AS & A Level · Practice questions · Populations, samples and the distribution of the sample mean

From one sample to the pattern of all samples.

Six original Cambridge-style Paper 6 questions on sampling and the distribution of the sample mean, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Centre \(\mu\), spread \(\sigma/\sqrt{n}\), shape normal.

Random sampling: number the population, select with random numbers, ignore repeats and out-of-range values.
Criticise in context: who is left out, and which way would that bias the result?
Sample mean: \(\bar{X} \sim N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\), sd \(\dfrac{\sigma}{\sqrt{n}}\).
Shape: exact if \(X\) is normal; by the CLT if \(n\) is large; say which.
Finding \(n\): set up an inequality in \(\sqrt{n}\) and round UP.

Recall

[2 marks]

Explain the difference between a population and a sample, and state why a sample should be chosen randomly.

The population is every member of the group being studied; a sample is a selection of members from it. ✓
Random selection gives every member an equal chance of being chosen, so the sample is unbiased and likely to be representative. ✓

Analysis

[2 marks]

A sports club has 800 members. Describe how random numbers could be used to choose a random sample of 20 members.

List the members and number them 001 to 800. ✓
Generate random 3-digit numbers, ignoring 000, numbers above 800 and repeats, until 20 are found; choose those members. ✓

Analysis

[2 marks]

To estimate the proportion of a city's adults who use public transport, a student questions people waiting at the main bus station. Give two reasons why this sample is unsatisfactory.

People who never use buses (or who use only trains) cannot be chosen, so the sample is not random. ✓
Everyone sampled is using public transport at that moment, so the proportion will be overestimated; also only people there at that time of day are included. ✓

Calculation

[3 marks]

\(X \sim N(120, 20^{2})\). A random sample of 16 values is taken. Find \(\text{P}(115 < \bar{X} < 125)\).

\(\bar{X} \sim N\!\left(120, \dfrac{400}{16}\right) = N(120, 25)\), sd 5. ✓
\(z = \dfrac{\pm 5}{5} = \pm 1\). ✓
\(\Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.683\) (3 s.f.). ✓

Calculation

[4 marks]

\(X \sim N(\mu, 25)\). Find the smallest sample size \(n\) such that \(\text{P}(\bar{X} - \mu > 1) < 0.01\).

\(\bar{X} - \mu \sim N\!\left(0, \dfrac{25}{n}\right)\). ✓
Need \(\dfrac{1}{5/\sqrt{n}} > 2.326\) (the \(z\) value with upper tail 0.01). ✓
\(\sqrt{n} > 11.63\), so \(n > 135.3\). ✓
Smallest \(n = 136\). ✓

Analysis

[5 marks]

The masses of grapes in punnets have mean 252 g and standard deviation 6 g. The distribution of masses is not normal.

(a) A random sample of 40 punnets is taken. Explain why the sample mean mass is approximately normally distributed.

(b) Find the probability that the sample mean is less than 250 g.

(a) The sample is large (\(n = 40\)), so by the Central Limit Theorem \(\bar{X}\) is approximately normal, even though the population is not. ✓
(b) \(\bar{X} \approx N\!\left(252, \dfrac{36}{40}\right)\), sd \(0.9487\). \(z = \dfrac{250 - 252}{0.9487} = -2.108\). ✓
\(\text{P}(\bar{X} < 250) = 1 - 0.9825 = 0.0175\) (3 s.f.). ✓
(c) With \(n = 4\) the sample is small and the population is not normal, so the CLT does not apply and \(\bar{X}\) need not be normal. ✓

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