Mathematics 9709/62 — October/November 2021

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

questions

marks

minutes

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

Q1Sampling and EstimationLinear Combinations of Random VariablesFree sample

The mass, in kilograms, of a block of cheese sold in a supermarket is denoted by the random variable $M$ .
The masses of a random sample of 40 blocks are summarised as follows.

n = 40 \quad \Sigma m = 20.50 \quad \Sigma m^2 = 10.7280

(a)

Calculate unbiased estimates of the population mean and variance of $M$ .

DifficultyMedium-Easy

Worked solution

Approach

We estimate the population mean by the sample mean. For the population variance we use the unbiased formula with denominator $n-1$ .

Working

The sample mean is

\bar m = \frac{\Sigma m}{n} = \frac{20.50}{40} = 0.5125

The unbiased estimate of the population variance is

s^2 = \frac{1}{n-1}\left(\Sigma m^2 - \frac{(\Sigma m)^2}{n}\right)

Substitute the given values:

s^2 = \frac{1}{39}\left(10.7280 - \frac{20.50^2}{40}\right)

= \frac{1}{39}(10.7280 - 10.50625)

= \frac{0.22175}{39} = 0.0056859

So to 3 significant figures, $s^2 = 0.00569$ .

Answer

Population mean estimate: $0.5125\text{ kg}$ .

Population variance estimate: $0.00569\text{ kg}^2$ (3 sf).

Final answer

Mean = 0.5125 kg, unbiased variance estimate = 0.00569 kg^2 (3 sf)

Detailed explanation

Walkthrough

We are given summary statistics for a random sample of 40 blocks of cheese: the total mass and the total of squared masses. The population mean is estimated by the sample mean, so we divide the total mass by the number of blocks.

For the variance, the formula

\frac{1}{n-1}\left(\Sigma m^2 - \frac{(\Sigma m)^2}{n}\right)

is used because dividing by $n-1$ gives an unbiased estimate of the population variance. The term $\frac{(\Sigma m)^2}{n}$ corrects for the fact that we are using the sample mean rather than the true population mean.

We first compute $\frac{20.50^2}{40} = 10.50625$ , subtract it from $10.7280$ to get $0.22175$ , and then divide by $39$ . This gives $0.0056859$ , which rounds to $0.00569$ at 3 significant figures.

Key Takeaways

The sample mean is an unbiased estimate of the population mean.
The unbiased variance estimate uses denominator $n-1$ , not $n$ .
Summary statistics $\Sigma m$ and $\Sigma m^2$ are enough to compute both estimates.

Common Mistakes

Using $n$ instead of $n-1$ in the variance formula gives the biased variance and scores no method or accuracy marks here.
Forgetting to subtract $\frac{(\Sigma m)^2}{n}$ before dividing.
Rounding intermediate values too early, which can change the final 3 significant figures.

Things to Be Careful About

Keep the full value $0.0056859$ for use in part (b).
The final variance should be given to 3 significant figures as $0.00569$ .
The mean can be written as $0.5125$ , $0.513$ , or $\frac{41}{80}$ .

Techniques used

calculate sample meanapply unbiased variance formula with n - 1round to 3 significant figures

(b)

The price, $P, of a block of cheese of mass $M\text{ kg}$ is found using the formula $P = 11M + 0.50$ .

Find estimates of the population mean and variance of $P$ .

DifficultyMedium-Easy

Worked solution

Approach

Since $P = 11M + 0.50$ , use the linear transformation rules for the mean and variance of a random variable:

\mathrm{E}(aX + b) = a\,\mathrm{E}(X) + b, \qquad \mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X)

Apply these to the estimates found in part (a).

Working

From part (a), the estimates are

\bar m = 0.5125, \qquad s^2 = 0.0056859

Estimate of the mean of $P$ :

\bar p = 11\bar m + 0.50 = 11(0.5125) + 0.50 = 5.6375 + 0.50 = 6.1375

Estimate of the variance of $P$ :

s_p^2 = 11^2 s^2 = 121 \times 0.0056859 = 0.6879936 \approx 0.688

The constant $0.50$ does not affect the variance.

Answer

Mean estimate of $P$ : $6.1375$ (or $6.14$ to 3 sf).

Variance estimate of $P$ : $0.688$ (3 sf).

Final answer

Mean = 6.1375, variance = 0.688 (3 sf)

Detailed explanation

Walkthrough

The price $P$ is a linear function of the mass $M$ : $P = 11M + 0.50$ . For any random variable $X$ , the mean of $aX+b$ is $a\mathrm{E}(X)+b$ , and the variance of $aX+b$ is $a^2\mathrm{Var}(X)$ . The constant $b$ shifts the mean but does not change the spread, so it does not appear in the variance.

We use the estimates from part (a) as our best guesses for the population mean and variance of $M$ . Multiplying the mean estimate by $11$ and adding $0.50$ gives the mean estimate of $P$ . For the variance, we multiply the variance estimate by $11^2 = 121$ ; we do not add anything for the $0.50$ .

Key Takeaways

A linear transformation $aX+b$ scales the mean by $a$ and shifts it by $b$ .
The variance is scaled by $a^2$ and is unaffected by adding a constant.
Estimates of transformed variables use the same transformation rules.

Common Mistakes

Adding $0.50$ to the variance.
Multiplying the variance by $11$ instead of $11^2$ .
Using the standard deviation instead of the variance when scaling.
Using the biased variance from part (a) instead of the unbiased estimate.

Things to Be Careful About

Use the unrounded variance $0.0056859$ from part (a) in the calculation.
The final variance should be rounded to 3 significant figures: $0.688$ .
The mean estimate can be given as $6.1375$ , $\frac{491}{80}$ , or $6.14$ to 3 sf.

Techniques used

apply linear transformation to meanapply variance scaling by square of coefficientround to 3 significant figures

The rest of this paper

6 more questions

Q2Sampling and Estimation3M
Q3Sampling and Estimation6M
Q4Sampling and Estimation · Hypothesis Tests7M
Q5The Poisson Distribution · Linear Combinations of Random Variables9M
Q6Hypothesis Tests10M
Q7Continuous Random Variables9M

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