Unit 05 — How to Solve Anything
Four topics. The first two are vocabulary you can write from memory; the last two are Q1 (a) and Q5 (b) — 7 marks between them, and both are the same calculation with one extra step bolted onto the normal distribution you drilled in Unit 4.
◆ The one sentence this whole unit rests on
A sample mean \(\bar{x}\) is itself a random variable. Take a different sample, get a different \(\bar{x}\). The distribution of all those possible \(\bar{x}\) values is the sampling distribution of the mean, and it is narrower than the population — by exactly \(\sqrt{n}\).
Sampling Distribution vs Population
\[ \mu_{\bar{x}} = \mu, \qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}, \qquad z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]
So every Unit 5 calculation is a Unit 4 normal question with \(\sigma\) replaced by \(\sigma/\sqrt{n}\). If you can do Q2 (b), you can already do Q1 (a) and Q5 (b) — you just have to remember to divide by \(\sqrt{n}\) first.
Topic 1 of 4 · Levin §6.1 · p. 270 · 7 textbook + 5 exam-shaped
Why We Sample — Judgment vs Probability Sampling
Pure theory, and reliably worth marks because the answers are lists you can rehearse. Every question here is answered from two vocabularies: the reasons to sample rather than count everyone, and the difference between a sample chosen by judgment and one chosen by chance.
How to know a question belongs to this topic
It asks you to
Argue or explain, not compute. Justify sampling · criticise someone's reasoning · name an advantage or drawback.
Give-away words
advantages of sampling over a census
judgment sampling
is there any defence for…
is a census possible?
statistic or parameter?
explain
What it hands you
A scenario about how data was collected, with an implied error in someone's reasoning. Numbers, if any, are there to be commented on rather than processed.
What a full answer contains
The named principle plus why it applies here. "No — sample size matters far more than the fraction sampled; a random 15% of a large workforce is ample."
Don't confuse it with
| If the question… | then it is… |
|---|---|
| asks which sampling method to use | Topic 2 — random sampling methods |
| gives μ, σ and \(n\) and asks for a probability | Topic 4 — the exam shape |
| asks what "standard error" measures | Topic 3 |
| asks whether a figure is a statistic or a parameter | Here — and the answer is decided by who was measured |
Master Method — Topic 1
The two lists to memorise
Why Sample (6 Reasons)
1. Cost: Far cheaper than census.
2. Time: Census takes too long.
3. Accuracy: Small well-supervised sample beats sloppy census.
4. Feasibility: Infinite/changing populations.
5. Destructive Testing: Testing destroys item (e.g. bulb life).
6. Accessibility: Unreachable members.
2. Time: Census takes too long.
3. Accuracy: Small well-supervised sample beats sloppy census.
4. Feasibility: Infinite/changing populations.
5. Destructive Testing: Testing destroys item (e.g. bulb life).
6. Accessibility: Unreachable members.
Judgment vs Probability Sampling
Judgment: Chosen by expert opinion \(\to\) Cannot measure error/bias.
Probability: Known chance for every element \(\to\) Error can be quantified objectives.
Probability: Known chance for every element \(\to\) Error can be quantified objectives.
Traps
- Confusing sample size with sampling fraction. The precision of \(\bar{x}\) depends on \(n\), not on \(n/N\). This is the single most common managerial error, and Levin sets a whole question on it.
- Saying judgment sampling is "wrong". It is legitimate and often useful — its drawback is specifically that the error cannot be measured.
- Calling a census figure a statistic. Coverage decides the label, not the size.
Textbook bank — every Levin exercise on this topic
§6.1, p. 270 · all 7, answered
7 exercises
Levin §6.1 Introduction to sampling pp. 267–270
Q. What is the major drawback of judgment sampling?
AnswerThe error cannot be measured. Because the elements were chosen by the investigator's opinion rather than by a known chance mechanism, no probability can be attached to the selection — so none of the machinery of statistical inference (standard error, confidence intervals, significance) can be applied. The sample may well be accurate; the problem is that you can never demonstrate that it is, or say by how much it might be off.
Q. Are judgment sampling and probability sampling necessarily mutually exclusive? Explain.
AnswerNo. Judgment routinely operates inside a probability design: an investigator uses judgment to define the population, to choose the strata in stratified sampling, to decide the sample size, and to interpret the result — and then uses a random mechanism to pick the actual elements. Stratified sampling is the clearest case: which strata to form is a judgment call, but within each stratum the selection is random. The two are best seen as complementary rather than opposed.
Q. List the advantages of sampling over complete enumeration, or census.
Why: a list question — marks are awarded per distinct advantage, so answer as a numbered list with a clause each, never as a paragraph.
Answer
1. Lower cost. Measuring a fraction of the population costs a fraction of the money.
2. Greater speed. Results arrive while the decision is still open.
3. Greater accuracy is possible. A small sample can be supervised, its enumerators trained, its measurements checked — a census of millions cannot be. Non-sampling error often exceeds sampling error.
4. Feasibility. Where the population is infinite, or is being produced continuously, a census is impossible in principle.
5. It permits destructive testing. Burn-life of bulbs, breaking strength of cable, taste-testing food — a census would destroy the product.
6. Greater scope. With fewer units to cover, you can ask more questions and measure each one more thoroughly.
The cost of all this is sampling error — \(\bar{x}\) will differ from \(\mu\). Units 5 and 6 exist to measure exactly how much.
2. Greater speed. Results arrive while the decision is still open.
3. Greater accuracy is possible. A small sample can be supervised, its enumerators trained, its measurements checked — a census of millions cannot be. Non-sampling error often exceeds sampling error.
4. Feasibility. Where the population is infinite, or is being produced continuously, a census is impossible in principle.
5. It permits destructive testing. Burn-life of bulbs, breaking strength of cable, taste-testing food — a census would destroy the product.
6. Greater scope. With fewer units to cover, you can ask more questions and measure each one more thoroughly.
The cost of all this is sampling error — \(\bar{x}\) will differ from \(\mu\). Units 5 and 6 exist to measure exactly how much.
Q. What are some disadvantages of probability sampling versus judgment sampling?
Answer
1. Cost and time. A proper random design needs a sampling frame, a selection mechanism and often wider geographic coverage.
2. It requires a complete list of the population (a sampling frame). For many populations no such list exists, or it is out of date.
3. It cannot exploit expert knowledge. A specialist who knows which units are genuinely representative is not allowed to use that knowledge.
4. Randomness can produce an obviously unrepresentative sample by chance — for a small \(n\), a random draw may miss an important subgroup entirely, where judgment would have included it deliberately.
5. It needs more technical skill to design and to analyse.
2. It requires a complete list of the population (a sampling frame). For many populations no such list exists, or it is out of date.
3. It cannot exploit expert knowledge. A specialist who knows which units are genuinely representative is not allowed to use that knowledge.
4. Randomness can produce an obviously unrepresentative sample by chance — for a small \(n\), a random draw may miss an important subgroup entirely, where judgment would have included it deliberately.
5. It needs more technical skill to design and to analyse.
Q. At the annual meeting, to which all shareholders are invited, the president asks the shareholders present whether they approve of a merger. 85% approve. Is this a sample statistic or a population parameter?
Why: the trap is the phrase "to which all shareholders are invited". Being invited is not being measured. Only those who attended and answered were measured.
AnswerA sample statistic. Only the shareholders who actually attended were asked, and they are a subset of all shareholders — so 85% is \(\hat{p}\), not \(p\).
And worse than an ordinary sample: attendance is self-selected. Shareholders who care enough to travel to an annual meeting are unlikely to be typical of the whole register, and those with strong feelings about the merger are the most likely to attend. So this is a biased sample, and the 85% cannot safely be generalised to all shareholders — which matters, because shareholder approval is exactly what the merger needs.
And worse than an ordinary sample: attendance is self-selected. Shareholders who care enough to travel to an annual meeting are unlikely to be typical of the whole register, and those with strong feelings about the merger are the most likely to attend. So this is a biased sample, and the 85% cannot safely be generalised to all shareholders — which matters, because shareholder approval is exactly what the merger needs.
Q. Jean Mason sampled a little over 15% of employees. Management dismissed her findings: "Everyone knows you have to check 50 percent to have any idea of the outcome." Is there any defence for Jean's position?
Why: this is the conceptual question of the unit. The precision of \(\bar{x}\) depends on \(n\) — the absolute number sampled — not on \(n/N\), the fraction. Management has confused the two.
Why we do itWhat to write
State the principle first, in its own sentence.
Yes, Jean is right. The accuracy of a sample estimate depends on the absolute sample size \(n\), not on the proportion of the population it represents.
Show it with the formula — the standard error contains \(n\) and does not contain \(N\).
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \] There is no \(N\) anywhere in it. A random 400 out of 2,000 and a random 400 out of 2,000,000 give essentially the same precision.
Give the concrete comparison that makes it obvious.
National opinion polls estimate the views of hundreds of millions from samples of about 1,000 — a sampling fraction far below 0.001%. Nobody proposes polling 50% of a country.
Concede the one thing that would be a fair criticism, which is what makes the answer complete.
The fraction matters only in the opposite direction: when \(n/N > 0.05\), the finite population correction makes the estimate more precise, not less (Unit 6, Topic 1). So a 15% fraction actually helps Jean. The legitimate questions are whether her sample was random and whether \(n\) was large enough in absolute terms — not whether 15% is "enough of" the workforce.
