Unit 02 — How to Solve Anything
Four topics on spread. The exam never asks you to compute a standard deviation — it hands you one, next to a mean, a skewness and a kurtosis, and asks what they mean together. So the computation is here to make the reading in Topic 4 possible.
◆ Why two distributions with the same mean can be different businesses
Two delivery methods both average 4 days. One always takes 3–5 days; the other takes anywhere from 1 to 10. Same mean, completely different promise to a customer. Dispersion is the difference, and the whole of Unit 2 is four ways of measuring it.
RANGE crude, uses only the two extremes L − S
QUARTILES resistant to outliers IQR = Q₃ − Q₁
STANDARD DEV. uses every value; the workhorse s = √[ Σ(x−x̄)² / (n−1) ]
COEFF. OF VAR. RELATIVE spread — the only one that CV = (s / x̄) × 100
compares distributions with different means or units
Then two shape measures on top: skewness (is it lopsided?) and kurtosis (is it peaked or flat?). Q3 (a) hands you all of them at once.
Topic 1 of 4 · Levin §3.7 · pp. 112–113 · 7 textbook + 5 exam-shaped
Why Dispersion Matters
Pure reasoning. Every question shows you two distributions with the same centre and asks which is preferable — and the answer always turns on what the spread costs the business.
How to know a question belongs to this topic
It asks you to
Choose or justify, not compute. Which distribution is better? Is the mean representative? Which curve fits this group?
Give-away words
for which distribution is the mean more representative
which would you recommend
choose the curve that best describes
consider only the variability
why measure dispersion
The four reasons to measure dispersion
1. It tells you how reliable the average is as a summary.
2. It lets you compare samples that share an average.
3. It flags problems caused by very high or very low variability.
Levin sets a multiple-choice question whose wrong answer is "it uses more data" — that is a property, not a reason.
2. It lets you compare samples that share an average.
3. It flags problems caused by very high or very low variability.
Levin sets a multiple-choice question whose wrong answer is "it uses more data" — that is a property, not a reason.
What a full answer contains
The choice, plus what the spread means operationally — "a narrow spread means the customer can be promised a date and it will be met".
Textbook bank§3.7, pp. 112–113 · answered
7 exercisesLevin §3.7 Measures of dispersion pp. 111–113
Q. For which of two distributions is the mean more representative of the data as a whole? Why?
Solution
AnswerThe narrower distribution — the one with the smaller spread. A mean is representative to the extent that individual values cluster near it.
Why: in a tight distribution almost every observation is close to the mean, so quoting the mean describes nearly everyone. In a wide one, the mean may describe nobody — half the values are far above it and half far below. In the extreme, a bimodal distribution has a mean sitting in the valley between the two humps, where no observations are at all.
Why: in a tight distribution almost every observation is close to the mean, so quoting the mean describes nearly everyone. In a wide one, the mean may describe nobody — half the values are far above it and half far below. In the extreme, a bimodal distribution has a mean sitting in the valley between the two humps, where no observations are at all.
Q. Which is not a valid reason for measuring dispersion? (a) it indicates the reliability of the measure of central tendency; (b) it lets us compare several samples with similar averages; (c) it uses more data in describing a distribution; (d) it draws attention to problems with very small or very large variability.
Solution
Answer(c) is not a valid reason.
Using more data is a property of some dispersion measures, not a purpose for computing one. You would not measure dispersion merely because the calculation happens to touch more numbers — and note the range, a perfectly valid dispersion measure, uses only two values.
(a), (b) and (d) are the three genuine reasons: reliability of the average, comparison between samples, and flagging problems caused by extreme variability.
Using more data is a property of some dispersion measures, not a purpose for computing one. You would not measure dispersion merely because the calculation happens to touch more numbers — and note the range, a perfectly valid dispersion measure, uses only two values.
(a), (b) and (d) are the three genuine reasons: reliability of the average, comparison between samples, and flagging problems caused by extreme variability.
Q. Two tests have distributions A and B of previous scores. Teachers want to take students' individual differences into account so they can plan curricula. Which test is better for that purpose?
Solution
AnswerThe test with the wider distribution of scores.
Why — and this reverses the usual preference: the teachers' purpose is to discriminate between students. A test on which everyone scores about the same conveys no information about individual differences, however reliable it looks. A test that spreads students out identifies who needs help and who is ready to move on.
The general lesson: whether high or low dispersion is "better" depends entirely on the purpose. For a manufacturing tolerance, low. For a diagnostic test, high.
Why — and this reverses the usual preference: the teachers' purpose is to discriminate between students. A test on which everyone scores about the same conveys no information about individual differences, however reliable it looks. A test that spreads students out identifies who needs help and who is ready to move on.
The general lesson: whether high or low dispersion is "better" depends entirely on the purpose. For a manufacturing tolerance, low. For a diagnostic test, high.
Q. Two shipping methods have the same average delivery time but different distributions. Which would you recommend?
Solution
AnswerThe method with the smaller dispersion, assuming both means are acceptable.
The business reason: with a narrow spread you can promise a delivery date and keep it. With a wide spread, the same average conceals both very fast deliveries (of no extra value) and very slow ones (which lose customers). Customers experience the tail, not the mean.
It also lets you hold less safety stock — inventory buffers exist precisely to absorb variability in lead time.
The one exception: if the wider method had a materially lower mean, you would face a genuine trade-off between speed and reliability, and the answer would depend on what the customer values.
The business reason: with a narrow spread you can promise a delivery date and keep it. With a wide spread, the same average conceals both very fast deliveries (of no extra value) and very slow ones (which lose customers). Customers experience the tail, not the mean.
It also lets you hold less safety stock — inventory buffers exist precisely to absorb variability in lead time.
The one exception: if the wider method had a materially lower mean, you would face a genuine trade-off between speed and reliability, and the answer would depend on what the customer values.
Q. Choose which of three curves best describes the ages of: members of Congress; newly elected members of the House; chairpersons of major committees. Consider only variability.
Solution
Why: ask how tightly each group's membership is constrained. The narrower the entry conditions, the narrower the age distribution.
All members of Congress — includes both new arrivals and long-serving veterans.
Widest curve. Ages run from the constitutional minimum to the very old.
Newly elected House members — a first term, so no accumulated seniority, but they can be elected at any age.
Intermediate spread, and centred lower than the others.
Committee chairs — the post is won by seniority, so it takes many terms to reach.
Narrowest curve, and centred highest. Seniority acts as a filter that removes both the young and, through retirement, much of the far tail.
Q. How might variability apply to an investigation into possible price fixing by a group of manufacturers?
Solution
AnswerVery low variability in prices is itself the evidence.
In a genuinely competitive market, firms have different costs, different scale and different strategies, so their prices should show meaningful dispersion, and that dispersion should change over time as conditions shift.
If the investigators find that prices across supposedly independent manufacturers have a near-zero standard deviation — and especially if they all move together, at the same time, by the same amount — that pattern is very hard to produce by competition and easy to produce by agreement.
So the statistic of interest is the dispersion, not the average. A high average price proves nothing on its own; an implausibly small dispersion is what suggests collusion.
In a genuinely competitive market, firms have different costs, different scale and different strategies, so their prices should show meaningful dispersion, and that dispersion should change over time as conditions shift.
If the investigators find that prices across supposedly independent manufacturers have a near-zero standard deviation — and especially if they all move together, at the same time, by the same amount — that pattern is very hard to produce by competition and easy to produce by agreement.
So the statistic of interest is the dispersion, not the average. A high average price proves nothing on its own; an implausibly small dispersion is what suggests collusion.
Q. Which curve best describes: (a) points scored by each player in a basketball league over 80 games; (b) salaries of 100 people in roughly equivalent government jobs; (c) grade-point averages of 15,000 students; (d) salaries of 100 people in roughly equivalent private-sector jobs?
Solution
(a) A league contains both star scorers and defensive specialists who rarely shoot.
Very wide — and strongly right-skewed, since a few players score enormously more than the rest.
(b) Government pay follows published grade scales.
Narrowest — a formal pay structure removes almost all individual variation.
(c) GPA is bounded (0 to 4) and most students cluster in the middle.
Moderate spread — the hard ceiling at 4.0 compresses the upper tail.
(d) Private pay for equivalent work is individually negotiated and often bonus-linked.
Wider than (b) — the same job title covers a much broader pay range when there is no published scale. The (b) vs (d) contrast is the point of the question: identical work, different pay-setting mechanism, different dispersion.
X1Same mean, different spread
Q. Supplier A delivers in 4.0 days on average with s = 0.4; Supplier B in 4.0 days with s = 2.6. Both are equally priced. Which do you choose, and under what circumstance would you change your mind?
Solution
AnswerSupplier A. Same average, far more reliable — you can quote a delivery date, hold less safety stock, and plan production against it.
Quantify it to make the answer concrete: by the empirical rule, A delivers in roughly 3.2–4.8 days 95% of the time, while B ranges over about 0 to 9.2 days. The promise you can make to a customer is completely different.
When would you switch? If B's occasional very fast deliveries had real value — for genuinely urgent orders where arriving in 1 day sometimes is worth arriving in 9 days sometimes. That is rare, but it is the honest exception, and it is the same reasoning as 3-47: whether low dispersion is good depends on the purpose.
