Mathematics 9709/61 — October/November 2022

Cambridge A-Level · Probability & Statistics 2 · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Sampling and Estimation · Hypothesis Tests · Linear Combinations of Random Variables · The Poisson Distribution · Continuous Random Variables

Q13MSampling and EstimationFree sample

The heights, in metres, of a random sample of 10 mature trees of a certain variety are given below.

5.9 \quad 6.5 \quad 6.7 \quad 5.9 \quad 6.9 \quad 6.0 \quad 6.4 \quad 6.2 \quad 5.8 \quad 5.8

Find unbiased estimates of the population mean and variance of the heights of all mature trees of this variety.

DifficultyMedium-Easy

Worked solution

Approach

Use the sample mean as an unbiased estimate of the population mean. For the unbiased estimate of the population variance, use the formula with divisor $n - 1$ :

s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)

First find $\sum x$ and $\sum x^2$ .

Working

There are $n = 10$ trees.

\sum x = 5.9 + 6.5 + 6.7 + 5.9 + 6.9 + 6.0 + 6.4 + 6.2 + 5.8 + 5.8 = 62.1

So the unbiased estimate of the population mean is

\bar{x} = \frac{62.1}{10} = 6.21

The sum of squares is

\sum x^2 = 5.9^2 + 6.5^2 + 6.7^2 + 5.9^2 + 6.9^2 + 6.0^2 + 6.4^2 + 6.2^2 + 5.8^2 + 5.8^2 = 387.05

Therefore the unbiased estimate of the population variance is

\begin{aligned} s^2 &= \frac{1}{9}\left(387.05 - \frac{62.1^2}{10}\right) \\ &= \frac{1}{9}\left(387.05 - 385.641\right) \\ &= \frac{1}{9}(1.409) \\ &= 0.156555\ldots \\ &= 0.157 \quad (3 \text{ sf}) \end{aligned}

Answer

Unbiased estimate of the population mean: $6.21$ m.

Unbiased estimate of the population variance: $0.157$ m $^2$ (3 sf).

Final answer

Mean = 6.21, unbiased variance = 0.157 (3 sf)

Detailed explanation

Walkthrough

We are given a random sample of 10 tree heights and need unbiased estimates of the population mean and variance.

The sample mean $\bar{x}$ is an unbiased estimator of the population mean $\mu$ , so we first add all the heights and divide by 10. This gives $6.21$ m.

For the variance, an unbiased estimate uses divisor $n - 1$ rather than $n$ . The formula is

s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)

We need both $\sum x$ and $\sum x^2$ . We already found $\sum x = 62.1$ . Squaring each height and adding gives $\sum x^2 = 387.05$ .

Substituting into the formula:

\frac{1}{9}\left(387.05 - \frac{62.1^2}{10}\right) = \frac{1}{9}(387.05 - 385.641) = \frac{1}{9}(1.409) = 0.156555\ldots

Rounded to 3 significant figures, this is $0.157$ .

Key Takeaways

The sample mean is an unbiased estimate of the population mean.
The unbiased estimate of the population variance uses divisor $n - 1$ , not $n$ .
To use the variance formula, you need both the sum of the observations and the sum of their squares.

Common Mistakes

Dividing by $n = 10$ instead of $n - 1 = 9$ gives the biased variance $0.1409$ , which is not accepted as an unbiased estimate.
Forgetting to subtract $\frac{(\sum x)^2}{n}$ before dividing by $n - 1$ .
Rounding intermediate values too early, which can change the final answer.

Things to Be Careful About

Keep enough decimal places in intermediate working; only round the final answer to 3 significant figures.
Remember the units: the mean is in metres, and the variance is in square metres.
The mark scheme allows the variance formula to be implied, but showing the substitution clearly avoids losing method marks.

Techniques used

calculate the sample meancalculate the sum of squared observationsapply the unbiased variance formula with divisor n - 1

The rest of this paper

6 more questions

Q2Hypothesis Tests8M
Q3The Poisson Distribution7M
Q4Linear Combinations of Random Variables8M
Q5Sampling and Estimation7M
Q6Linear Combinations of Random Variables · Continuous Random Variables7M
Q7Hypothesis Tests10M

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