Q. A consumer organisation is conducting a census of everyone injured by a space heater, asking each victim about the heater's behaviour before it malfunctioned — information available only from the victim. Several victims have since died. Is any census of the victims now possible?
Why: a census requires that every member of the population be measured. Ask whether that is still physically possible.
AnswerNo — a complete census is no longer possible. The information exists only in the victims' own accounts, and some victims can no longer be interviewed, so those observations are permanently unobtainable. What remains achievable is a census of the surviving victims, which is a sample of all victims — and a biased one, because the deceased were disproportionately elderly, and elderly victims may have used the heater differently or been less able to react to a malfunction. Any conclusion drawn will therefore understate the severity of the hazard for the most vulnerable group, and the report must say so.
Five more, shaped like the exam
theory questions in this unit are answered as lists — practise producing them fast
X1Definitions
Q. Distinguish between a population and a sample, and between a parameter and a statistic. Give the symbols for the mean, standard deviation and proportion in each case.
Solution
Answer
A population is the entire set of elements about which a conclusion is wanted; a sample is a subset of it selected to represent it.
A descriptive measure computed on a population is a parameter; one computed on a sample is a statistic.
Parameters (Greek): mean \(\mu\), standard deviation \(\sigma\), proportion \(p\), size \(N\).
Statistics (Roman): mean \(\bar{x}\), standard deviation \(s\), proportion \(\hat{p}\), size \(n\).
The whole of inferential statistics is using the Roman letters to estimate the Greek ones.
A descriptive measure computed on a population is a parameter; one computed on a sample is a statistic.
Parameters (Greek): mean \(\mu\), standard deviation \(\sigma\), proportion \(p\), size \(N\).
Statistics (Roman): mean \(\bar{x}\), standard deviation \(s\), proportion \(\hat{p}\), size \(n\).
The whole of inferential statistics is using the Roman letters to estimate the Greek ones.
X2Sampling error
Q. What is sampling error? Distinguish it from non-sampling error, and say which one a larger sample fixes.
Solution
Answer
Sampling error is the difference between a sample statistic and the population parameter that arises purely because a sample, not the whole population, was measured. It is unavoidable, random, and measurable — the standard error \(\sigma/\sqrt{n}\) is exactly its size.
Non-sampling error is everything else: a biased sampling frame, badly worded questions, non-response, measurement mistakes, data-entry errors. It is systematic and not measurable from the data.
A larger sample fixes only the first. Increasing \(n\) shrinks the standard error, but it does nothing for bias — a bigger biased sample is simply a more precisely wrong answer. This is why Levin's third advantage of sampling is that a small, well-supervised sample can beat a large careless census.
Non-sampling error is everything else: a biased sampling frame, badly worded questions, non-response, measurement mistakes, data-entry errors. It is systematic and not measurable from the data.
A larger sample fixes only the first. Increasing \(n\) shrinks the standard error, but it does nothing for bias — a bigger biased sample is simply a more precisely wrong answer. This is why Levin's third advantage of sampling is that a small, well-supervised sample can beat a large careless census.
X3Statistic or parameter
Q. Label each and give the symbol: (a) the mean salary of all 480 employees, from payroll records; (b) the mean satisfaction score of 60 customers surveyed out of 9,000; (c) the proportion defective in a batch of 5,000, every one inspected; (d) the proportion of 200 sampled voters favouring a candidate.
Solution
Why: coverage decides the label — was everyone measured, or a subset?
Everyone measured ⇒ parameter ⇒ Greek.
(a) \(\mu\) — parameter · (c) \(p\) — parameter
Subset measured ⇒ statistic ⇒ Roman.
(b) \(\bar{x}\) — statistic · (d) \(\hat{p}\) — statistic
Note (a): 480 is small, but it is all of them — size is irrelevant.
Note (a): 480 is small, but it is all of them — size is irrelevant.
X4Critique
Q. A magazine prints a questionnaire and reports that "of the 4,200 readers who replied, 78% support the proposal". Criticise this as a basis for claiming public support.
Solution
Answer
1. Self-selection. Readers chose whether to reply. People with strong opinions reply far more often, so the respondents are not representative even of the readership.
2. The frame is wrong. Even a perfect sample of this magazine's readers would not represent the public — readers of any one publication are a distinct group.
3. A large \(n\) does not repair bias. 4,200 makes the sampling error small, which only means the biased figure is estimated precisely. Precision is not accuracy.
4. The response rate is unstated. 4,200 out of 10,000 is one thing; out of 900,000 is another entirely.
Conclusion: the 78% describes the people who chose to reply and nothing more. To claim public support you would need a probability sample of the public.
2. The frame is wrong. Even a perfect sample of this magazine's readers would not represent the public — readers of any one publication are a distinct group.
3. A large \(n\) does not repair bias. 4,200 makes the sampling error small, which only means the biased figure is estimated precisely. Precision is not accuracy.
4. The response rate is unstated. 4,200 out of 10,000 is one thing; out of 900,000 is another entirely.
Conclusion: the 78% describes the people who chose to reply and nothing more. To claim public support you would need a probability sample of the public.
X5Judgment vs probability
Q. A quality manager selects "the machines I know give trouble" for inspection. A colleague selects machines using random numbers. Compare the two approaches, and say when each is appropriate.
Solution
Answer
The first is judgment sampling, the second probability sampling.
Judgment is faster, cheaper, and uses real expertise — and for finding faults quickly it is the better tool, because it deliberately targets where problems live.
Probability sampling is the only one that supports a general claim — "the defect rate across all machines is 2% ± 0.4%" — because only it lets you compute a standard error.
So the choice follows the purpose: use judgment sampling to diagnose and fix; use probability sampling to estimate and report. The manager's approach would be indefensible if he then claimed his findings described the whole plant.
Judgment is faster, cheaper, and uses real expertise — and for finding faults quickly it is the better tool, because it deliberately targets where problems live.
Probability sampling is the only one that supports a general claim — "the defect rate across all machines is 2% ± 0.4%" — because only it lets you compute a standard error.
So the choice follows the purpose: use judgment sampling to diagnose and fix; use probability sampling to estimate and report. The manager's approach would be indefensible if he then claimed his findings described the whole plant.
Topic 2 of 4 · Levin §6.2 · pp. 280–282 · 12 textbook questions
Random Sampling Methods — Which One, and Why
Four probability designs, and one rule that chooses between the two that are always confused. Nearly every question here is "which method would you use?", and the answer turns on a single comparison: is the variation mostly within the natural groups, or mostly between them?
How to know a question belongs to this topic
It asks you to
Choose a sampling design, or judge whether someone's design will actually produce a random sample.
Give-away words
which sampling method is appropriate
stratified, systematic, cluster or simple random
will this produce a random sample?
use the random digits table
every tenth report
subgroups
The one rule that decides stratified vs cluster
Variation is BETWEEN groups, groups are internally similar → STRATIFIED (sample from every group).
Variation is WITHIN groups, groups resemble each other → CLUSTER (sample a few whole groups).
In stratified sampling each stratum is a miniature of nothing; in cluster sampling each cluster is a miniature of the whole population.
Variation is WITHIN groups, groups resemble each other → CLUSTER (sample a few whole groups).
In stratified sampling each stratum is a miniature of nothing; in cluster sampling each cluster is a miniature of the whole population.
The systematic-sampling trap
Systematic sampling fails when the list has a periodic pattern whose cycle matches your interval. Ten machines in sequence and "every fifth lot" means you sample only machines 5 and 10 — forever.
What a full answer contains
The method named + the reason in terms of the variation structure + any bias you can foresee.
Master Method — Topic 2
The four designs
| Method | How it works | Use it when | Weakness |
|---|---|---|---|
| Simple random | Every element has an equal chance; pick with a random-digit table or generator | The population is homogeneous and a full list exists | Needs a complete frame; may miss small subgroups |
| Systematic | Pick a random start, then every \(k\)th element, \(k=N/n\) | The list is long and in no meaningful order | Fails if the list is periodic with period \(k\) |
| Stratified | Divide into strata, then sample randomly within every stratum | Groups differ from each other but are internally similar | Needs to know the strata in advance |
| Cluster | Divide into clusters, randomly pick whole clusters, measure everything in them | Groups each resemble the whole population; wide internal variation | Less precise than stratified for the same \(n\) |
Traps
- Swapping stratified and cluster. Ask where the variation lives — that single question settles it.
- Assuming "random selection of houses" gives a random sample of people. If you call between noon and 5 p.m. on weekdays you systematically miss everyone at work.
- Using a phone book or any incomplete list as the frame. Everyone not on the list has zero probability of selection, so the sample is not random over the intended population.
- Systematic sampling on a list sorted by the variable of interest — or on any list with a repeating cycle.
Textbook bank
§6.2, pp. 280–282 · all 12, answered
12 exercises
Levin §6.2 Random sampling pp. 271–282
Q. Probability distributions for three natural subgroups of a larger population are shown. For which situation would you recommend stratified sampling?
Why: a picture question with one rule. Look at where the three curves sit relative to each other, and how wide each one is.
AnswerRecommend stratified sampling for the situation where the three subgroup distributions are separated from one another but each is narrow — i.e. large variation between groups, small variation within each group.
Why: when groups are internally homogeneous, a small sample from each stratum pins that stratum down accurately, and combining them gives a very precise overall estimate. If instead the three curves overlap heavily and are each wide, stratifying gains nothing — the groups are not telling you anything the whole population does not, and cluster or simple random sampling is the better choice.