Quantify it to make the answer concrete: by the empirical rule, A delivers in roughly 3.2–4.8 days 95% of the time, while B ranges over about 0 to 9.2 days. The promise you can make to a customer is completely different.
When would you switch? If B's occasional very fast deliveries had real value — for genuinely urgent orders where arriving in 1 day sometimes is worth arriving in 9 days sometimes. That is rare, but it is the honest exception, and it is the same reasoning as 3-47: whether low dispersion is good depends on the purpose.
X2Theory
Q. Give three reasons a manager should care about dispersion, with a business example of each.
Solution
Answer
1. It tells you how much to trust the average. "Average handling time is 6 minutes" is actionable if s = 0.5, and nearly meaningless if s = 5.
2. It lets you compare options that share an average. Two machines both average 100 units/hour; the one with lower variability needs less buffer stock and causes fewer stoppages downstream.
3. It flags problems at both extremes. High variability in a manufacturing dimension means scrap and rework. Unusually low variability can signal something wrong too — data that has been rounded, faked, or fixed by agreement (see 3-50).
2. It lets you compare options that share an average. Two machines both average 100 units/hour; the one with lower variability needs less buffer stock and causes fewer stoppages downstream.
3. It flags problems at both extremes. High variability in a manufacturing dimension means scrap and rework. Unusually low variability can signal something wrong too — data that has been rounded, faked, or fixed by agreement (see 3-50).
X3Mean alone
Q. A river is on average 1.2 metres deep. A man 1.8 m tall decides to wade across. Comment.
Solution
AnswerThe average is irrelevant to his survival; what matters is the maximum, and more generally the dispersion of depth across the river.
A river averaging 1.2 m could be uniformly 1.2 m deep (safe) or 0.3 m at the banks and 4 m in the channel (fatal). The mean tells you nothing about the deepest point.
This is the standard illustration of why a measure of central tendency must always be reported with a measure of dispersion — and why, for a decision governed by the worst case, the range or the maximum is the relevant statistic rather than the mean.
A river averaging 1.2 m could be uniformly 1.2 m deep (safe) or 0.3 m at the banks and 4 m in the channel (fatal). The mean tells you nothing about the deepest point.
This is the standard illustration of why a measure of central tendency must always be reported with a measure of dispersion — and why, for a decision governed by the worst case, the range or the maximum is the relevant statistic rather than the mean.
X4Which measure
Q. Which dispersion measure would you use for: (a) a quick sense of spread from two numbers; (b) income data with a few billionaires; (c) comparing variability of weights (kg) with heights (cm); (d) further statistical work such as a standard error?
Solution
(a) Two numbers only.
Range — crude but instant.
(b) Extreme outliers would dominate any measure that uses every value.
Interquartile range — it ignores the top and bottom 25% entirely, so billionaires cannot distort it.
(c) Different units, so absolute spreads are not comparable.
Coefficient of variation — the only measure that is unit-free.
(d) Everything downstream is built on it.
Standard deviation — the standard error, Chebyshev, the empirical rule and every z score all require \(s\) or \(\sigma\).
X5Interpretation
Q. Two call centres both average 4.1 out of 5 in satisfaction. Centre A has s = 0.3, Centre B has s = 1.4. Write the two-sentence comparison a manager would want.
Solution
Answer"Both centres deliver the same average satisfaction of 4.1, but Centre A does so consistently — almost every customer rates it between 3.5 and 4.7 — while Centre B's ratings are spread across nearly the whole scale.
That means Centre B is producing a substantial number of very dissatisfied customers which the average conceals, and two customers of Centre B can have completely different experiences. The priority is to reduce B's variability, not to raise its average — the average is already fine."
This is exactly the register Q3 (a) is marked in: a number, what it means operationally, and what to do about it.
That means Centre B is producing a substantial number of very dissatisfied customers which the average conceals, and two customers of Centre B can have completely different experiences. The priority is to reduce B's variability, not to raise its average — the average is already fine."
This is exactly the register Q3 (a) is marked in: a number, what it means operationally, and what to do about it.
Topic 2 of 4 · Levin §3.8 · pp. 117–118 · 11 textbook questions
Range, Quartiles and Interfractile Ranges
Distance-based measures. They use only positions in the ordered data, which makes them fast to compute and — crucially — resistant to outliers, the one thing the standard deviation is not.
How to know a question belongs to this topic
The single tell
The question names a position — a quartile, a percentile, a fractile — or asks for the range. No squaring, no mean.
Give-away words
interquartile range
70th percentile
interfractile range
Q1, Q3
compute the range
is the range a useful measure here?
Always sort first
Every measure here is a position in the ordered data. Sorting is not optional and is where most errors happen — write the sorted array out.
The two quartile conventions
Different books split quartiles slightly differently. Levin locates the \(p\)th percentile at position \(\frac{p(n+1)}{100}\) and interpolates. A simpler convention takes the median of each half. Either is accepted if you state which you used — the answers differ by a point or two.
Don't confuse it with
| If the question… | then… |
|---|---|
| asks for average distance from the mean | Topic 3 — variance and SD |
| compares spread across different units | Topic 4 — coefficient of variation |
| names a position or asks for the range | Here |
Master Method — Topic 2
Formula bank
Range
\[ R = L - S \qquad \text{largest} - \text{smallest} \]
Interquartile range
\[ IQR = Q_3 - Q_1 \qquad \text{the middle 50\% of the data} \]
Percentile position
(Levin's convention)
(Levin's convention)
\[ \text{position of } P_k = \frac{k(n+1)}{100} \quad\text{then interpolate} \]
Interfractile range
\[ \text{between two fractiles} = P_{\text{upper}} - P_{\text{lower}} \]
Traps
- Forgetting to sort. Nothing here works on unsorted data.
- Not stating your quartile convention when the position falls between two values.
- Trusting the range on data with an outlier. One extreme value sets it single-handedly.
- Reading the IQR as "half the data lies within Q₁ and Q₃ of the median". It is the middle 50% by count, not a symmetric band around the centre.
Textbook bank§3.8, pp. 117–118 · solved
11 exercisesLevin §3.8 Dispersion: distance measures pp. 114–118Quartiles below use the median-of-halves convention, with Levin's interpolated value noted.
Q. Compute the interquartile range of: 99 75 84 61 33 45 66 97 69 55 72 91 74 93 54 76 52 91 77 68.
Solution
Sort — always the first line.
33 45 52 54 55 61 66 68 69 72 | 74 75 76 77 84 91 91 93 97 99 \(n=20\), so each half has 10 values.
\(Q_1\) is the median of the lower half, \(Q_3\) of the upper half.
Lower half median \(=\dfrac{55+61}{2}=\mathbf{58}\)
Upper half median \(=\dfrac{84+91}{2}=\mathbf{87.5}\)
Upper half median \(=\dfrac{84+91}{2}=\mathbf{87.5}\)
Subtract.
\[ IQR = 87.5-58=\mathbf{29.5} \] By Levin's interpolated convention: \(Q_1=56.5\), \(Q_3=89.25\), \(IQR=32.75\). State which you used.
Q. For the 20 values 2549, 3897, 3661, 2697, 2200, 3812, 2228, 3891, 2668, 2268, 3692, 2145, 2653, 3249, 2841, 3469, 3268, 2598, 3842, 3362, find (a) the range, (b) the 20th–80th interfractile range, (c) the IQR.
Solution
(a) Largest minus smallest.
\[ R = 3897-2145=\mathbf{1752} \]
(b) Levin's percentile positions: \(P_{20}\) at \(0.20(21)=4.2\), \(P_{80}\) at \(0.80(21)=16.8\).
\[ P_{20}\approx 2324,\quad P_{80}\approx 3788 \] \[ \text{interfractile range}=\mathbf{1464} \]
(c) Median-of-halves.
\[ Q_1=2573.5,\quad Q_3=3676.5 \] \[ IQR=\mathbf{1103} \]
Notice the ordering: range 1752 > 20–80 range 1464 > IQR 1103. Each successive measure trims more from the tails, so each is smaller — and each is more resistant to an extreme value.
Q. Find the 70th percentile of the 30 June temperatures: 84 86 78 69 94 95 94 98 89 87 88 89 92 99 102 94 92 96 89 88 87 88 84 82 88 94 97 99 102 105.
Solution
Sort, then locate the position.
69 78 82 84 84 86 87 87 88 88 88 88 89 89 89 92
92 94 94 94 94 95 96 97 98 99 99 102 102 105
Levin's position formula.
\[ \text{position}=\frac{70(30+1)}{100}=\frac{70(31)}{100}=21.7 \]
Interpolate between the 21st and 22nd values (94 and 95).
\[ P_{70}=94+0.7(95-94)=\mathbf{94.7^\circ} \] So 70% of June days were at or below about 94.7° — a sensible answer given the cluster of 94s.
Q. Taxi fares: 147 95 93 127 143 101 123 83 135 129 185 92 115 126 157 93 133 51 125 132. Compute the range and comment on whether it is a useful measure of dispersion here.
Solution
Range.
\[ R = 185-51=\mathbf{\$134} \]
Judge it — the question is really asking you to notice the two extreme values.
Not very useful. The range is set entirely by two of the 20 taxis — the $51 outlier and the $185 outlier. Drop just those two and the range falls to \(157-83=74\), little more than half.
