Mathematics 9709/61 — 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 · The Poisson Distribution · Hypothesis Tests · Continuous Random Variables · Linear Combinations of Random Variables

Q1Sampling and EstimationFree sample

It is known that the height $H$ , in metres, of trees of a certain kind has the distribution $N(12.5, 10.24)$ .
A scientist takes a random sample of 25 trees of this kind and finds the sample mean, $\bar{H}$ , of the heights.

(a)

State the distribution of $\bar{H}$ , giving the values of any parameters.

DifficultyMedium-Easy

Worked solution

Approach

Since the population is normal, the sample mean of $25$ independent observations is also normal. Its mean is the population mean, and its variance is the population variance divided by the sample size.

Working

\bar{H} \sim N\left(12.5, \frac{10.24}{25}\right)

E(\bar{H}) = 12.5

and

\text{Var}(\bar{H}) = \frac{10.24}{25} = 0.4096

Answer

\bar{H} \sim N(12.5, 0.4096)

Final answer

N(12.5, 0.4096)

Detailed explanation

Walkthrough

We are told the population of heights is normally distributed with mean $12.5$ m and variance $10.24$ m². When we take a random sample of size $25$ , the sample mean is a random variable. For a normal population, the sample mean is exactly normal, not just approximately normal. Its mean is the same as the population mean, $12.5$ . Its variance is the population variance divided by the sample size, so $\frac{10.24}{25} = 0.4096$ . Therefore the distribution is $N(12.5, 0.4096)$ .

Key Takeaways

For a normal population, the sample mean is exactly normally distributed.
The mean of the sample mean is the population mean.
The variance of the sample mean is the population variance divided by $n$ .
Always state the parameters clearly: mean and variance.

Common Mistakes

Writing the variance as $10.24$ instead of $\frac{10.24}{25}$ .
Writing the standard deviation as $10.24$ instead of $\sqrt{0.4096}$ .
Using the Central Limit Theorem unnecessarily; it is not needed because the population is already normal.

Things to Be Careful About

The distribution must be stated with variance, not standard deviation, unless the question asks for standard deviation.
The mark scheme accepts $0.410$ (3sf) or $\frac{10.24}{25}$ as the variance, so an unsimplified fraction is acceptable.
Remember the sample size is $25$ , not $1$ .

Techniques used

apply the sampling distribution of the sample meandivide the population variance by the sample sizestate the normal distribution with its parameters

(b)

Find $P(12 < \bar{H} < 13)$ .

DifficultyMedium

Worked solution

Approach

Use the distribution from part (a). Standardise the sample mean to a standard normal variable, then find the probability that Z lies between the two corresponding z-values.

Working

From part (a), $\bar{H} \sim N(12.5, 0.4096)$ , so the standard deviation is

\sqrt{0.4096} = 0.64

Standardise:

z = \frac{13 - 12.5}{0.64} = 0.78125

and by symmetry the lower boundary is

z = \frac{12 - 12.5}{0.64} = -0.78125

Therefore

\begin{aligned} P(12 < \bar{H} < 13) &= \Phi(0.78125) - \Phi(-0.78125) \\ &= \Phi(0.78125) - \left(1 - \Phi(0.78125)\right) \\ &= 2\Phi(0.78125) - 1 \\ &= 0.565 \quad (3\text{ sf}) \end{aligned}

Answer

P(12 < \bar{H} < 13) = 0.565

Final answer

0.565 (3sf)

Detailed explanation

Walkthrough

We need the probability that the sample mean lies between $12$ and $13$ . Since the sample mean is normal with mean $12.5$ and standard deviation $0.64$ , we convert the boundaries to z-scores. The upper boundary $13$ gives $z = \frac{13 - 12.5}{0.64} = 0.78125$ . The lower boundary $12$ gives $z = \frac{12 - 12.5}{0.64} = -0.78125$ . The probability is the area under the standard normal curve between $-0.78125$ and $0.78125$ . By symmetry, this is $2\Phi(0.78125) - 1$ . Using tables, $\Phi(0.78125) \approx 0.7827$ , so the probability is about $0.565$ .

Key Takeaways

To find a probability for a normal sample mean, standardise using the sample mean's own mean and standard deviation.
The standard deviation of the sample mean is $\sqrt{\text{Var}(\bar{H})}$ .
Symmetry lets us write $P(-a < Z < a) = 2\Phi(a) - 1$ .
Always show the standardisation step; it is a method mark.

Common Mistakes

Standardising with the population variance $10.24$ instead of the sample mean variance $0.4096$ .
Forgetting to take the square root of the variance before standardising.
Using $1 - \Phi(0.781)$ for both tails instead of subtracting the two tail probabilities correctly.
Giving an unsupported final answer; the mark scheme requires the standardisation and central-area method to be shown.

Things to Be Careful About

The mark scheme accepts standardising with either $13$ or $12$ , as long as the correct central area is then found.
Use enough decimal places when reading the table; $0.78125$ rounds to $0.781$ for table use.
The final answer should be given to 3 significant figures: $0.565$ .
Check whether the question asks for the probability that the sample mean is between two values, not the population height.

Techniques used

standardise the sample mean using its mean and standard deviationuse symmetry of the normal distribution to find a central probabilityread standard normal probabilities from tables

The rest of this paper

6 more questions

Q2The Poisson Distribution4M
Q3Sampling and Estimation5M
Q4Continuous Random Variables9M
Q5The Poisson Distribution · Hypothesis Tests9M
Q6Sampling and Estimation · Linear Combinations of Random Variables8M
Q7Hypothesis Tests · Sampling and Estimation10M

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