Why: when groups are internally homogeneous, a small sample from each stratum pins that stratum down accurately, and combining them gives a very precise overall estimate. If instead the three curves overlap heavily and are each wide, stratifying gains nothing — the groups are not telling you anything the whole population does not, and cluster or simple random sampling is the better choice.
Q. Sample 15 pages from this textbook using the random digits table, and count the words in italics on each page.
Method
1. Establish the frame: the book has pages numbered 1 to \(N\) — say 1 to 800. Since \(N\) is a 3-digit number, read the table in 3-digit groups.
2. Choose a random starting point in the table (close your eyes and point) and a fixed direction of travel; state both, because a chosen start is what makes it random rather than convenient.
3. Read 3-digit groups. Discard any number above 800 and any repeat (sampling without replacement).
4. Continue until 15 usable page numbers are obtained; those are your sample.
5. Count italicised words on each and report the 15 counts, with \(\bar{x}\) and \(s\).
The examinable content is steps 1–3: the frame, the digit width matching \(N\), and the rules for discarding out-of-range and duplicate numbers.
2. Choose a random starting point in the table (close your eyes and point) and a fixed direction of travel; state both, because a chosen start is what makes it random rather than convenient.
3. Read 3-digit groups. Discard any number above 800 and any repeat (sampling without replacement).
4. Continue until 15 usable page numbers are obtained; those are your sample.
5. Count italicised words on each and report the 15 counts, with \(\bar{x}\) and \(s\).
The examinable content is steps 1–3: the frame, the digit width matching \(N\), and the rules for discarding out-of-range and duplicate numbers.
Q. Using a calendar, systematically sample every eighteenth day of the year, beginning with January 6.
Why: mechanical — but notice what the interval does to the days of the week, because that is the point of the exercise.
Answer
Starting at day 6 (6 Jan) with \(k = 18\), the sampled days of the year are
6, 24, 42, 60, 78, 96, 114, 132, 150, 168, 186, 204, 222, 240, 258, 276, 294, 312, 330, 348, 366
— i.e. 6 Jan, 24 Jan, 11 Feb, 1 Mar, 19 Mar, 6 Apr, … giving about 21 days across the year.
The point: \(18 = 2\times 7 + 4\), so the day of the week advances by 4 each time and cycles through all seven — this interval does not lock onto a single weekday. An interval of 7, 14 or 21 would have sampled the same day of the week all year, which for anything with a weekly rhythm (sales, absenteeism, traffic) would be badly biased. That is the periodicity trap in systematic sampling.
6, 24, 42, 60, 78, 96, 114, 132, 150, 168, 186, 204, 222, 240, 258, 276, 294, 312, 330, 348, 366
— i.e. 6 Jan, 24 Jan, 11 Feb, 1 Mar, 19 Mar, 6 Apr, … giving about 21 days across the year.
The point: \(18 = 2\times 7 + 4\), so the day of the week advances by 4 each time and cycles through all seven — this interval does not lock onto a single weekday. An interval of 7, 14 or 21 would have sampled the same day of the week all year, which for anything with a weekly rhythm (sales, absenteeism, traffic) would be badly biased. That is the periodicity trap in systematic sampling.
Q. A population is made up of groups with wide variation within each group but little variation from group to group. The appropriate sampling method is (a) stratified, (b) systematic, (c) cluster, (d) judgment.
Why: this is the rule stated as a question. Wide within, little between ⇒ each group is already a miniature of the population.
Answer(c) Cluster sampling.
Because the groups barely differ from one another, any single group is a fair representation of the whole — so you can take a few complete groups and save the cost of reaching every group. And because variation within each group is wide, each cluster already contains the full range of values.
Stratified would be the wrong answer here: stratifying pays off only when groups differ from each other, and these do not.
Because the groups barely differ from one another, any single group is a fair representation of the whole — so you can take a few complete groups and save the cost of reaching every group. And because variation within each group is wide, each cluster already contains the full range of values.
Stratified would be the wrong answer here: stratifying pays off only when groups differ from each other, and these do not.
Q. In the random digits table, what is the probability that a 4 appears as the leftmost digit of each set of 10 digits? A 7? A 2? How many times would you expect to see each? How many are actually found? Explain any difference.
Why we do itWhat to write
In a table of random digits every digit 0–9 is equally likely in every position.
\[ P(4)=P(7)=P(2)=\frac{1}{10}=\mathbf{0.10} \] — the same for any digit, in any position.
Expected count is just \(np\) — a binomial mean.
With \(N\) sets of digits, expect \(0.10N\) of each. For Levin's table of 100 sets: 10 of each.
Explain the discrepancy — this is what the question is really asking.
The observed counts will not be exactly 10 each, and that is expected: the count is binomial with \(n=100,\ p=0.1\), so \(\mu=10\) and \(\sigma=\sqrt{100(0.1)(0.9)}=3\). Counts anywhere in roughly 4 to 16 (\(\mu\pm2\sigma\)) are entirely consistent with randomness. A table in which every digit appeared exactly 10 times would be evidence that it was not random.
Q. A cable company has 20,000 subscribers: 35% college students, 45% white-collar, 15% blue-collar, 5% other — but it believes there is much variation within these groups. Which method: random, systematic, stratified or cluster?
Why: the phrase "much variation within these groups" is the whole answer. It is the exact condition that kills stratification.
Why we do itWhat to write
Test the stratified condition first, since the neat percentage breakdown is designed to tempt you into it.
Stratified needs groups that are internally similar. Here the company explicitly says variation within each group is large — so stratifying by occupation would not reduce the standard error.
Choose on that basis and justify.
Simple random sampling (or systematic, which is easier to administer over a 20,000-name subscriber list). Since the occupational groups do not predict channel preference any better than the population as a whole does, there is nothing to gain from stratifying — and a straightforward random sample of all subscribers is cheaper and simpler.
Q. A door-to-door poll on day-care centres uses a random scheme for selecting houses, and is conducted on weekdays from noon to 5 p.m. Will this produce a random sample?
AnswerNo. The houses are selected randomly, but the people reached are not. Anyone employed outside the home on weekday afternoons has zero chance of being interviewed, and a random sample requires every member of the population to have a known, non-zero probability of selection.
Why it matters especially here: the topic is day-care centres, and the people systematically excluded — working parents — are precisely the group with the strongest interest in the subject. The bias is not merely present, it points directly at the variable being measured.
Fix: include evenings and weekends, make callbacks at houses where nobody answers, and record the response rate.
Why it matters especially here: the topic is day-care centres, and the people systematically excluded — working parents — are precisely the group with the strongest interest in the subject. The bias is not merely present, it points directly at the variable being measured.
Fix: include evenings and weekends, make callbacks at houses where nobody answers, and record the response rate.
Q. Mohan Adhikari plans a telephone survey, selecting random numbers from the telephone book with addresses in the company's service area. Will his sample be random?
AnswerNo — the sampling frame is incomplete. The telephone book omits households with no telephone, with unlisted numbers, with mobile-only service, and those who moved in since publication. Every one of those has zero probability of selection, so the sample is random over the phone book, not over the company's customers.
Why it biases this particular study: the campaign is about energy consciousness, and the excluded groups skew toward lower-income and more mobile households — plausibly different in both energy use and in exposure to the advertising. Adhikari would be measuring the campaign's effect on a settled, listed, higher-income subset and reporting it as the effect on all customers.
Fix: sample from the company's own customer billing list, which is the actual population of interest and is complete by construction.
Why it biases this particular study: the campaign is about energy consciousness, and the excluded groups skew toward lower-income and more mobile households — plausibly different in both energy use and in exposure to the advertising. Adhikari would be measuring the campaign's effect on a settled, listed, higher-income subset and reporting it as the effect on all customers.
Fix: sample from the company's own customer billing list, which is the actual population of interest and is complete by construction.
Q. Ten machines stamp coins in lots of 50, arranged sequentially on one conveyor past an inspection station. Should the inspector sample every fifth lot or every seventh? Why?
Why: the classic periodicity question. Compare the sampling interval with the cycle length of the list — here the cycle is 10, because the machines repeat every 10 lots.
Why we do itWhat to write
Identify the cycle in the list. Lots arrive machine 1, 2, …, 10, 1, 2, … so the period is 10.
The list has a period of 10 — one cycle per pass through the machines.
Test every fifth. Since 5 divides 10, the sample locks onto a fixed pair of positions.
Every 5th lot ⇒ positions 5, 10, 15, 20, … ⇒ only machines 5 and 10, forever. Eight of the ten machines are never inspected.
Test every seventh. Since 7 and 10 share no factor, the position advances and cycles through all ten.
Every 7th lot ⇒ positions 7, 14, 21, 28, 35, … ⇒ machines 7, 4, 1, 8, 5, 2, 9, 6, 3, 10, then repeats — all ten machines covered, equally.
Answer and state the general rule.
Every seventh. General rule: the sampling interval \(k\) must be coprime with the period of the list — if \(k\) shares a factor with the cycle, systematic sampling degenerates into sampling a fixed subset.
Q. 250 of 2,500 work-related accident reports are to be sampled. Reports are filed by date. Marsha proposes taking every tenth report. Is systematic sampling appropriate here?
AnswerBroadly yes, with one caution.