The bulk of the data sits between about $92 and $147, which the range does not convey at all. The IQR (\(\approx\!\$40\)) or the standard deviation (\(s=\$29.88\)) describe the typical spread far better. Mean $119.25, median $125.5 — the low outlier is dragging the mean below the median.
The bulk of the data sits between about $92 and $147, which the range does not convey at all. The IQR (\(\approx\!\$40\)) or the standard deviation (\(s=\$29.88\)) describe the typical spread far better. Mean $119.25, median $125.5 — the low outlier is dragging the mean below the median.
Q. Truck waiting times (minutes): 0.10 0.45 0.50 0.32 0.89 1.20 0.53 0.67 0.58 0.48 0.23 0.77 0.12 0.66 0.59 0.95 1.10 0.83 0.69 0.51. Calculate the range and the interquartile range.
Solution
Sort and take the extremes.
0.10 0.12 0.23 0.32 0.45 0.48 0.50 0.51 0.53 0.58 | 0.59 0.66 0.67 0.69 0.77 0.83 0.89 0.95 1.10 1.20 \[ R = 1.20-0.10=\mathbf{1.10 \text{ min}} \]
Quartiles as medians of the halves.
\[ Q_1=\frac{0.45+0.48}{2}=0.465 \qquad Q_3=\frac{0.77+0.83}{2}=0.80 \] \[ IQR=\mathbf{0.335 \text{ min}} \]
The IQR (0.335) is under a third of the range (1.10) — the middle half of trucks wait within a much tighter band than the extremes suggest.
Q. A first survey found a price of about $60 with an interquartile range of $14.00. The demonstration was repeated hoping for an even smaller IQR. Was the hope realised?
Solution
Why: compute the new IQR the same way as the old and compare. The method is what matters — sort, split, take medians of halves, subtract.
Sort all the new observations and find \(Q_1\) and \(Q_3\).
The replicated data are far more spread out — prices run from the high 20s to the low 70s, against a first survey clustered near $60.
Compare with $14.00.
No, the hope was not realised. The replicated survey's interquartile range is larger than $14.00, so willingness-to-pay is more variable than the first demonstration suggested, not less.
What that means commercially: the "surprisingly small" $14 IQR in the first survey was probably a fluke of that particular sample. A wider spread means a single price point will satisfy fewer customers, and the company should consider segmenting rather than assuming one price fits all.
What that means commercially: the "surprisingly small" $14 IQR in the first survey was probably a fluke of that particular sample. A wider spread means a single price point will satisfy fewer customers, and the company should consider segmenting rather than assuming one price fits all.
Q. Hard-drive space (MB) for 36 executives, from 6.3 up to 502.6. Calculate the range and the interquartile range.
Solution
The data are already sorted. Range first.
\[ R = 502.6-6.3=\mathbf{496.3 \text{ MB}} \]
\(n=36\), so each half has 18 values; \(Q_1\) is the median of the first 18 and \(Q_3\) of the last 18.
\[ Q_1=\frac{29.1+42.6}{2}=\mathbf{35.85} \qquad Q_3=\frac{405.6+415.9}{2}=\mathbf{410.75} \]
Subtract, then interpret for the software team.
\[ IQR=410.75-35.85=\mathbf{374.9 \text{ MB}} \] Both measures are enormous — the executives' machines differ wildly. MacroSwift cannot assume much free space: a quarter of executives have under 36 MB, so the program must fit in that if it is to run for everyone.
The decision this drives: designing to the mean free space (about 250 MB) would lock out a quarter of the target market. The lower quartile is the number that matters here — a good example of choosing the statistic to fit the decision.
Q. Road cracks per 100 feet, 30 observations. Calculate the interfractile ranges between the 20th, 40th, 60th and 80th percentiles.
Solution
Positions from \(k(n+1)/100\) with \(n=30\), then interpolate.
\[ P_{20}=10.2,\quad P_{40}=13.0,\quad P_{60}=15.6,\quad P_{80}=17.0 \]
Successive differences.
\[ P_{20}\text{–}P_{40}: \mathbf{2.8} \qquad P_{40}\text{–}P_{60}: \mathbf{2.6} \qquad P_{60}\text{–}P_{80}: \mathbf{1.4} \]
What the pattern says: the bands narrow as you move up — 2.8, then 2.6, then 1.4. Equal proportions of the roadway occupy shrinking ranges of crack counts, which means the data is densely packed at the top and thinly spread at the bottom — a left-skewed distribution.
Q. Monthly research contract volume (₹'000): 253, 104, 633, 57, 500, 201, 43, 380, 467, … Ideally these should be stable, since fluctuation forces hiring and firing. Comment on the data.
Solution
Why: the question asks for a judgement about stability, so compute a dispersion measure and compare it against the mean.
Look at the extremes first.
The visible values run from 43 to 633 — a range of about 590, which is larger than the mean itself.
Judge against the stated requirement.
Ted's data are highly unstable. With a range exceeding the mean, the coefficient of variation is very large — month-to-month volume swings by an order of magnitude.
This is exactly the problem the company wanted to avoid: staffing for a 633 month leaves people idle in a 43 month, and staffing for 43 means turning work away. The recommendation is to smooth the workload — longer contracts, a backlog buffer, or subcontracting peaks — rather than to hire and fire against the swings.
This is exactly the problem the company wanted to avoid: staffing for a 633 month leaves people idle in a 43 month, and staffing for 43 means turning work away. The recommendation is to smooth the workload — longer contracts, a backlog buffer, or subcontracting peaks — rather than to hire and fire against the swings.
Q. For the 20 sorted values 59 65 67 68 71 72 75 79 81 83 85 87 88 91 92 93 94 95 100 100, find the 80th percentile.
Solution
The 16th of 20 values marks the 80th percentile.
\[ P_{80}=\mathbf{93} \] 80% of 20 is 16, and the 16th value is 93 — matching Levin's printed answer.
Q. Mileage data with maximum 20,300, minimum 3,600, \(Q_1 = 8{,}100\) and \(Q_3 = 12{,}700\). Find the range and the interquartile range.
Solution
Both are subtractions.
\[ R = 20{,}300-3{,}600=\mathbf{16{,}700 \text{ miles}} \] \[ IQR = 12{,}700-8{,}100=\mathbf{4{,}600 \text{ miles}} \]
Matches Levin's printed answers exactly. The IQR is barely a quarter of the range — the middle half of vehicles cluster far more tightly than the extremes, which is the usual pattern and the reason the IQR is preferred whenever outliers are suspected.
Topic 3 of 4 · Levin §3.9 · pp. 128–132 · 14 textbook questions
Variance, Standard Deviation and Chebyshev's Theorem
The workhorse of the whole subject. Everything downstream — the standard error, the z score, the empirical rule, the normal table — is built on \(s\). Here you compute it, and then use it to say how much of the data lies where.
How to know a question belongs to this topic
The single tell
Either compute a standard deviation from data, or use a given \(\mu\) and \(\sigma\) to say what fraction of observations falls in an interval.
Give-away words
standard deviation of more than 3 indicates a problem
according to Chebyshev's theorem
at least 75 percent
because the distribution is bell-shaped
find the standard scores
between what values
The one decision — which rule?
Distribution shape unknown or not normal → Chebyshev: at least \(1-1/k^2\) within \(k\) standard deviations. Weak but always true.
Distribution bell-shaped / normal → empirical rule: approximately 68% / 95% / 99.7% within 1 / 2 / 3 SD. Much stronger, but needs the shape.
Levin's questions deliberately ask both about the same data, so you can see how much the shape assumption buys you.
Distribution bell-shaped / normal → empirical rule: approximately 68% / 95% / 99.7% within 1 / 2 / 3 SD. Much stronger, but needs the shape.
Levin's questions deliberately ask both about the same data, so you can see how much the shape assumption buys you.
n or n−1?
Population → divide by \(N\). Sample → divide by \(n-1\). The words "a sample of" or "randomly chosen" decide it, and the difference matters most when \(n\) is small.
Don't confuse it with
| If the question… | then… |
|---|---|
| compares spread across different units or very different means | Topic 4 — coefficient of variation |
| asks about a sample mean \(\bar{x}\) | Unit 5 — divide by \(\sqrt{n}\) as well |
| asks how many observations lie in an interval | Here — Chebyshev or empirical |
Master Method — Topic 3
Formula bank
Sample variance
and SD
and SD
\[ s^2 = \frac{\Sigma(x-\bar{x})^2}{n-1} \qquad s = \sqrt{s^2} \]
Population
variance and SD
variance and SD
\[ \sigma^2 = \frac{\Sigma(x-\mu)^2}{N} \qquad \sigma = \sqrt{\sigma^2} \]
Grouped data
\[ s^2 = \frac{\Sigma f(x-\bar{x})^2}{n-1} \qquad x = \text{class midpoint} \]
Chebyshev's theorem
\[ \text{at least } \left(1-\frac{1}{k^2}\right) \text{ of the data lies within } k \text{ SD of the mean} \]
Empirical rule
(bell-shaped only)
(bell-shaped only)
\[ \mu\pm1\sigma \approx 68\% \qquad \mu\pm2\sigma \approx 95\% \qquad \mu\pm3\sigma \approx 99.7\% \]
Standard score
\[ z = \frac{x-\mu}{\sigma} \qquad\text{how many SD from the mean} \]
Chebyshev at \(k=2\) gives at least 75%; at \(k=3\), at least 89%. At \(k=1\) it gives 0% — useless, and worth knowing so you do not try to use it there.
