table, but they are properly taught in Unit 02, Concepts 3–4 — read those two concepts straight after this page.

Pattern	Wording cue	Formula to reach for
Plain list of numbers	"the marks of 12 students are…"	\(\bar{x}=\Sigma x/n\). Check whether it is a sample or a population, and use the matching symbol.
Class intervals given	a table of classes and frequencies	\(\bar{x}=\Sigma fx/\Sigma f\) with midpoints.
Large / awkward midpoints	midpoints in hundreds or thousands, equal widths	Step-deviation — it turns big numbers into −3 … +3.
Items have different importance	"credits", "weights", "quantities purchased"	Weighted mean \(\Sigma wx/\Sigma w\). Never the simple mean.
Two or more groups	"40 men averaged … and 60 women averaged …"	Combined mean. Weight by the group sizes.
A value was misread	"it was later found that 25 was actually 52"	Rebuild \(\Sigma x\) from the old mean, swap the value, re-divide.
One frequency missing	"…and the mean is 38. Find the missing frequency."	Set \(\Sigma fx = \bar{x}\,\Sigma f\) and solve for the unknown.
Cumulative table given	"less than 10 → 8, less than 20 → 25 …"	Difference consecutive \(cf\) to get \(f\) first, then the ordinary grouped mean.

Source	Notation and layout it uses
Class notes	Writes \(\mu = \dfrac{\Sigma f_i x_i}{\Sigma f_i}\) and lays the work out as four columns — CI · xᵢ · fᵢ · fᵢxᵢ — with \(\Sigma f_i\) and \(\Sigma f_ix_i\) on the bottom row. Copy this layout exactly notes p. 4
Class notes	The red rule: inclusive series need NOT be converted for the mean; conversion IS required for median and mode notes p. 3
Levin	§3.2 arithmetic mean, §3.3 weighted mean — and the caution that the mean is pulled by extreme values pp. 77–95
Exam	Never asks you to compute a grouped mean. It hands you a mean and asks what it means next to the median and mode — Topic 5.

Pattern	Wording cue	What to do
Compute the median	"find the median of the following distribution"	cf column → locate → \(M = L+\frac{n/2-cf}{f_m}h\).
Compute the mode	"find the modal wage / most typical value"	Highest \(f\) → \(Z = L+\frac{f_1-f_0}{2f_1-f_0-f_2}h\).
One average missing	"mean is 48 and mode is 54; find the median"	Empirical relation \(Z = 3M-2\bar{x}\), rearranged.
Name the skew	"what does the difference between mean, median and mode signify?"	Order the three, name the direction, describe the tail in the story's own language.
Consistency comparison	"what do skewness and kurtosis imply about consistency?"	SD first (spread), then kurtosis (peakedness). Low SD + high K = consistent.
Recommend actions	"what strategies should the firm implement?"	Each recommendation must name the statistic it attacks. That link is the mark.
Which average?	"income data with a few very high earners"	Median — the mean is dragged by extremes; the mode for categorical / most-common questions.