\(k = N/n = 2500/250 = 10\), and the file is ordered by date — a variable with no reason to cycle with period 10. Days of the week cycle with period 7, not 10, so every tenth report will not lock onto a single weekday. The design is therefore acceptable and much easier to administer than drawing 250 random numbers.
The caution: the study wants to examine time of day among its variables. If accident reports happen to be filed in a repeating within-day order (say, each day's reports filed shift by shift), a period could appear. Marsha should check the file for any repeating structure before committing, and must pick her starting point at random between 1 and 10 — not simply start at the first report.
\(k = N/n = 2500/250 = 10\), and the file is ordered by date — a variable with no reason to cycle with period 10. Days of the week cycle with period 7, not 10, so every tenth report will not lock onto a single weekday. The design is therefore acceptable and much easier to administer than drawing 250 random numbers.
The caution: the study wants to examine time of day among its variables. If accident reports happen to be filed in a repeating within-day order (say, each day's reports filed shift by shift), a period could appear. Marsha should check the file for any repeating structure before committing, and must pick her starting point at random between 1 and 10 — not simply start at the first report.
Q. Using the random digits table, starting at the top of the third column and taking the last 4 digits of each number, read across rows to obtain a sample.
AnswerThe resulting sample is
0892 1652 2963 2913 3181 9348 4959 7695 7712 8136 9659 2526 6988 1781 7652 8559 2204 4339 6299 3397 The examinable points, which the numbers themselves do not show: the starting point and direction must be fixed before reading (otherwise you are choosing your sample after seeing it), the digit width must match the size of the population, and any number out of range or already drawn is discarded.
0892 1652 2963 2913 3181 9348 4959 7695 7712 8136 9659 2526 6988 1781 7652 8559 2204 4339 6299 3397 The examinable points, which the numbers themselves do not show: the starting point and direction must be fixed before reading (otherwise you are choosing your sample after seeing it), the digit width must match the size of the population, and any number out of range or already drawn is discarded.
Q. Which sampling method is more appropriate when the population appears to contain two very dissimilar groups?
AnswerStratified sampling. Two very dissimilar groups means the variation between groups is large while the variation within each group is comparatively small — exactly the condition stratification exploits.
Why it helps: sampling separately within each stratum guarantees both groups are represented in proportion, removes between-group variation from the sampling error entirely, and so produces a smaller standard error than simple random sampling of the same total size. A simple random sample might, by chance, draw mostly from one group and misrepresent the population badly.
Why it helps: sampling separately within each stratum guarantees both groups are represented in proportion, removes between-group variation from the sampling error entirely, and so produces a smaller standard error than simple random sampling of the same total size. A simple random sample might, by chance, draw mostly from one group and misrepresent the population badly.
Topic 3 of 4 · Levin §6.3 · pp. 288–289 · 8 textbook + 4 exam-shaped
Sampling Distributions and the Standard Error — the Concepts
The idea that makes Topic 4 computable. Almost every question here is a trap about what \(\bar{x}\) is: a single sample mean differing from \(\mu\) is not evidence of anything wrong, because \(\bar{x}\) is supposed to vary. Learning to say that sentence correctly is most of this topic.
How to know a question belongs to this topic
It asks you to
Interpret a sampling distribution, or judge a claim about one — not to compute a probability.
Give-away words
is the conclusion correct?
is this a sampling distribution of the mean?
what type of distribution did the sample come from?
what does "error" refer to?
could the company's claim be correct?
The three distributions to keep apart
1. The population — all individual values, mean \(\mu\), s.d. \(\sigma\).
2. A sample — \(n\) individual values drawn from it, mean \(\bar{x}\), s.d. \(s\).
3. The sampling distribution of \(\bar{x}\) — all the possible sample means, mean \(\mu\), s.d. \(\sigma/\sqrt{n}\).
Most questions here are testing whether you can tell #2 from #3.
2. A sample — \(n\) individual values drawn from it, mean \(\bar{x}\), s.d. \(s\).
3. The sampling distribution of \(\bar{x}\) — all the possible sample means, mean \(\mu\), s.d. \(\sigma/\sqrt{n}\).
Most questions here are testing whether you can tell #2 from #3.
The recurring trap
Someone observes one \(\bar{x} \ne \mu\) and concludes something is wrong. It usually is not: the standard error tells you how far \(\bar{x}\) should routinely wander. Always compute \(\sigma/\sqrt{n}\) and compare the gap against it.
Master Method — Topic 3
Formula bank
Standard error
of the mean
of the mean
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
Mean of the
sampling distribution
sampling distribution
\[ \mu_{\bar{x}} = \mu \qquad\text{(unbiased — same centre as the population)} \]
What "error" means here. It does not mean a mistake. It is the natural, unavoidable variability of \(\bar{x}\) from sample to sample — a measure of dispersion, not of blunder. Levin sets a question on exactly this wording.
The routine for "is this conclusion correct?" questions
- Identify which distribution the number came from — population, one sample, or the sampling distribution.
- Compute the standard error \(\sigma/\sqrt{n}\) if \(n\) is given.
- Measure the gap between the observed \(\bar{x}\) and the claimed \(\mu\), in standard errors: \(z = (\bar{x}-\mu)/(\sigma/\sqrt{n})\).
- Judge it. Within about \(\pm2\) standard errors is ordinary variation; beyond \(\pm3\) is genuine evidence against the claim.
- Say so in words, and name what would change your mind.
Traps
- Treating one \(\bar{x}\ne\mu\) as proof of a problem. Without \(n\) you cannot even judge it.
- Confusing \(s\) with \(\sigma_{\bar{x}}\). \(s\) describes the spread of individual values in one sample; \(\sigma_{\bar{x}}\) describes the spread of means across samples.
- Thinking the standard error can be driven to zero. \(\sigma/\sqrt{n}\) only reaches 0 when \(n\) reaches the whole population — and even then only for a finite population.
- Calling a distribution of daily means a sample from the population. It is a sample from the sampling distribution.
Textbook bank
§6.3, pp. 288–289 · all 8, answered
8 exercises
Levin §6.3 Introduction to sampling distributions pp. 283–289
Q. Sampling from a population with mean 2.15, what sample size will guarantee that (a) the sample mean is 2.15? (b) the standard error of the mean is zero?
Why: both parts are trick questions with the same underlying point — the only way to remove sampling variability entirely is to stop sampling.
Why we do itWhat to write
(a) A sample mean is a random variable; no sample size makes a random variable take a chosen value with certainty.
No sample size guarantees it. Even \(n = N\) guarantees \(\bar{x}=\mu\) only because you have then measured everyone — that is a census, not a sample. For any \(n
(b) Set the standard error to zero and see what \(n\) must be.
\[ \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} = 0 \;\Rightarrow\; n \to \infty \] For a finite population the finite population correction \(\sqrt{\tfrac{N-n}{N-1}}\) does reach 0 — at \(n = N\), again a census. So: no finite sample size. Only a complete enumeration.
Q. The term error, in "standard error of the mean", refers to what type of error?
AnswerIt refers to sampling error — the natural, unavoidable variability of \(\bar{x}\) from one sample to the next, caused simply by the fact that different samples contain different elements.
It does not mean a mistake, a blunder, or bad measurement. The standard error is a measure of dispersion — it is the standard deviation of the sampling distribution of the mean, and it would be present even if every measurement were made perfectly.
The word is historical and misleading; "standard deviation of the sample means" would describe it better.
It does not mean a mistake, a blunder, or bad measurement. The standard error is a measure of dispersion — it is the standard deviation of the sampling distribution of the mean, and it would be present even if every measurement were made perfectly.
The word is historical and misleading; "standard deviation of the sample means" would describe it better.
Q. A cereal company claims 2.0 cups of raisins per box on average, with a standard deviation of 0.2 cup. Your box contained 1.9 cups. Could the company's claim be correct?
Why: note \(n = 1\) — this is a single individual value, not a sample mean. So compare it against \(\sigma\) itself, not against a standard error.
Why we do itWhat to write
Identify which distribution the observation comes from. One box is one individual observation.
This is a single value from the population, so the relevant spread is \(\sigma = 0.2\) cup — not \(\sigma/\sqrt{n}\).
Measure the gap in standard deviations.
\[ z = \frac{1.9-2.0}{0.2} = -0.50 \]
Judge it.
Yes, the claim could easily be correct. The box is only half a standard deviation below the claimed mean — around 31% of all boxes would contain less than 1.9 cups even if the claim is exactly true. One box is no evidence at all.
To test the claim properly you would need a sample of boxes and would compare \(\bar{x}\) against \(\sigma/\sqrt{n}\), which is much narrower.
To test the claim properly you would need a sample of boxes and would compare \(\bar{x}\) against \(\sigma/\sqrt{n}\), which is much narrower.
Q. An interviewer records viewing times from passersby for 20 days, computing the mean of each day's interviews, then builds a frequency distribution of those 20 daily means. Is this a sampling distribution of the mean?
AnswerIt is an empirical approximation to one, but not the sampling distribution itself — and it has a serious flaw.
What is right: each point plotted is a sample mean, not an individual observation, so the distribution being built is of \(\bar{x}\) values. That is the right kind of object.
Two reservations:
1. A true sampling distribution is the distribution of \(\bar{x}\) over all possible samples of a fixed size. Twenty days gives 20 observations from it — a sample of the sampling distribution, not the distribution.
2. The daily samples are not of equal size, and \(\sigma_{\bar{x}} = \sigma/\sqrt{n}\) depends on \(n\). Days with more interviews produce less variable means, so the 20 values do not come from a single common distribution.