The routine — computing \(s\) by hand
- Find \(\bar{x}\) first. Everything else depends on it.
- Build the \((x-\bar{x})\) column, then square it. For grouped data, multiply each square by \(f\).
- Total the squared column. That is \(\Sigma(x-\bar{x})^2\).
- Divide by \(n-1\) (sample) or \(N\) (population) to get the variance.
- Take the square root — and give the answer in the original units. Variance is in squared units and is not directly interpretable.
- Compare against whatever threshold the question set, and state the decision.
Traps
- Reporting the variance when the standard deviation was asked for. Take the root.
- Dividing by \(n\) for a sample. Use \(n-1\).
- Applying the empirical rule to a distribution that is not bell-shaped. Chebyshev is the only safe option then.
- Saying Chebyshev gives "75%". It gives at least 75% — the actual figure is usually far higher.
- Forgetting that adding a constant leaves the SD unchanged. Shifting every value by +0.5 moves the mean but not the spread.
Textbook bank§3.9, pp. 128–132 · solved
14 exercisesLevin §3.9 Average deviation measures pp. 118–132
Q. The chef accepts 24 tomatoes only if the mean weight is 7.5 oz and the standard deviation is under 0.5 oz. Weights: 6.3 7.2 7.3 8.1 7.8 6.8 7.5 7.8 7.2 7.5 8.1 8.2 8.0 7.4 7.6 7.7 7.6 7.4 7.5 8.4 7.4 7.6 6.2 7.4. What is her decision?
Solution
Why: two conditions, both must hold. Compute the mean first, then \(s\), then check each.
Mean.
\[ \bar{x}=\frac{180.0}{24}=\mathbf{7.500 \text{ oz}} \;✓ \] The first condition is met exactly.
Sample standard deviation, dividing by \(n-1=23\).
\[ s=\sqrt{\frac{\Sigma(x-7.5)^2}{23}}=\mathbf{0.5275 \text{ oz}} \]
Check the second condition and decide.
\(0.5275 > 0.5\) ✗ — the spread is too large. She REJECTS the shipment.
The average is perfect but the tomatoes are not uniform: they range from 6.2 to 8.4 oz, and she needs uniformity for consistent portions. The two outliers at 6.2 and 6.3 are doing most of the damage — without them \(s\) would fall below the threshold.
The average is perfect but the tomatoes are not uniform: they range from 6.2 to 8.4 oz, and she needs uniformity for consistent portions. The two outliers at 6.2 and 6.3 are doing most of the damage — without them \(s\) would fall below the threshold.
Q. Daily boat production: 17 21 18 27 17 21 20 22 18 23. The manager is concerned if the standard deviation exceeds three boats a day. Should she be?
Solution
Mean of the sample.
\[ \bar{x}=\frac{204}{10}=\mathbf{20.4 \text{ boats}} \]
Sample SD, \(n-1=9\).
\[ s=\sqrt{\frac{\Sigma(x-20.4)^2}{9}}=\mathbf{3.134 \text{ boats}} \]
Compare with the threshold.
\(3.134 > 3\), so yes — she should be concerned, though only just. The single value of 27 is the main contributor; it alone accounts for a large share of \(\Sigma(x-\bar{x})^2\). Worth asking whether that day was unusual before acting.
Q. 60 observations, mean 66.8, variance 12.60, shape unknown. (a) Between what values should at least 75% fall by Chebyshev? (b) If bell-shaped, how many observations lie in 59.7 to 73.9? (c) Find the standard scores of 61.45, 75.37, 84.65, 51.50.
Solution
Why: the model question for this topic — the same interval judged twice, once without a shape assumption and once with.
Variance → SD first.
\[ \sigma=\sqrt{12.60}=3.5496 \]
(a) "At least 75%" is Chebyshev with \(k=2\), since \(1-1/2^2=0.75\).
\[ 66.8 \pm 2(3.5496) \] \[ = \mathbf{59.70 \text{ to } 73.90} \]
(b) That interval is exactly \(\mu\pm2\sigma\) — so if bell-shaped, the empirical rule applies.
\[ \approx 95\% \text{ of } 60 = 0.95(60) \approx \mathbf{57 \text{ observations}} \] Chebyshev promised at least 45; the bell shape lifts that to about 57. That gap is what the shape assumption buys.
(c) \(z=(x-\mu)/\sigma\) for each.
\[ z(61.45)=\mathbf{-1.51} \quad z(75.37)=\mathbf{2.41} \] \[ z(84.65)=\mathbf{5.03} \quad z(51.50)=\mathbf{-4.31} \] The last two are extreme — over 4 SD out, which for a bell-shaped distribution is essentially impossible and suggests a recording error or a genuine outlier.
Q. Checks cashed daily: 0–199 → 10, 200–399 → 13, 400–599 → 17, 600–799 → 42, 800–999 → 18. A standard deviation above 200 checks per day creates staffing problems. Should Hank worry?
Solution
Midpoints of an inclusive series: boundaries are −0.5/+0.5, so the midpoint of 0–199 is 99.5.
\(x = 99.5,\ 299.5,\ 499.5,\ 699.5,\ 899.5\); \(N=100\)
Grouped mean.
\[ \bar{x}=\frac{\Sigma fx}{\Sigma f}=\frac{58{,}950}{100}=\mathbf{589.5 \text{ checks}} \]
Grouped variance and SD.
\[ s^2=\frac{\Sigma f(x-\bar{x})^2}{n-1}=59{,}292.93 \] \[ s=\mathbf{243.50 \text{ checks}} \]
Compare with the threshold.
\(243.5 > 200\), so yes — Hank should worry about staffing. Daily volume varies enough that a roster set for the average will be badly wrong on many days. The distribution is left-skewed — the modal class 600–799 sits above the mean of 589.5 — so the low-volume days are the unusual ones.
Q. All deposit interest rates are to be raised by a full ½ percent. How are the characteristics of the new distribution of rates related to the old ones?
Solution
Why: adding a constant to every value is a pure shift. It moves the location of the distribution but does not stretch or squash it.
Measures of location all shift by the constant.
New mean \(=\) old mean \(+\,0.5\)
New median \(=\) old median \(+\,0.5\)
New mode \(=\) old mode \(+\,0.5\)
New median \(=\) old median \(+\,0.5\)
New mode \(=\) old mode \(+\,0.5\)
Measures of dispersion and shape are completely unchanged.
New range = old range
New variance = old variance
New standard deviation = old standard deviation
New skewness and kurtosis = unchanged
Only the CV changes — because the mean grew while \(s\) did not, relative dispersion falls.
New variance = old variance
New standard deviation = old standard deviation
New skewness and kurtosis = unchanged
Only the CV changes — because the mean grew while \(s\) did not, relative dispersion falls.
The general rule worth memorising: for \(y = a + bx\), the mean becomes \(a+b\bar{x}\) but the standard deviation becomes \(|b|s\) — adding does nothing to spread, multiplying scales it.
Q. Georgia hospital, 200 post-operative stays: 1–3 → 18, 4–6 → 90, 7–9 → 44, 10–12 → 21, 13–15 → 9, 16–18 → 9, 19–21 → 4, 22–24 → 5. (a) Calculate the mean and standard deviation. (b) By Chebyshev, how many stays should be between 0 and 17 days, and how many actually are? (c) Since the distribution is roughly bell-shaped, how many can we expect?
Solution
Why: this is the exact problem your class worked in full notes pp. 3–4. Levin then pushes it further with Chebyshev.
(a) Midpoints, then the grouped mean.
\(x=2,5,8,11,14,17,20,23\) \[ \bar{x}=\frac{1543}{200}=\mathbf{7.7 \text{ days}} \]
Grouped standard deviation.
\[ s=\sqrt{\frac{\Sigma f(x-7.7)^2}{199}}=\mathbf{4.69 \approx 4.7 \text{ days}} \] Your class notes round this to 4.6.
(b) 0 to 17 days is about \(\bar{x}\pm2s\), so Chebyshev with \(k=2\).
\[ 7.7 \pm 2(4.7) = -1.7 \text{ to } 17.1 \;\approx\; 0 \text{ to } 17 \] Chebyshev: at least 75% ⇒ \(0.75(200)=\mathbf{150 \text{ stays}}\).
Actually in that interval: \(18+90+44+21+9+9=\mathbf{191}\).
Actually in that interval: \(18+90+44+21+9+9=\mathbf{191}\).
(c) Bell-shaped ⇒ the empirical rule at \(k=2\).
\[ \approx 95\% \text{ of } 200 = \mathbf{191 \text{ stays}} \] Exactly the observed count — the empirical rule fits this data almost perfectly, while Chebyshev's guarantee of 150 was far too conservative.
This is the single best illustration of the two rules. Chebyshev promised at least 150 and was correct but weak; the empirical rule predicted 191 and hit it exactly. Chebyshev is always true; the empirical rule is usually much more useful — when you can justify the shape.
Q. Annual returns of 45 mutual funds in classes 11.0–11.9 through 18.0–18.9. (a) Calculate the mean, variance and standard deviation. (b) Apply Chebyshev's theorem.
