Paper 6. We rarely measure a whole population, so we take a sample and use its mean. Different random samples give different means, and that variation follows a pattern: centred on the population mean, narrower as the sample grows, and close to normal for large samples whatever the population looks like.
A random sample gives every member of the population an equal chance of selection, so it avoids bias. The sample mean \(\bar{X}\) of \(n\) independent observations has \(\text{E}(\bar{X}) = \mu\) and \(\text{Var}(\bar{X}) = \dfrac{\sigma^{2}}{n}\). If \(X\) is normal, \(\bar{X}\) is exactly normal; if \(n\) is large, the Central Limit Theorem says \(\bar{X}\) is approximately normal anyway.
Fig. 1: one value from N(50, 12²) (dashed) and the mean of a random sample of 9 (blue), drawn to scale. Both are centred on 50, but the sample mean has standard deviation 12/3 = 4; the shaded area is P(X̄ > 53) = 0.2266
Section 01
Populations and random samples.
The population is every item of interest; a sample is the part you actually observe. A sample is only useful if it represents the population, and the way to make that likely is to choose it at random.
Random sample: every member of the population is equally likely to be chosen, and the choices are independent.
Random numbers: number the population (e.g. 001 to 600), then use a random number generator or table and take numbers in turn, ignoring any outside the range and any repeats.
Unsatisfactory methods leave some members with no chance, or a smaller chance, of selection: only people at one place or time, only volunteers, only the first names on a list.
Explain in context: say which group is left out and why that group might differ on the thing being measured.
Worked example Choosing with random numbers
A school has 600 students. Describe how to use random numbers to choose a random sample of 30 students.
Obtain a list of all 600 students and number them 001 to 600.
Generate random three-digit numbers (calculator or table).
Ignore numbers 000 and above 600, and ignore repeats.
Continue until 30 different numbers are obtained; the students with those numbers form the sample.
Answer: Number the list, select by random numbers, discard out-of-range values and repeats.
Worked example Why a method is unsatisfactory
To estimate how much time residents of a town spend exercising each week, a researcher asks the first 50 people to enter a gym on a Monday morning. Explain why this sample is unsatisfactory.
Only people who use that gym can be chosen; residents who never go to a gym have no chance of selection.
People entering a gym are likely to exercise more than the average resident.
Only Monday-morning users are included, so people who work then are left out.
Answer: The sample is not random and is biased towards people who exercise more, so it will overestimate the mean.
Examiner trap
"It is too small" or "it is not random" alone rarely scores. Say WHO is excluded and HOW that would push the answer up or down.
Section 02
The sample mean is a random variable.
Each new random sample gives a new value of \(\bar{X}\). Its distribution is centred on \(\mu\), but because highs and lows in a sample partly cancel, it is less spread out than a single value: \(\text{Var}(\bar{X}) = \dfrac{\sigma^{2}}{n}\), standard deviation \(\dfrac{\sigma}{\sqrt{n}}\). If \(X \sim N(\mu, \sigma^{2})\), then \(\bar{X} \sim N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\) exactly, for any \(n\).
Worked example Probability for a sample mean
\(X \sim N(50, 12^{2})\). A random sample of 9 values is taken. Find \(\text{P}(\bar{X} > 53)\).
\(\bar{X} \sim N\!\left(50, \dfrac{144}{9}\right) = N(50, 16)\), standard deviation 4.
\(\text{Var}(\bar{X}) = \dfrac{\sigma^{2}}{n}\), so the standard deviation is \(\dfrac{\sigma}{\sqrt{n}}\), not \(\dfrac{\sigma}{n}\). With \(\sigma = 12\) and \(n = 9\) that is 4, not \(1.33\).
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
\(X\) has variance 36. Find the variance of the mean of a random sample of 4 values.
Quick check+10 XP
A survey on how often residents use the local library interviews people leaving the library. Why is this unsatisfactory?
Quick check+10 XP
What does the Central Limit Theorem say?
Quick check+10 XP
Random samples of size \(n\) are taken from a population with mean \(\mu\). What is \(\text{E}(\bar{X})\)?
Section 03
The Central Limit Theorem.
Even when the population is not normal (a Poisson count, a skewed waiting time), the mean of a LARGE random sample is approximately normal: \(\bar{X} \approx N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\). Cambridge needs only this informal statement; \(n \ge 30\) or so is usually treated as large. The same idea gives the total: \(X_1 + \cdots + X_n \approx N(n\mu, n\sigma^{2})\).
Worked example Mean of Poisson counts
The number of emails a worker receives per hour has distribution \(\text{Po}(4)\). For a random sample of 50 hours, find the probability that the mean number of emails per hour is less than 3.8.
\(\mu = 4\), \(\sigma^{2} = 4\). The population is not normal, but \(n = 50\) is large, so by the CLT \(\bar{X} \approx N\!\left(4, \dfrac{4}{50}\right) = N(4, 0.08)\).
A random variable has mean 2.5 and variance 1.25 (not normal). Find, approximately, the probability that the total of 40 independent values exceeds 108.
By the CLT, \(T \approx N(40 \times 2.5,\ 40 \times 1.25) = N(100, 50)\).
\(z = \dfrac{108 - 100}{\sqrt{50}} = 1.131\).
\(\text{P}(T > 108) = 1 - 0.8710 = 0.1290\).
Answer: \(0.129\) (3 s.f.)
Examiner trap
Use the CLT only when the population is NOT normal and \(n\) is large, and say so. If the population is normal, \(\bar{X}\) is exactly normal and the CLT is not needed; if the population is not normal and \(n\) is small, neither applies.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
\(X \sim N(80, 15^{2})\). A random sample of 25 values is taken. Find \(\text{P}(\bar{X} < 77)\).
Exam style+20 XP
The number of faults per metre of wire has distribution \(\text{Po}(2.5)\). Use a suitable approximation to find the probability that the mean number of faults per metre in a random sample of 60 one-metre lengths exceeds 2.7.
Exam style+20 XP
\(X \sim N(\mu, 4^{2})\). Find the smallest sample size for which \(\text{P}(|\bar{X} - \mu| < 1) > 0.95\).
Exam style+20 XP
A random variable, not normally distributed, has mean 3.2 and standard deviation 1.5. Find, approximately, the probability that the total of 50 independent values exceeds 170.
Skill unlocked
Samples understood, means distributed.
This skill is now recorded on your profile. Keep the chain going.