And the sampling method is not random: interviewing passersby on a street is convenience sampling, so each daily mean may be biased as well.
What is right: each point plotted is a sample mean, not an individual observation, so the distribution being built is of \(\bar{x}\) values. That is the right kind of object.
Two reservations:
1. A true sampling distribution is the distribution of \(\bar{x}\) over all possible samples of a fixed size. Twenty days gives 20 observations from it — a sample of the sampling distribution, not the distribution.
2. The daily samples are not of equal size, and \(\sigma_{\bar{x}} = \sigma/\sqrt{n}\) depends on \(n\). Days with more interviews produce less variable means, so the 20 values do not come from a single common distribution.
And the sampling method is not random: interviewing passersby on a street is convenience sampling, so each daily mean may be biased as well.
Q. Rehana Ali graphs weekly sales frequency distributions for the year before and the year after a competitor appeared, on identical axes (1–6 thousand cartons per week). What has been the effect on average weekly sales?
Why: a comparison-of-shapes question. Read three things in order — centre, spread, shape — and say what each means commercially.
Answer
Centre. Compare where the bulk of each distribution sits. If the "after" distribution has shifted left, average weekly sales have fallen, and the size of the shift estimates by how much.
Spread. If the "after" distribution is also wider, sales have become less predictable as well as lower — a separate and equally important problem for production planning.
Shape. A distribution that becomes bimodal after the competitor arrives would suggest two regimes — good weeks and bad weeks — rather than a uniform decline.
The caution that earns the mark: these are observational data, not an experiment. Sales could have fallen for reasons unrelated to the competitor — the economy, seasonality, a price change, a lost distributor. The graphs establish that sales are associated with the competitor's arrival in time; they cannot establish that the competitor caused the fall.
Spread. If the "after" distribution is also wider, sales have become less predictable as well as lower — a separate and equally important problem for production planning.
Shape. A distribution that becomes bimodal after the competitor arrives would suggest two regimes — good weeks and bad weeks — rather than a uniform decline.
The caution that earns the mark: these are observational data, not an experiment. Sales could have fallen for reasons unrelated to the competitor — the economy, seasonality, a price change, a lost distributor. The graphs establish that sales are associated with the competitor's arrival in time; they cannot establish that the competitor caused the fall.
Q. A CEO randomly selects 50 regional managers; each randomly selects 5 supervisors; each supervisor randomly surveys 30 customers. What type of distribution did the sample come from?
AnswerThis is multistage cluster sampling — three stages: managers, then supervisors, then customers.
The results reaching the CEO are summaries of groups of 30 customers, so what he receives is a sample from the sampling distribution of the mean for samples of size 30 — not a sample from the customer population itself.
The consequence: the spread the CEO sees among those figures is \(\sigma/\sqrt{30}\), which is much narrower than the variation among individual customers. If he interprets it as customer-level variation he will badly understate how differently customers actually experience the service. Multistage designs also inflate the true standard error above \(\sigma/\sqrt{n}\), because customers within a supervisor's area resemble each other.
The results reaching the CEO are summaries of groups of 30 customers, so what he receives is a sample from the sampling distribution of the mean for samples of size 30 — not a sample from the customer population itself.
The consequence: the spread the CEO sees among those figures is \(\sigma/\sqrt{30}\), which is much narrower than the variation among individual customers. If he interprets it as customer-level variation he will badly understate how differently customers actually experience the service. Multistage designs also inflate the true standard error above \(\sigma/\sqrt{n}\), because customers within a supervisor's area resemble each other.
Q. A bottle-filling machine has mean 125 g and standard deviation 20 g. A quality manager takes a random sample of filled bottles, finds a sample mean of 130, and concludes the sample must not be representative. Is the conclusion correct?
Why: the textbook question of the topic. A sample mean is expected to differ from \(\mu\); the question is whether 130 is further away than sampling variation can explain.
Why we do itWhat to write
Name the error in the reasoning first.
No, the conclusion is not correct. A sample mean is a random variable — it is expected to differ from \(\mu\). Finding \(\bar{x}\ne\mu\) is not evidence that anything is wrong.
Show what would be needed to judge it — and note it is missing.
The gap must be measured against the standard error: \[ z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \frac{130-125}{20/\sqrt{n}} \] The sample size \(n\) was never given, so the manager cannot judge his own result.
Show how much \(n\) changes the verdict — this is the persuasive part.
\(n=4\): \(\sigma_{\bar{x}}=10\), \(z=0.5\) — completely ordinary.
\(n=25\): \(\sigma_{\bar{x}}=4\), \(z=1.25\) — still unremarkable.
\(n=100\): \(\sigma_{\bar{x}}=2\), \(z=2.5\) — now worth investigating.
Same \(\bar{x}\), same \(\sigma\), opposite conclusions. The sample size decides everything, and the manager ignored it.
\(n=25\): \(\sigma_{\bar{x}}=4\), \(z=1.25\) — still unremarkable.
\(n=100\): \(\sigma_{\bar{x}}=2\), \(z=2.5\) — now worth investigating.
Same \(\bar{x}\), same \(\sigma\), opposite conclusions. The sample size decides everything, and the manager ignored it.
Q. Information is gathered on mean flossings per day for groups of 50 patients. What is this a sample from?
AnswerIt is a sample from the sampling distribution of the mean for samples of size 50, drawn from the patient population — not a sample from the patient population itself.
Why the distinction matters: the values recorded are group averages, so their spread is \(\sigma/\sqrt{50}\) — about one-seventh of the variation between individual patients. Anyone treating these figures as patient-level data would conclude that flossing habits are far more uniform than they really are.
Why the distinction matters: the values recorded are group averages, so their spread is \(\sigma/\sqrt{50}\) — about one-seventh of the variation between individual patients. Anyone treating these figures as patient-level data would conclude that flossing habits are far more uniform than they really are.
Four more, shaped like the exam
the "is this conclusion correct?" shape, drilled with numbers
X1Standard error
Q. A population has \(\mu = 500\) and \(\sigma = 40\). Find the standard error of the mean for \(n = 25\), \(n = 100\) and \(n = 400\). What does the pattern show, and what does it cost to halve the standard error?
Solution
Why: three substitutions into \(\sigma/\sqrt{n}\), then read the pattern — this is the "diminishing returns" idea that Unit 6 builds on.
Substitute.
\[ n=25:\ \tfrac{40}{5}=\mathbf{8.00} \quad n=100:\ \tfrac{40}{10}=\mathbf{4.00} \quad n=400:\ \tfrac{40}{20}=\mathbf{2.00} \]
Read the pattern and state its cost.
Each time \(n\) is quadrupled, the standard error is halved — because \(n\) enters through \(\sqrt{n}\). To double your precision you must quadruple your sample, so precision gets rapidly more expensive: this is the law of diminishing returns in sampling.
X2σ vs σ/√n
Q. Bag weights have \(\mu = 50\) kg, \(\sigma = 3\) kg. (a) What proportion of individual bags weigh over 51 kg? (b) What proportion of samples of 36 bags have a mean over 51 kg? Explain why the two answers differ so much.
Solution
Why: the same threshold applied to an individual and to a mean. (a) uses \(\sigma\); (b) uses \(\sigma/\sqrt{n}\).
(a) One bag ⇒ the population spread.
\[ z=\frac{51-50}{3}=0.33 \] \[ P=0.5-A(0.33)=0.5-0.1293=\mathbf{0.3707} \]
(b) A mean of 36 ⇒ the standard error.
\[ \sigma_{\bar{x}}=\frac{3}{\sqrt{36}}=0.5 \qquad z=\frac{51-50}{0.5}=2.00 \] \[ P=0.5-A(2.00)=0.5-0.4772=\mathbf{0.0228} \]
Explain the gap.
37% of individual bags exceed 51 kg, but only 2.3% of samples of 36 average above it. Averaging cancels out the high and low bags, so \(\bar{x}\) clusters far more tightly around \(\mu\) — by a factor of \(\sqrt{36}=6\). A value that is unremarkable for an individual can be extreme for a mean.
X3Judge the claim
Q. A supplier claims its components last \(\mu = 1{,}200\) hours with \(\sigma = 90\). A buyer tests 36 and gets \(\bar{x} = 1{,}165\). Is the claim credible?
Solution
Why: measure the gap in standard errors, then judge against the ±2 rule of thumb.
Standard error.
\[ \sigma_{\bar{x}}=\frac{90}{\sqrt{36}}=15 \]
The gap, in standard errors.
\[ z=\frac{1165-1200}{15}=-2.33 \] \[ P(\bar{x}\le1165)=0.5-A(2.33)=0.5-0.4901=\mathbf{0.0099} \]
Judge and state what follows.
If the claim were true, a sample mean this low would occur under 1% of the time — beyond 2 standard errors. That is evidence against the supplier's claim, not ordinary sampling variation. The buyer is entitled to challenge it (formally, this is a hypothesis test — Unit 9).
X4Name the distribution
Q. For each, say whether it describes the population, a single sample, or the sampling distribution of the mean: (a) the heights of all 4,000 recruits; (b) the heights of 40 recruits measured on Tuesday; (c) the set of all possible mean heights from samples of 40; (d) the standard deviation of the values in (c).
Solution
(a) Every element.
Population — mean \(\mu\), s.d. \(\sigma\).
(b) One subset actually measured.