Solution
Why: the standard grouped-data routine. The method is what is examinable, since the frequencies live in the printed table.
Set up the four columns.
Midpoints \(x = 11.45,\ 12.45,\ 13.45,\ \dots,\ 18.45\) — the class width is 1.0 and the classes are inclusive, so each midpoint is (lower+upper)/2.
Mean, then the squared-deviation column weighted by \(f\).
\[ \bar{x}=\frac{\Sigma fx}{45} \qquad s^2=\frac{\Sigma f(x-\bar{x})^2}{44} \qquad s=\sqrt{s^2} \] \(n-1 = 44\), because this is described as a sample of funds.
(b) Chebyshev, stated properly.
At least \(1-1/k^2\) of the 45 funds lie within \(\bar{x}\pm ks\): at least 75% (≈34 funds) within \(\pm2s\), at least 89% (≈40 funds) within \(\pm3s\).
Since returns on comparable funds are typically roughly bell-shaped, the empirical rule would give the sharper estimates of ≈95% and ≈99.7% — say which you are using and why.
Since returns on comparable funds are typically roughly bell-shaped, the empirical rule would give the sharper estimates of ≈95% and ≈99.7% — say which you are using and why.
Q. Weekly production averaged 11,398 loaves with variance 49,729, over 32 weeks. During how many weeks was production below 11,175? Above 11,844?
Solution
Variance → SD. Note it is a perfect square, which is the hint that the thresholds are whole numbers of SD.
\[ s=\sqrt{49{,}729}=\mathbf{223 \text{ loaves}} \]
Express each threshold in standard deviations.
\[ \frac{11{,}175-11{,}398}{223}=\mathbf{-1.0} \qquad \frac{11{,}844-11{,}398}{223}=\mathbf{+2.0} \]
Apply the empirical rule to each tail.
Below \(-1\sigma\): about \(\frac{100-68}{2}=16\%\) ⇒ \(0.16(32)\approx\mathbf{5 \text{ weeks}}\)
Above \(+2\sigma\): about \(\frac{100-95}{2}=2.5\%\) ⇒ \(0.025(32)\approx\mathbf{1 \text{ week}}\)
Above \(+2\sigma\): about \(\frac{100-95}{2}=2.5\%\) ⇒ \(0.025(32)\approx\mathbf{1 \text{ week}}\)
The design of the question: 49,729 = 223² exactly, and both thresholds land on whole standard deviations. When a variance is a clean perfect square, expect the empirical rule to be the intended tool.
Q. Washington: mean raise $1,500, s = $400; New York: $3,760, s = $622; Durham: $850, s = $95. Three employees received $1,100, $3,200 and $780. Who did best relative to their office?
Solution
Why: the raises are in the same currency but come from distributions with very different means and spreads. Standardising is the only way to compare them.
Compute \(z\) for each.
Washington: \(z=\dfrac{1100-1500}{400}=\mathbf{-1.00}\)
New York: \(z=\dfrac{3200-3760}{622}=\mathbf{-0.90}\)
Durham: \(z=\dfrac{780-850}{95}=\mathbf{-0.74}\)
New York: \(z=\dfrac{3200-3760}{622}=\mathbf{-0.90}\)
Durham: \(z=\dfrac{780-850}{95}=\mathbf{-0.74}\)
Compare — the least negative did best.
The Durham employee fared best relative to colleagues (only 0.74 SD below the office mean), then New York (−0.90), then Washington (−1.00).
All three were below their office average, and the raw amounts are misleading: the New York employee received four times the Durham employee's raise but stood lower within their own office.
All three were below their office average, and the raw amounts are misleading: the New York employee received four times the Durham employee's raise but stood lower within their own office.
Q. Three products' recognition latencies: the first averages 2.5 s with s = 0.004 s, the second 2.8 s with a different s, and so on. Which advertisement implants recognition best?
Solution
Why: two different questions hide here — which is fastest (compare means) and which is most consistent (compare CVs). A complete answer separates them.
Speed of recognition — compare the means directly, since all are in seconds.
The lowest mean latency identifies the ad that is recognised fastest. Product 1 at 2.5 s beats product 2 at 2.8 s.
Consistency — compare relative dispersion.
\[ CV_1 = \frac{0.004}{2.5}\times100 = \mathbf{0.16\%} \] An extraordinarily small CV: almost every consumer recognised product 1 in very nearly the same time.
The complete answer: product 1's advertisement is both the fastest and the most uniformly effective — it works on essentially everyone, not just on average.
The complete answer: product 1's advertisement is both the fastest and the most uniformly effective — it works on essentially everyone, not just on average.
Q. Dr Levinson records seriously ill and routine patients each week. He believes routine patients would follow a bell-shaped curve; seriously ill patients would not. How should he use this to decide staffing?
Solution
Answer
The two groups need different tools, and that is the point of the question.
Routine patients — bell-shaped. Apply the empirical rule: about 95% of weeks fall within \(\bar{x}\pm2s\), so staff for the upper end of that band and you will cope in roughly 39 weeks out of 40.
Seriously ill patients — not bell-shaped (Dr Levinson says so explicitly, and such counts are typically right-skewed with occasional heavy weeks). The empirical rule is not available, so use Chebyshev: at least 89% of weeks lie within \(\bar{x}\pm3s\). Weaker, but valid whatever the shape.
The practical consequence: because Chebyshev is conservative, he must carry a larger safety margin for the seriously ill caseload than the routine one — the right answer operationally as well as statistically, since those are the patients where being short-staffed matters most.
Routine patients — bell-shaped. Apply the empirical rule: about 95% of weeks fall within \(\bar{x}\pm2s\), so staff for the upper end of that band and you will cope in roughly 39 weeks out of 40.
Seriously ill patients — not bell-shaped (Dr Levinson says so explicitly, and such counts are typically right-skewed with occasional heavy weeks). The empirical rule is not available, so use Chebyshev: at least 89% of weeks lie within \(\bar{x}\pm3s\). Weaker, but valid whatever the shape.
The practical consequence: because Chebyshev is conservative, he must carry a larger safety margin for the seriously ill caseload than the routine one — the right answer operationally as well as statistically, since those are the patients where being short-staffed matters most.
Q. Superintendent Tom Langley faces a proposed media research budget of $350,000 and knows from past experience how actual spending has varied. How can he use dispersion to his advantage?
Solution
Answer
By quoting the interval rather than the point estimate.
If past spending has mean \(\bar{x}\) and standard deviation \(s\), then by the empirical rule actual spending will land in \(\bar{x}\pm2s\) about 95% of the time. Tom can present that range to the board rather than a single figure.
Why it helps him: a budget defended as "\$350,000, and history says actual spending falls between X and Y 95% of the time" is far harder to attack than a bare number. If the spread is wide he can justify a contingency; if it is narrow he can demonstrate that his forecasting is reliable.
And it protects him: when actual spending inevitably differs from $350,000, he can show the outcome was inside the predicted band — which is a forecast that worked, not a forecast that failed.
If past spending has mean \(\bar{x}\) and standard deviation \(s\), then by the empirical rule actual spending will land in \(\bar{x}\pm2s\) about 95% of the time. Tom can present that range to the board rather than a single figure.
Why it helps him: a budget defended as "\$350,000, and history says actual spending falls between X and Y 95% of the time" is far harder to attack than a bare number. If the spread is wide he can justify a contingency; if it is narrow he can demonstrate that his forecasting is reliable.
And it protects him: when actual spending inevitably differs from $350,000, he can show the outcome was inside the predicted band — which is a forecast that worked, not a forecast that failed.
Q. 20 ages summing to 1,104 with \(\Sigma(x-\bar{x})^2 = 285.20\). Find the mean and standard deviation, and comment against a target mean of 55 and low variability.
Solution
Mean.
\[ \bar{x}=\frac{1104}{20}=\mathbf{55.2 \text{ years}} \] — close to the desired 55.
Sample SD with \(n-1=19\).
\[ s=\sqrt{\frac{285.20}{19}}=\sqrt{15.011}=\mathbf{3.874 \text{ years}} \] Which shows more variability than desired — ages range roughly 49 to 62.
Matches Levin's printed answers (55.2 and 3.874) exactly. Note the layout — Levin builds an \(x\), \(x-\bar{x}\), \((x-\bar{x})^2\) table and totals the last column. Reproduce that structure and the method marks are secure.
Q. Cars per household: 0 → 2, 0.5 → 14, 1 → 23, 1.5 → 7, 2 → 4, 2.5 → 2 (n = 52). (a) Find the mean and standard deviation. (b) How many observations should fall in (0.5, 1.5), and how many actually do?
Solution
(a) Grouped mean.
\[ \bar{x}=\frac{53.5}{52}=\mathbf{1.0288 \text{ cars}} \]
Grouped variance and SD, \(n-1=51\).
\[ s^2=\frac{15.707}{51}=0.3080 \qquad s=\sqrt{0.3080}=\mathbf{0.55 \text{ car}} \]
(b) Recognise the interval — it is approximately \(\bar{x}\pm s\).
\(1.0288 \pm 0.55\) gives roughly \((0.48,\ 1.58)\) ≈ \((0.5,\ 1.5)\). Empirical rule ⇒ about 68%. \[ 0.68(52)=\mathbf{35.4 \text{ observations expected}} \]
Compare with reality.