A single sample — mean \(\bar{x}\), s.d. \(s\).
(c) All possible values of the statistic.
The sampling distribution of the mean — centred on \(\mu\).
(d) The spread of that distribution.
The standard error, \(\sigma_{\bar{x}} = \sigma/\sqrt{40}\). Note it is not \(s\), and not \(\sigma\).
Topic 4 of 4 · Levin §6.4 · pp. 299–301 · Q1 (a) and Q5 (b) — 7 marks
The Central Limit Theorem and Probabilities for \(\bar{x}\)
The unit's payoff. Every question is a Unit 4 normal problem with one extra line at the top — compute \(\sigma_{\bar{x}} = \sigma/\sqrt{n}\) — and then everything proceeds identically. Write that line first, every time, and this topic is 7 nearly-free marks.
How to know a question belongs to this topic
The single tell
The question gives \(\mu\), \(\sigma\) and a sample size \(n\), then asks for a probability about \(\bar{x}\) — "the sample mean", "the average of the 36", "the mean yield per acre". The word "mean" or "average" attached to a group is the signal.
Give-away words
in a sample of n observations
the probability that the sample mean…
the mean of the 48 days
average yield per acre
what is P(x̄ < …)
how large a sample must be taken so that…
The one line that separates it from Unit 4
Unit 4: \(z = \dfrac{x-\mu}{\sigma}\) — about a single observation.
Here: \(z = \dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}\) — about a mean.
Forgetting the \(\sqrt{n}\) is the single most expensive error in the paper, and it is easy to make because everything else looks identical.
Here: \(z = \dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}\) — about a mean.
Forgetting the \(\sqrt{n}\) is the single most expensive error in the paper, and it is easy to make because everything else looks identical.
When can you use the normal curve?
Population already normal → \(\bar{x}\) is normal for any \(n\).
Population not normal, or unknown → \(\bar{x}\) is approximately normal once \(n \ge 30\), by the Central Limit Theorem. Say which case you are in — Levin sets a deliberately skewed population (6-29) to check.
Population not normal, or unknown → \(\bar{x}\) is approximately normal once \(n \ge 30\), by the Central Limit Theorem. Say which case you are in — Levin sets a deliberately skewed population (6-29) to check.
Watch for TOTALS
If the question asks about a total rather than a mean (total weight on a ferry, combined life of 3 monitors), use \(\mu_T = n\mu\) and \(\sigma_T = \sigma\sqrt{n}\) — the total spreads wider, not narrower. Two of Levin's exercises here are totals in disguise.
Don't confuse it with
| If the question… | then… |
|---|---|
| asks about one item's value | Unit 4 — divide by \(\sigma\), no \(\sqrt{n}\) |
| asks about a mean of \(n\) items | Here — divide by \(\sigma/\sqrt{n}\) |
| asks about a total of \(n\) items | Here, but \(\sigma_T=\sigma\sqrt{n}\) |
| samples more than 5% of a finite population | Unit 6 — multiply by the finite population correction |
Master Method — Topic 4
Formula bank
Standard error
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
Standardising a
sample mean
sample mean
\[ z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \]
Totals
\[ \mu_T = n\mu \qquad \sigma_T = \sigma\sqrt{n} \]
Solving for \(n\)
\[ n = \left(\frac{z\,\sigma}{\bar{x}-\mu}\right)^{2} \qquad\text{then round \textbf{up}} \]
Careful with the variance. Levin often gives \(\sigma^2\) rather than \(\sigma\) — "a variance of 256" means \(\sigma = 16\). Taking the root is a separate step and an easy one to skip.
The routine — five lines, every time
- Write \(\mu\), \(\sigma\), \(n\). If a variance was given, take the square root now and write \(\sigma\) explicitly.
- Justify the normal curve. Either "the population is normal" or "\(n = 75 \ge 30\), so by the CLT \(\bar{x}\) is approximately normal". One line, and it is a mark.
- Compute the standard error \(\sigma_{\bar{x}} = \sigma/\sqrt{n}\) on its own line.
- Standardise, sketch and read — exactly as in Unit 4, using \(\sigma_{\bar{x}}\) in place of \(\sigma\).
- Answer in the story's words, and if a decision was asked for, give it.
How the exam asks it
| Slot | The question |
|---|---|
| Q1 (a) · 2 marks | Probability that \(\bar{x}\) falls in a symmetric interval around \(\mu\). Both papers worked here |
| Q5 (b) · 5 marks | Compute the standard error, then a probability for \(\bar{x}\) worked here |
Traps
- Dividing by \(\sigma\) instead of \(\sigma/\sqrt{n}\). The whole unit, lost in one line.
- Using \(\sigma^2\) as \(\sigma\). "Variance 361" means \(\sigma=19\), not 361.
- Forgetting to justify normality when the population is skewed or unstated.
- Treating a total as a mean. Totals get \(\sigma\sqrt{n}\); means get \(\sigma/\sqrt{n}\). They move in opposite directions.
- Rounding \(n\) down. When solving for a sample size, always round up — rounding down breaks the precision you promised.
Textbook bank — every Levin exercise on the sampling distribution of the mean
§6.4, pp. 299–301 · all 14, solved in the A(z) convention
14 exercises
Levin §6.4 Sampling distributions in more detail pp. 290–301
This is the highest-value exercise set in Units 5–6.
Q. In a sample of 16 observations from a normal distribution with mean 150 and variance 256, find (a) \(P(\bar{x}<160)\), (b) \(P(\bar{x}>142)\). With 9 observations instead, find (c) and (d) the same two.
Why: note "variance 256" — take the root first, \(\sigma = 16\). Then the only thing that changes between (a)/(b) and (c)/(d) is \(n\), which shows directly how \(n\) controls the spread.
Why we do itWhat to write
Variance → standard deviation. The population is stated to be normal, so no CLT justification is needed.
\[ \sigma=\sqrt{256}=16, \qquad \mu=150 \]
Standard error for \(n=16\).
\[ \sigma_{\bar{x}}=\frac{16}{\sqrt{16}}=4 \]
(a) and (b) — standardise with 4.
\[ \text{(a) } z=\frac{160-150}{4}=2.50 \;\Rightarrow\; P=0.5+A(2.50)=0.5+0.4938=\mathbf{0.9938} \] \[ \text{(b) } z=\frac{142-150}{4}=-2.00 \;\Rightarrow\; P=0.5+A(2.00)=\mathbf{0.9772} \]
Now \(n=9\) — a smaller sample, so a wider standard error.
\[ \sigma_{\bar{x}}=\frac{16}{\sqrt{9}}=5.3333 \]
(c) and (d).
\[ \text{(c) } z=\frac{10}{5.3333}=1.88 \;\Rightarrow\; P=0.5+0.4699=\mathbf{0.9699} \] \[ \text{(d) } z=\frac{-8}{5.3333}=-1.50 \;\Rightarrow\; P=0.5+0.4332=\mathbf{0.9332} \]
What the comparison shows: dropping from \(n=16\) to \(n=9\) widened the standard error from 4 to 5.33, and every probability moved away from certainty. A smaller sample makes \(\bar{x}\) less reliable — visible here in the numbers rather than asserted.
Q. A sample of 19 from a normal distribution with mean 18 and standard deviation 4.8. (a) \(P(16<\bar{x}<20)\)? (b) \(P(16\le\bar{x}\le20)\)? (c) The same as (a) if \(n = 48\).
Why: this is precisely the Q1 (a) shape — a symmetric interval around \(\mu\). Part (b) checks whether you know that endpoints carry no probability for a continuous variable.
Why we do itWhat to write
Standard error for \(n=19\).
\[ \sigma_{\bar{x}}=\frac{4.8}{\sqrt{19}}=1.1012 \]
(a) The interval is symmetric about \(\mu=18\), so one \(z\) does both ends and the two areas add.
\[ z=\pm\frac{2}{1.1012}=\pm1.82 \] \[ P=2A(1.82)=2(0.4656)=\mathbf{0.9312} \]
(b) Continuous distribution ⇒ single points have zero probability.
\[ P(16\le\bar{x}\le20)=P(16<\bar{x}<20)=\mathbf{0.9312} \] Identical. For a continuous variable, \(<\) and \(\le\) are interchangeable — unlike the binomial in Unit 4.
(c) Larger \(n\) ⇒ narrower standard error ⇒ \(\bar{x}\) hugs \(\mu\) more tightly.
\[ \sigma_{\bar{x}}=\frac{4.8}{\sqrt{48}}=0.6928 \qquad z=\pm\frac{2}{0.6928}=\pm2.89 \] \[ P=2A(2.89)=2(0.4981)=\mathbf{0.9962} \]
Q. In a normal distribution with mean 56 and standard deviation 21, how large a sample is needed so there is at least a 90% chance that its mean exceeds 52?
Why: the backwards direction. You are given the probability and must find \(n\) — so read the table inwards for \(z\), then solve the standardising equation for \(n\).
Why we do itWhat to write
Turn the 90% into a \(z\). We want 90% above 52, so 52 lies below the mean with 40% between it and \(\mu\).
\[ A(z)=0.90-0.5=0.40 \;\Rightarrow\; z=-1.28 \]
Write the standardising equation with \(n\) unknown.
\[ -1.28 = \frac{52-56}{21/\sqrt{n}} = \frac{-4\sqrt{n}}{21} \]
Solve for \(\sqrt{n}\), then \(n\).
\[ \sqrt{n} = \frac{1.28 \times 21}{4} = 6.72 \] \[ n = 6.72^2 = 45.16 \]
Round up — 45 would give slightly less than 90%.
\[ n = \mathbf{46} \]
Q. Mean 375, standard deviation 48. How large a sample is needed so the probability is at least 0.95 that the sample mean falls between 370 and 380?
Why: same as 6-26 but two-sided — 95% between two limits means 47.5% on each side of the mean, giving the familiar \(z = 1.96\).
Why we do itWhat to write
Two-sided ⇒ split the 95% equally about the mean.
\[ A(z)=\frac{0.95}{2}=0.475 \;\Rightarrow\; z=1.96 \]
The interval is symmetric about 375, so the half-width is 5.
\[ 1.96 = \frac{380-375}{48/\sqrt{n}} = \frac{5\sqrt{n}}{48} \]
Solve and round up.
\[ \sqrt{n} = \frac{1.96 \times 48}{5} = 18.816 \] \[ n = 354.04 \;\Rightarrow\; n = \mathbf{355} \]
This is the same algebra as Q3 (b) sample-size determination (Unit 6, Topic 3), written in a different order. There the half-width is called the tolerable error \(E\), and the formula is stated as \(n=(z\sigma/E)^2\) — here \(E = 5\), and \((1.96\times48/5)^2 = 354.04\). Identical.
Q. Meteors appear at an average of 50 per hour with a variance of 9. (a) Watching for 4 hours, what is the probability that at least 48 meteors per hour appear? (b) Watching one more hour, will this probability rise or fall? Why?
Why: "48 meteors per hour" over 4 hours is a mean of 4 observations, so \(n = 4\). Part (b) asks for the reasoning, not just a number — but computing it makes the answer convincing.
Why we do itWhat to write
Variance → SD, and identify \(n\) as the number of hours.
\[ \sigma=\sqrt{9}=3, \quad \mu=50, \quad n=4 \] \[ \sigma_{\bar{x}}=\frac{3}{\sqrt{4}}=1.5 \]
(a) Standardise and read the upper region.
\[ z=\frac{48-50}{1.5}=-1.33 \] \[ P(\bar{x}\ge48)=0.5+A(1.33)=0.5+0.4082=\mathbf{0.9082} \]
(b) Recompute with \(n=5\) to show the direction rather than assert it.
\[ \sigma_{\bar{x}}=\frac{3}{\sqrt{5}}=1.3416 \qquad z=\frac{-2}{1.3416}=-1.49 \] \[ P=0.5+A(1.49)=0.5+0.4319=\mathbf{0.9319} \]
Answer the "why".
It rises, from 0.9082 to 0.9319. A larger \(n\) shrinks the standard error, so \(\bar{x}\) clusters more tightly around 50 — and since 48 lies below the mean, tighter clustering pushes more of the distribution above it.
Q. Test scores are not normal — they are skewed left — with mean 86 and standard deviation 16. In a sample of 75 applicants, what is the probability the mean score is less than 84 or greater than 90?
Why: the population is explicitly not normal, so you must invoke the CLT before using a normal curve. That justification is a mark, and this question exists to see whether you write it.
Why we do itWhat to write
Justify normality first — the population being skewed does not stop us, because the CLT is about \(\bar{x}\), not about individuals.
\(n = 75 \ge 30\), so by the Central Limit Theorem the sampling distribution of \(\bar{x}\) is approximately normal regardless of the skew in the population.
Standard error.
\[ \sigma_{\bar{x}}=\frac{16}{\sqrt{75}}=1.8475 \]
Two opposite tails — standardise each.
\[ z_1=\frac{84-86}{1.8475}=-1.08 \qquad z_2=\frac{90-86}{1.8475}=2.17 \]
Each tail is \(0.5-A(z)\); add them.
\[ P=(0.5-0.3599)+(0.5-0.4850) \] \[ =0.1401+0.0150=\mathbf{0.1551} \]
The point of the question: the population is skewed, but \(\bar{x}\) is not. That is the whole content of the Central Limit Theorem, and it is why this unit works at all — you almost never know the shape of a real population.
Q. A monitor lasts on average 4,300 hours with standard deviation 730. A refinery has a primary plus two standby units, each taking over when the previous fails; lives are independent. Find the probability that a set of monitors lasts (a) at least 13,000 hours, (b) at most 12,630 hours.
Why: the monitors run one after another, so the quantity of interest is the total of three lives, not their mean. Totals use \(\sigma\sqrt{n}\) — the spread grows.
Why we do itWhat to write
Recognise it as a sum. Three monitors in sequence give a combined life \(T = X_1+X_2+X_3\).
\[ \mu_T = 3(4300) = 12{,}900 \text{ hours} \] \[ \sigma_T = 730\sqrt{3} = 1264.40 \]
(a) Upper region.
\[ z=\frac{13000-12900}{1264.40}=0.08 \] \[ P=0.5-A(0.08)=0.5-0.0319=\mathbf{0.4681} \]
(b) Lower region.
\[ z=\frac{12630-12900}{1264.40}=-0.21 \] \[ P=0.5-A(0.21)=0.5-0.0832=\mathbf{0.4168} \]
The trap in one line: had you used \(\sigma/\sqrt{3}=421.5\) you would have got wildly different answers. Means narrow by \(\sqrt{n}\); totals widen by \(\sqrt{n}\). Ask whether the question is about an average or an accumulation.
Q. Contaminants are normal with mean 64 ppm and variance 17.6. For 35 lakes sampled, find the probability that the sample average is (a) above 72 ppm, (b) between 64 and 72, (c) exactly 64, (d) above 94. (e) If the sample gave \(\bar{x} = 100\), would you trust the EPA study?
Why: the standard error here is tiny (0.71), so ordinary-looking values become astronomically improbable. Part (e) is the payoff — using the sampling distribution to challenge a claim.
Why we do itWhat to write
Variance → SD, then the standard error.
\[ \sigma=\sqrt{17.6}=4.1952 \qquad \sigma_{\bar{x}}=\frac{4.1952}{\sqrt{35}}=0.7091 \]
(a) Standardise — and notice how large \(z\) is.
\[ z=\frac{72-64}{0.7091}=11.28 \] \[ P(\bar{x}>72)\approx\mathbf{0.0000} \] Off the end of any table — effectively impossible.
(b) From the mean up to 72 is therefore essentially the entire upper half.
\[ P(64<\bar{x}<72)=A(11.28)\approx\mathbf{0.5000} \]
(c) A single point of a continuous distribution.
\[ P(\bar{x}=64)=\mathbf{0} \] True for any exact value of a continuous variable.
(d) Even further out.
\[ z=\frac{94-64}{0.7091}=42.3 \;\Rightarrow\; P\approx\mathbf{0.0000} \]
(e) The judgement — this is the real question.
No. \(\bar{x}=100\) is over 50 standard errors above 64 — a probability so small it is indistinguishable from zero. Rather than accept an impossibly rare event, conclude that one of the study's assumptions is wrong: the true mean is higher than 64, the variance is larger than 17.6, the lakes were not randomly sampled, or contamination has increased since the study.
Q. Returns average 110 phones per day with standard deviation 64. Unless Mr Ensor can be at least 80% certain that the mean over the next 48 days will be no more than 120, he will order an overhaul. Will the overhaul be ordered?
Why: a threshold decision. Compute \(P(\bar{x}\le120)\) and compare it against the 80% requirement — the comparison is the answer.
Why we do itWhat to write
Standard error over 48 days.
\[ \sigma_{\bar{x}}=\frac{64}{\sqrt{48}}=9.2376 \] \(n=48\ge30\), so the CLT applies even though the shape of daily returns is unstated.
Standardise the threshold.
\[ z=\frac{120-110}{9.2376}=1.08 \]
120 is above the mean and we want everything below it.
\[ P(\bar{x}\le120)=0.5+A(1.08)=0.5+0.3599=\mathbf{0.8599} \]
Compare with the requirement and state the decision.
\(0.8599 > 0.80\), so he can be at least 80% certain. The overhaul will NOT be ordered. Margin: 86% against a required 80% — comfortable, but not enormous.
Q. Investment profits are normal with mean $7,500 and standard deviation $3,300. For last year's 1,800 investments, find the probability that the sample mean (a) exceeded $7,700, (b) was less than $7,400, (c) was between $7,275 and $7,650.
Why: a very large \(n\) makes the standard error small (77.78 against a population SD of 3,300), so differences of a couple of hundred dollars become highly significant. That contrast is the lesson.
Why we do itWhat to write
Standard error — note how much smaller it is than \(\sigma\).
\[ \sigma_{\bar{x}}=\frac{3300}{\sqrt{1800}}=77.78 \]
(a) Upper tail.
\[ z=\frac{7700-7500}{77.78}=2.57 \] \[ P=0.5-A(2.57)=0.5-0.4949=\mathbf{0.0051} \]
(b) Lower tail.
\[ z=\frac{7400-7500}{77.78}=-1.29 \] \[ P=0.5-A(1.29)=0.5-0.4015=\mathbf{0.0985} \]
(c) Band straddling the mean ⇒ add.
\[ z_1=\frac{7275-7500}{77.78}=-2.89 \qquad z_2=\frac{7650-7500}{77.78}=1.93 \] \[ P=0.4981+0.4732=\mathbf{0.9713} \]
What Clara can predict without looking: with 1,800 investments the mean is almost certainly within a few hundred dollars of $7,500 — 97% chance of landing between $7,275 and $7,650. A single investment, by contrast, has a standard deviation of $3,300. Aggregation is what makes the portfolio predictable even though each investment is not.