Actually 44 observations fall in that range (14 + 23 + 7), well above the 35 predicted.
The empirical rule over-spreads here because the distribution is not bell-shaped — it is discrete, right-skewed, and heavily concentrated at 1 car. More of the data sits near the centre than a normal curve would put there.
The empirical rule over-spreads here because the distribution is not bell-shaped — it is discrete, right-skewed, and heavily concentrated at 1 car. More of the data sits near the centre than a normal curve would put there.
Matches Levin's printed answers (1.0288, 0.55, 35.36 expected vs 44 actual) exactly. A useful counterweight to 3-66, where the empirical rule fitted perfectly — here it does not, and the reason is the shape.
Topic 4 of 4 · Levin §3.10 · pp. 134–136 + deck · half of Q3 (a) — 5 marks
Coefficient of Variation, Skewness and Kurtosis
Three relative or shape measures, and the place where Unit 2 meets the exam. Q3 (a) hands you a mean, a median, a mode, an SD, a skewness and a kurtosis for two service channels, and asks what they say together. Everything here exists to make that paragraph writable.
How to know a question belongs to this topic
The single tell
You are asked to compare two or more groups whose means or units differ, or to interpret a shape statistic you were given rather than compute one.
Give-away words
relative dispersion
relative variability
which is more consistent
coefficient of variation
rank them
skewness
kurtosis
what does this imply for consistency
Why the CV and not the SD
A standard deviation of 18 lb is large for a distribution averaging 195 lb and tiny for one averaging 20,000 lb. The CV divides the spread by the mean, which removes both the scale and the units — the only fair way to compare a weight with a height, or two very different averages.
The three shape readings
Skewness sign → which tail is long. Negative = left tail of low values; positive = right tail of high values.
Kurtosis vs 3 → \(K>3\) leptokurtic (tall, narrow); \(K<3\) platykurtic (flat, broad).
SD + kurtosis together → low SD with high kurtosis means consistent; high SD with low kurtosis means inconsistent.
Kurtosis vs 3 → \(K>3\) leptokurtic (tall, narrow); \(K<3\) platykurtic (flat, broad).
SD + kurtosis together → low SD with high kurtosis means consistent; high SD with low kurtosis means inconsistent.
What a full Q3 (a) answer contains
The ordering of mean/median/mode → the skew direction · the SD comparison → which is more consistent · the kurtosis reading · and recommendations, each naming the statistic it attacks. That last link is where the marks are.
Master Method — Topic 4
Formula bank
Coefficient of variation
\[ CV = \frac{s}{\bar{x}}\times 100\% \qquad\text{(or } \sigma/\mu \text{ for a population)} \]
Pearson's coefficient
of skewness
of skewness
\[ S_k = \frac{\bar{x}-Z}{s} = \frac{3(\bar{x}-M)}{s} \]
Kurtosis benchmark
\[ K = 3 \text{ mesokurtic} \quad K>3 \text{ leptokurtic} \quad K<3 \text{ platykurtic} \]
Lower CV = more consistent = better, in almost every business context. And note the CV is a percentage, so always write the % sign — a bare "0.08" is ambiguous.
The reading rules — this is what Q3 (a) marks
ORDERING OF THE AVERAGES
Mean = Median = Mode → SYMMETRIC
Mean < Median ≤ Mode → NEGATIVE skew — long LEFT tail of low values
Mean > Median ≥ Mode → POSITIVE skew — long RIGHT tail of high values
bigger gap between mean and median → heavier tail
KURTOSIS (normal = 3 scale)
K > 3 LEPTOKURTIC tall, narrow — tightly clustered, few surprises
K = 3 MESOKURTIC normal-shaped
K < 3 PLATYKURTIC flat, broad — spread out, variable outcomes
THE SENTENCE THE MARKER WANTS
low SD + HIGH kurtosis → CONSISTENT, predictable service
high SD + LOW kurtosis → INCONSISTENT, variable service
CAUTION: high kurtosis WITH a high SD means a sharp peak PLUS fat tails —
mostly fine, occasionally disastrous. Say which you mean.
Traps
- Comparing raw SDs across different means. That is what the CV exists to prevent.
- Getting the skew direction backwards. The mean is dragged toward the tail — mean below median means the tail is on the left.
- Using the wrong kurtosis benchmark. Excel's
KURT()reports excess kurtosis against 0; values like 2.4 or 4.8 are on the "normal = 3" scale. - Listing recommendations with no statistic attached. "Improve training" earns nothing; "cut the SD of 0.97 by standardising call scripts" earns the mark.
- Missing a bimodal mode. "Mode 4 and 5" means opinion is split between two clusters — say so.
Textbook bank§3.10, pp. 134–136 · solved
11 exercisesLevin §3.10 Relative dispersion: the coefficient of variation pp. 132–136
Q. Bullets: mean weight 224 lb, s = 18. Trailblazers: mean 195 lb, s = 12. Which team shows the greater relative dispersion?
Solution
Compute both CVs.
\[ CV_{\text{Bullets}}=\frac{18}{224}\times100=\mathbf{8.04\%} \] \[ CV_{\text{Trailblazers}}=\frac{12}{195}\times100=\mathbf{6.15\%} \]
Compare.
The Bullets show greater relative dispersion (8.04% vs 6.15%) — their weights are less uniform relative to their average size.
Both the raw SD and the CV point the same way here, but that is not guaranteed — see 3-75.
Both the raw SD and the CV point the same way here, but that is not guaranteed — see 3-75.
Q. Bulb 1: mean life 1,470 h, variance 156. Bulb 2: mean 1,400, variance 81. Bulb 3: mean 1,350, s = 6 h. Rank them by relative variability. Which is best?
Solution
Why: two are given as variances and one as a standard deviation — take the roots first, or the comparison is meaningless.
Convert everything to standard deviations.
Bulb 1: \(s=\sqrt{156}=12.49\)
Bulb 2: \(s=\sqrt{81}=9.00\)
Bulb 3: \(s=6.00\) (already given)
Bulb 2: \(s=\sqrt{81}=9.00\)
Bulb 3: \(s=6.00\) (already given)
CV for each.
\[ CV_1=\frac{12.49}{1470}\times100=\mathbf{0.850\%} \] \[ CV_2=\frac{9.00}{1400}\times100=\mathbf{0.643\%} \] \[ CV_3=\frac{6.00}{1350}\times100=\mathbf{0.444\%} \]
Rank, then answer "which is best" carefully.
Least to most variable: Bulb 3, Bulb 2, Bulb 1.
"Best" depends on the criterion, and a full answer says so: Bulb 1 lasts longest on average (1,470 h) but is the least consistent; Bulb 3 is the most reliable but has the shortest life. If you need predictable replacement scheduling, choose Bulb 3; if you want maximum average life, Bulb 1.
"Best" depends on the criterion, and a full answer says so: Bulb 1 lasts longest on average (1,470 h) but is the least consistent; Bulb 3 is the most reliable but has the shortest life. If you need predictable replacement scheduling, choose Bulb 3; if you want maximum average life, Bulb 1.
Q. Ages of students in the daytime and evening MBA programmes. If homogeneity helps learning, which group will be easier to teach?
Solution
Why: "homogeneity" is exactly low relative dispersion, and the two programmes have different mean ages — so the CV, not the SD, is the right comparison.
Compute \(\bar{x}\) and \(s\) for each sample, then the CV.
\[ CV=\frac{s}{\bar{x}}\times100 \text{ for each programme} \]
The group with the lower CV is the more homogeneous.
The daytime programme is typically the more homogeneous — students come more or less straight from undergraduate study, so ages cluster in a narrow band. The evening programme draws working professionals at very different career stages, giving a much wider age spread and a higher CV.
So the daytime group is easier to teach on the homogeneity criterion. Worth adding: the evening group's diversity of experience may be a teaching asset even though it is statistically less homogeneous.
So the daytime group is easier to teach on the homogeneity criterion. Worth adding: the evening group's diversity of experience may be a teaching asset even though it is statistically less homogeneous.
Q. Percentage of sales goal met by three salespeople over five years. (a) Who is the most consistent? (b) Comment on using consistency alongside percentage of goal met. (c) Suggest a better measure of consistency.
Solution
(a) Compute the CV for each salesperson's five percentages.
The lowest CV identifies the most consistent performer — the one whose results vary least relative to their own average.
(b) The catch — consistency alone can reward the wrong thing.
A salesperson who hits exactly 70% of goal every year is perfectly consistent and consistently failing. Consistency is only a virtue alongside an adequate level: you need both the mean percentage and the CV, and neither on its own tells you who to promote.
(c) A better measure.
Consider consistency in exceeding the goal rather than consistency as such — for example the proportion of years the goal was met, or the variability of the shortfall only (semi-deviation), which penalises misses but not over-performance.
An unusually good salesperson who has one spectacular year looks "inconsistent" on the CV, yet that variability is entirely upside.
An unusually good salesperson who has one spectacular year looks "inconsistent" on the CV, yet that variability is entirely upside.
Q. Company 1: average return 28.0%, s = 5.3%. Company 2: average 37.8%, s = 4.8%. If risk means greater relative dispersion, which is riskier?
Solution
CVs.
\[ CV_1=\frac{5.3}{28.0}\times100=\mathbf{18.93\%} \] \[ CV_2=\frac{4.8}{37.8}\times100=\mathbf{12.70\%} \]
Compare — and note that both criteria favour the same company here.
Company 1 is riskier (18.93% vs 12.70%).
Company 2 dominates outright: a higher average return (37.8% vs 28.0%) and lower relative risk. Normally return and risk trade off; here they do not, so Company 2 is the clear acquisition target on both counts.
Company 2 dominates outright: a higher average return (37.8% vs 28.0%) and lower relative risk. Normally return and risk trade off; here they do not, so Company 2 is the clear acquisition target on both counts.
Q. One machine doses 100 cc with s = 5.2 cc; another doses 180 cc with s = 8.6 cc. Which has the lower accuracy from the standpoint of relative dispersion?
Solution
Why: the raw SDs point one way (8.6 > 5.2, so the 180 cc machine looks worse) and the CVs point the other. This is precisely the situation the CV was invented for.
CVs.
\[ CV_{100}=\frac{5.2}{100}\times100=\mathbf{5.20\%} \] \[ CV_{180}=\frac{8.6}{180}\times100=\mathbf{4.78\%} \]
Answer — and flag the reversal.
The 100 cc machine has the lower accuracy in relative terms (5.20% vs 4.78%), despite having the smaller absolute standard deviation.
Relative to the dose it is supposed to deliver, the 100 cc machine is less precise. For medication, relative error is what matters clinically — being 5 cc out on a 100 cc dose is a bigger problem than being 8 cc out on 180 cc.
Relative to the dose it is supposed to deliver, the 100 cc machine is less precise. For medication, relative error is what matters clinically — being 5 cc out on a 100 cc dose is a bigger problem than being 8 cc out on 180 cc.
This is the exam-worthy question of the topic — the raw SD and the CV give opposite answers, so it tests whether you know which one the question is asking for. Read the words "relative dispersion" carefully.
Q. Four employees' data-entry speed records (correct entries per minute). Infotech needs someone quick and consistent. Who is best, based on relative dispersion?
Solution
Method
1. For each employee compute \(\bar{x}\) (speed) and \(s\) (variability) from their speed records.
2. Compute \(CV = (s/\bar{x})\times100\) for each.
3. The lowest CV is the most consistent — but the question asks for quick and consistent, so both criteria must be reported.
The complete answer names the trade-off: if one employee has both the highest \(\bar{x}\) and the lowest CV, they are the obvious hire. If the fastest is also the most erratic, say so explicitly and recommend on the basis of what Infotech values more — for data entry, where errors are costly and throughput is steady, consistency usually wins.
2. Compute \(CV = (s/\bar{x})\times100\) for each.
3. The lowest CV is the most consistent — but the question asks for quick and consistent, so both criteria must be reported.
The complete answer names the trade-off: if one employee has both the highest \(\bar{x}\) and the lowest CV, they are the obvious hire. If the fastest is also the most erratic, say so explicitly and recommend on the basis of what Infotech values more — for data entry, where errors are costly and throughput is steady, consistency usually wins.
Q. Wyatt Seed grades corn seed by consistency of germination. Laboratory results give germination counts out of packages of 100 for each grade. Does Wyatt's grading make sense?
Solution
Why: the grades claim to rank consistency, so test that claim by computing the CV of each grade and checking whether the ordering matches.
Compute the CV of germination counts for each grade.
\[ CV_{\text{grade}}=\frac{s}{\bar{x}}\times100 \]
Test whether the grading is consistent with the data.
The grading makes sense if and only if the CVs increase steadily as the grade falls — the top grade should have the smallest CV, the next the second smallest, and so on.
If the CVs are ordered correctly, Wyatt's grades convey real information and the premium price for the top grade is justified. If two grades have similar CVs, or the order is broken, the grading is not supported by the data and customers are paying for a distinction that does not exist.
If the CVs are ordered correctly, Wyatt's grades convey real information and the premium price for the top grade is justified. If two grades have similar CVs, or the order is broken, the grading is not supported by the data and customers are paying for a distinction that does not exist.
Q. Configuration I: mean 34.8 min, s = 4.8. Configuration II: mean 25.5, s = 7.5. Configuration III: mean 37.5, s = 3.8. Which has the least relative variation?
Solution
CVs.
\[ CV_{\text{I}}=\frac{4.8}{34.8}\times100=\mathbf{13.79\%} \] \[ CV_{\text{II}}=\frac{7.5}{25.5}\times100=\mathbf{29.41\%} \] \[ CV_{\text{III}}=\frac{3.8}{37.5}\times100=\mathbf{10.13\%} \]
Rank, then note the tension with speed.
Configuration III has the least relative variation (10.13%), then I (13.79%), then II (29.41%).
But Configuration II is much the fastest (25.5 min vs 37.5). So the real choice is between II — fast but erratic and III — slow but predictable. For a production line feeding downstream stations, predictability usually wins; for a bottleneck operation, speed may. Configuration I is dominated: slower than II and less consistent than III.
But Configuration II is much the fastest (25.5 min vs 37.5). So the real choice is between II — fast but erratic and III — slow but predictable. For a production line feeding downstream stations, predictability usually wins; for a bottleneck operation, speed may. Configuration I is dominated: slower than II and less consistent than III.
Q. Program A: average training time 32.11 hours, variance 68.09. Program B: average 19.75, variance 71.14. Which has less relative variability?
Solution
Why: the variances are nearly equal (68 vs 71) but the means differ by a factor of 1.6 — so the CVs are far apart. Take the square roots first.
Variances → standard deviations.
\[ s_A=\sqrt{68.09}=8.2517 \qquad s_B=\sqrt{71.14}=8.4345 \]
CVs.
\[ CV_A=\frac{8.2517}{32.11}\times100=\mathbf{25.70\%} \] \[ CV_B=\frac{8.4345}{19.75}\times100=\mathbf{42.71\%} \] Program A has less relative variability.
The insight: the two programmes have almost the same absolute spread, but because Program B is much faster on average, that spread is a far larger share of its training time. Program B trains people quicker but far less predictably — a real problem for scheduling.
Q. Lee: \(\bar{x}=62.42\), \(s=0.7497\). Forrest: \(\bar{x}=62.18\), \(s=0.9257\). Davis: \(\bar{x}=62.46\), \(s=0.9762\). Based on relative dispersion, who is the best customer?
Solution
CVs.
\[ CV_{\text{Lee}}=\frac{0.7497}{62.42}\times100=\mathbf{1.20\%} \] \[ CV_{\text{Forrest}}=\frac{0.9257}{62.18}\times100=\mathbf{1.49\%} \] \[ CV_{\text{Davis}}=\frac{0.9762}{62.46}\times100=\mathbf{1.56\%} \]
Rank — and then say the honest thing.
Lee is the best customer on relative dispersion, but there really isn't much difference among the three — all three CVs lie between 1.2% and 1.6%, and all three means are within 0.3 of each other.
Saying that is part of the answer. A difference of 0.36 percentage points in CV is unlikely to justify treating these customers differently.
Saying that is part of the answer. A difference of 0.36 percentage points in CV is unlikely to justify treating these customers differently.
Matches Levin's printed answers (1.20%, 1.49%, 1.56%) exactly — including his own conclusion that the differences are not meaningful. Knowing when a difference is too small to act on is a real skill, and the exam rewards it.
X1The Q3 (a) shape — full
Q. A retailer measures satisfaction (1–5) on two channels.
| Statistic | Online Chat | Walk-In Desk |
|---|---|---|
| Mean | 3.8 | 4.6 |
| Median | 4 | 5 |
| Mode | 4 and 5 | 5 |
| Standard deviation | 1.05 | 0.54 |
| Skewness | −0.42 | −1.62 |
| Kurtosis | 2.3 | 4.6 |
(a) What does the difference between mean, median and mode signify? (b) What do skewness and kurtosis imply for consistency? (c) What should the retailer do about Online Chat? [5 marks]
Solution
Why: answer the three sub-parts under the question's own headings. No arithmetic is wanted — every mark is for a reading.
Why we do itWhat to write
(a) Order the three averages for each channel.
Chat: \(3.8 < 4 \le 4/5\). Desk: \(4.6 < 5 = 5\). Both show Mean < Median ≤ Mode ⇒ negatively skewed — a left tail of low ratings pulling each mean down.
Size the gaps and note the bimodality.
Chat's gap is 0.2 with skew −0.42; the Desk's is 0.4 with skew −1.62 — the Desk is far more concentrated at the top of the scale. Chat is bimodal (4 and 5): opinion is split between two clusters rather than settling on one typical experience.
(b) SD first — that is what consistency means.
SD 1.05 vs 0.54: the Desk's ratings vary about half as much.
Then kurtosis on the "normal = 3" scale.
Chat 2.3 < 3 ⇒ platykurtic, flat and spread out. Desk 4.6 > 3 ⇒ leptokurtic, tall and tightly clustered.
Low SD + high kurtosis ⇒ the Walk-In Desk is consistent and predictable. High SD + low kurtosis ⇒ Online Chat is inconsistent: two customers can have completely different experiences.
Low SD + high kurtosis ⇒ the Walk-In Desk is consistent and predictable. High SD + low kurtosis ⇒ Online Chat is inconsistent: two customers can have completely different experiences.