Q. Yield per acre is normal with mean 120 bushels and standard deviation 12. Farmer Braun has 60 acres. Find (a) the expected mean yield, (b) the standard deviation of the sample mean, (c) \(P(\bar{x}>123.8)\), (d) \(P(117<\bar{x}<122)\).
Why: the cleanest possible statement of the whole unit — parts (a) and (b) are the two facts about the sampling distribution, and (c), (d) apply them.
Why we do itWhat to write
(a) The sampling distribution is centred on the population mean — it is unbiased.
\[ \mu_{\bar{x}} = \mu = \mathbf{120 \text{ bushels}} \]
(b) And its spread is the standard error.
\[ \sigma_{\bar{x}}=\frac{12}{\sqrt{60}}=\mathbf{1.5492 \text{ bushels}} \]
(c) Upper tail.
\[ z=\frac{123.8-120}{1.5492}=2.45 \] \[ P=0.5-A(2.45)=0.5-0.4929=\mathbf{0.0071} \]
(d) Band across the mean ⇒ add.
\[ z_1=\frac{117-120}{1.5492}=-1.94 \qquad z_2=\frac{122-120}{1.5492}=1.29 \] \[ P=0.4738+0.4015=\mathbf{0.8753} \]
For Farmer Braun's planning: his 60-acre average will land between 117 and 122 bushels about 88% of the time, and will exceed 123.8 less than once in 140 years. Individual acres vary by \(\sigma=12\); his farm average varies by only 1.55.
Q. A ferry carries 25 passengers, each of weight normal with mean 168 lb and variance 361. Safety rules say total passenger weight should exceed 4,250 lb no more than 5% of the time. Find (a) \(P(\text{total} > 4{,}250)\), (b) the 95th percentile of total weight. Is the ferry complying?
Why: "total weight" — a sum, so \(\sigma_T = \sigma\sqrt{n}\). Part (b) runs the table backwards, and the two together answer the compliance question.
Why we do itWhat to write
Variance → SD, then the parameters of the total.
\[ \sigma=\sqrt{361}=19 \] \[ \mu_T=25(168)=4200 \qquad \sigma_T=19\sqrt{25}=95 \]
(a) Upper tail of the total.
\[ z=\frac{4250-4200}{95}=0.53 \] \[ P=0.5-A(0.53)=0.5-0.2019=\mathbf{0.2981} \]
(b) Backwards — 95% below means \(A(z)=0.45\).
\[ z=1.64 \;\Rightarrow\; x = 4200+1.64(95)=\mathbf{4355.8 \text{ lb}} \] (with \(z=1.645\): 4356.3 lb)
Answer the compliance question by comparing the two results.
No — the ferry is not complying. The regulation allows the total to exceed 4,250 lb at most 5% of the time, but it actually happens 29.8% of the time. Equivalently, the 95th percentile is 4,356 lb, well above the 4,250 lb limit. To comply, the operator must either raise the limit to about 4,356 lb or reduce the passenger count.
Q. \(\mu = 98.6\), \(\sigma = 17.2\). Find \(P(92<\bar{x}<102)\) for (a) \(n = 25\), (b) \(n = 36\).
(a) \(n=25\).
\[ \sigma_{\bar{x}}=\frac{17.2}{5}=3.44 \] \[ z_1=\frac{92-98.6}{3.44}=-1.92 \qquad z_2=\frac{102-98.6}{3.44}=0.99 \] \[ P=0.4726+0.3389=\mathbf{0.8115} \]
(b) \(n=36\) — narrower standard error, so more of the distribution falls inside the same interval.
\[ \sigma_{\bar{x}}=\frac{17.2}{6}=2.8667 \] \[ z_1=-2.30 \qquad z_2=\frac{3.4}{2.8667}=1.186 \] \[ P=0.4893+0.3810=\mathbf{0.8703} \]
A rounding note worth knowing. In (b), \(z_2 = 1.186\) sits between table rows. Levin rounds it down to 1.18 (\(A = 0.3810\)) giving 0.8703; conventional rounding gives 1.19 (\(A = 0.3830\)) and 0.8723. Both are accepted — state which you used. The Re-Exam creates this situation deliberately, so practise saying "taking \(z \approx 1.19\)".
Q. \(\mu = 112\), \(\sigma = 56\), \(n = 50\). Find (a) \(P(\bar{x}<100)\), (b) \(P(100<\bar{x}<130)\).
Justify the CLT, then the standard error.
\(n=50\ge30\), so \(\bar{x}\) is approximately normal. \[ \sigma_{\bar{x}}=\frac{56}{\sqrt{50}}=7.920 \]
(a) Lower tail.
\[ z=\frac{100-112}{7.920}=-1.52 \] \[ P=0.5-A(1.52)=0.5-0.4357=\mathbf{0.0643} \]
(b) Band across the mean ⇒ add.
\[ z_1=-1.52 \qquad z_2=\frac{130-112}{7.920}=2.27 \] \[ P=0.4357+0.4884=\mathbf{0.9241} \]
Both match Levin's printed answers exactly.
Four more, shaped like Q1 (a) and Q5 (b)
the two slots this topic is worth 7 marks in
X1Q1 (a) shape
Q. Parcel weights have \(\mu = 12.5\) kg and \(\sigma = 2.4\) kg. For a random sample of 36 parcels, find the probability that the sample mean lies between 12 and 13 kg.
Solution
Why: a symmetric interval about \(\mu\) — exactly Q1 (a). One \(z\) does both ends and the areas add.
Standard error first.
\[ \sigma_{\bar{x}}=\frac{2.4}{\sqrt{36}}=0.40 \]
Symmetric ⇒ \(2A(z)\).
\[ z=\pm\frac{0.5}{0.40}=\pm1.25 \] \[ P=2A(1.25)=2(0.3944)=\mathbf{0.7888} \]
X2Q5 (b) shape
Q. Monthly office-supply spend has \(\mu = ₹1{,}800\) and \(\sigma = ₹300\). For a sample of 36 months, (a) find the standard error, (b) \(P(1750<\bar{x}<1850)\), (c) \(P(\bar{x}<1750)\).
Solution
Why: the Q5 (b) pattern — standard error asked for explicitly, then probabilities. Notice the interval in (b) is exactly \(\mu\pm1\sigma_{\bar{x}}\).
(a) Standard error.
\[ \sigma_{\bar{x}}=\frac{300}{\sqrt{36}}=\mathbf{₹50} \]
(b) The limits sit one standard error either side — the empirical rule's 68%.
\[ z=\pm\frac{50}{50}=\pm1.00 \] \[ P=2A(1.00)=2(0.3413)=\mathbf{0.6826} \]
(c) Lower tail.
\[ z=-1.00 \;\Rightarrow\; P=0.5-0.3413=\mathbf{0.1587} \]
X3Solve for n
Q. A population has \(\sigma = 15\). How large a sample is needed so that the sample mean is within 2 units of \(\mu\) with 95% probability?
Solution
Why: "within 2 units" is a symmetric interval of half-width 2, so 95% two-sided gives \(z=1.96\).
Get \(z\) from the two-sided probability.
\[ A(z)=0.475 \;\Rightarrow\; z=1.96 \]
Rearrange and round up.
\[ n=\left(\frac{z\sigma}{E}\right)^2=\left(\frac{1.96\times15}{2}\right)^2=216.09 \] \[ n=\mathbf{217} \]
X4Effect of n
Q. \(\mu = 500\), \(\sigma = 100\). Find \(P(\bar{x}>530)\) for \(n = 25\) and for \(n = 64\), and explain the change.
Solution
\(n=25\).
\[ \sigma_{\bar{x}}=\frac{100}{5}=20 \qquad z=\frac{30}{20}=1.50 \] \[ P=0.5-A(1.50)=0.5-0.4332=\mathbf{0.0668} \]
\(n=64\).
\[ \sigma_{\bar{x}}=\frac{100}{8}=12.5 \qquad z=\frac{30}{12.5}=2.40 \] \[ P=0.5-A(2.40)=0.5-0.4918=\mathbf{0.0082} \]
Explain.
The probability falls by a factor of about eight. A bigger sample makes \(\bar{x}\) cluster more tightly around 500, so a mean as far out as 530 becomes far less likely. The same deviation is more surprising in a larger sample — which is exactly why larger samples give stronger evidence.
THE 90-SECOND VERSION — WHAT TO CARRY INTO THE HALL
Unit 5 Summary Rules
\[ \text{Standard Error: } \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}, \qquad z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \]
\[ \text{Central Limit Theorem: } n \ge 30 \implies \bar{x} \sim N(\mu, \sigma/\sqrt{n}) \]
\[ \text{Sample Size Formula: } n = \left(\frac{z \cdot \sigma}{E}\right)^2 \quad (\text{round UP}) \]
◆ Where this sits in the rest of the site
Concepts: Unit 05 notes. The two real exam questions from this unit — Q1 (a) and Q5 (b) — are worked in both versions at Solved Papers.
Unit 6 continues directly from here: the same \(\sigma/\sqrt{n}\), plus the finite population correction for when you sample more than 5% of a population, the same machinery applied to a proportion, and sample-size determination — Q3 (b), another 5 marks.