(c) Each recommendation must name the statistic it attacks.
• Cut the SD of 1.05 — standardise chat scripts and escalation rules so every agent handles a query the same way.
• Attack the −0.42 left tail — diagnose the specific low ratings (wait times, unresolved first contacts) rather than trying to lift the average generally.
• Raise kurtosis from 2.3 toward the Desk's shape — continuous training and post-chat feedback, rewarding consistent agents rather than peak performers.
• Investigate the bimodality — two clusters at 4 and 5 suggest two different service experiences are being delivered; identify which agents or shifts produce which.
• Track all four monthly: mean up, SD down, skewness toward 0, kurtosis up.
• Attack the −0.42 left tail — diagnose the specific low ratings (wait times, unresolved first contacts) rather than trying to lift the average generally.
• Raise kurtosis from 2.3 toward the Desk's shape — continuous training and post-chat feedback, rewarding consistent agents rather than peak performers.
• Investigate the bimodality — two clusters at 4 and 5 suggest two different service experiences are being delivered; identify which agents or shifts produce which.
• Track all four monthly: mean up, SD down, skewness toward 0, kurtosis up.
This is the Final and Re-Exam Q3 (a) with the numbers changed. The real versions are worked at Solved Papers → Q3 (a), and the official key is structured under exactly these three headings.
X2Skew from the averages
Q. Name the skew and describe the tail: (a) mean 52, median 50, mode 47; (b) mean 30, median 33, mode 35; (c) mean 18, median 18, mode 18. For (a), if s = 10, find Pearson's coefficient.
Solution
Order each triple.
(a) \(52>50>47\) ⇒ positively skewed — a right tail of high values pulls the mean up.
(b) \(30<33<35\) ⇒ negatively skewed — a left tail of low values.
(c) all equal ⇒ symmetric.
(b) \(30<33<35\) ⇒ negatively skewed — a left tail of low values.
(c) all equal ⇒ symmetric.
Pearson's coefficient for (a).
\[ S_k=\frac{\bar{x}-Z}{s}=\frac{52-47}{10}=\mathbf{+0.50} \] Positive confirms the right tail; \(|S_k|<1\) means the skew is moderate, not extreme.
Cross-check with the other form: \(3(\bar{x}-M)/s = 3(52-50)/10 = 0.60\) — the two versions agree closely, as they should.
Cross-check with the other form: \(3(\bar{x}-M)/s = 3(52-50)/10 = 0.60\) — the two versions agree closely, as they should.
X3Kurtosis with a high SD
Q. A courier reports delivery times with mean 22 min, SD 11 min, skewness +1.8, kurtosis 5.4. A colleague says "kurtosis above 3 means leptokurtic, so the service is consistent". Is that right?
Solution
AnswerNo — the colleague has applied the rule without its condition.
"Leptokurtic ⇒ consistent" is only safe when the SD is also low. Here the SD is 11 minutes against a mean of 22 — a CV of 50%, which is enormous.
High kurtosis with a high SD means a sharp peak plus fat tails: most deliveries cluster tightly near the typical time, but extreme delays occur far more often than a normal curve would predict. Combined with the strong positive skew (+1.8), the picture is mostly fine, occasionally disastrous.
The correct reading: the service is usually predictable but has a genuine tail of very late deliveries — and it is that tail, not the average, that customers will complain about. The fix targets the tail, not the mean.
"Leptokurtic ⇒ consistent" is only safe when the SD is also low. Here the SD is 11 minutes against a mean of 22 — a CV of 50%, which is enormous.
High kurtosis with a high SD means a sharp peak plus fat tails: most deliveries cluster tightly near the typical time, but extreme delays occur far more often than a normal curve would predict. Combined with the strong positive skew (+1.8), the picture is mostly fine, occasionally disastrous.
The correct reading: the service is usually predictable but has a genuine tail of very late deliveries — and it is that tail, not the average, that customers will complain about. The fix targets the tail, not the mean.
X4CV across units
Q. A quality team measures bolt length (mean 40 mm, s = 1.2 mm) and bolt weight (mean 15 g, s = 0.6 g). Which characteristic is under tighter control?
Solution
Different units — the SDs cannot be compared directly at all.
\[ CV_{\text{length}}=\frac{1.2}{40}\times100=\mathbf{3.0\%} \] \[ CV_{\text{weight}}=\frac{0.6}{15}\times100=\mathbf{4.0\%} \]
Compare.
Length is under tighter control (3.0% vs 4.0%) — even though its absolute SD (1.2) is twice the weight's (0.6). Comparing millimetres with grams is meaningless; comparing percentages is not. The weight process is the one to improve.
X5Which average to report
Q. A firm's salaries have mean ₹9.2 lakh, median ₹5.8 lakh, mode ₹4.9 lakh, SD ₹7.4 lakh. Describe the distribution and say which figure the HR head should quote to a candidate.
Solution
Order the averages.
\(9.2 > 5.8 > 4.9\) ⇒ strongly positively skewed — a right tail of a few very high salaries.
Quantify how strong.
\[ S_k=\frac{9.2-4.9}{7.4}=+0.58 \] and the mean is 59% above the median — a very large gap. The CV is \(7.4/9.2 = 80\%\), extremely high.
Answer the practical question.
Quote the median, ₹5.8 lakh. The mean of ₹9.2 lakh is inflated by a small number of senior salaries and describes almost nobody — well over half of employees earn less than the mean.
Quoting the mean to a candidate would be technically true and materially misleading. The honest presentation gives the median and the range for the relevant grade.
Quoting the mean to a candidate would be technically true and materially misleading. The honest presentation gives the median and the range for the relevant grade.
X6Full comparison
Q. Two machines: A has mean output 500 units/day, s = 25, skewness −0.1, kurtosis 3.1. B has mean 620, s = 62, skewness +1.4, kurtosis 5.8. Compare them for a manager who must guarantee a daily minimum to a customer.
Solution
Relative dispersion first.
\[ CV_A=\frac{25}{500}=\mathbf{5.0\%} \qquad CV_B=\frac{62}{620}=\mathbf{10.0\%} \] B is twice as variable in relative terms.
Shape.
A: skew −0.1, kurtosis 3.1 ⇒ essentially symmetric and normal — predictable.
B: skew +1.4, kurtosis 5.8 ⇒ strongly right-skewed with fat tails — occasional very high days, and a longer left side to the bulk of the distribution.
B: skew +1.4, kurtosis 5.8 ⇒ strongly right-skewed with fat tails — occasional very high days, and a longer left side to the bulk of the distribution.
Answer the manager's actual question — a guaranteed minimum.
B produces more on average (620 vs 500), but the guarantee depends on the low end, not the average.
For A, \(\bar{x}-2s = 450\) units — a defensible daily guarantee.
For B, \(\bar{x}-2s = 496\) units, but the positive skew means the bulk of B's distribution sits below its own mean, so low days are more common than the SD alone suggests.
Recommend A if the contract penalises shortfalls: its output is predictable and the guarantee is safe. Recommend B only if the contract rewards total volume and tolerates variable daily output.
For A, \(\bar{x}-2s = 450\) units — a defensible daily guarantee.
For B, \(\bar{x}-2s = 496\) units, but the positive skew means the bulk of B's distribution sits below its own mean, so low days are more common than the SD alone suggests.
Recommend A if the contract penalises shortfalls: its output is predictable and the guarantee is safe. Recommend B only if the contract rewards total volume and tolerates variable daily output.
THE 90-SECOND VERSION — WHAT TO CARRY INTO THE HALL
FOUR MEASURES OF SPREAD
Range R = L − S crude, set by two values
IQR Q₃ − Q₁ resistant to outliers
Std dev s = √[Σ(x−x̄)²/(n−1)] sample: n−1 population: N
grouped: s = √[Σf(x−x̄)²/(n−1)]
CV (s / x̄) × 100 % the ONLY one that compares
different units or different means
HOW MUCH DATA LIES WHERE
shape UNKNOWN → CHEBYSHEV at least 1 − 1/k² within k SD
k=2 → ≥75% k=3 → ≥89%
shape BELL → EMPIRICAL ≈68% / 95% / 99.7% within 1 / 2 / 3 SD
LINEAR CHANGE y = a + bx → ȳ = a + bx̄ , s_y = |b| s_x
ADDING a constant does NOT change the spread
READ THE SHAPE — Q3 (a)
x̄ = M = Z symmetric
x̄ < M ≤ Z NEGATIVE skew — left tail of low values
x̄ > M ≥ Z POSITIVE skew — right tail of high values
K > 3 leptokurtic (tall, narrow) K < 3 platykurtic (flat, broad)
low SD + K > 3 → CONSISTENT
high SD + K < 3 → INCONSISTENT
high SD + K > 3 → sharp peak BUT fat tails — say which you mean
EVERY RECOMMENDATION MUST NAME THE STATISTIC IT ATTACKS.
◆ Where this sits in the rest of the site
Concepts: Unit 02 notes. Q3 (a) is worked in both versions at Solved Papers, and the averages it reads are computed in Unit 01, Topic 5.
The standard deviation you computed here is the input to everything that follows: the normal distribution in Unit 4, and the standard error \(\sigma/\sqrt{n}\) in Units 5–6.