9709/62 May/June 2024 — Mathematics Solved Past Paper

Q1MediumThe Poisson Distribution

A random variable $X$ has the distribution $\text{Po}(145)$ .

(a)

Use a suitable approximating distribution to calculate $P(X \leq 150)$ .

DifficultyMedium

Worked solution

Approach

Since $\lambda = 145$ is large, approximate the Poisson distribution $\text{Po}(145)$ by a normal distribution with the same mean and variance:

Y \sim \text{N}(145, 145)

Because $\text{Po}(145)$ is discrete, apply a continuity correction to the upper boundary $150$ , giving $150.5$ . Then standardise and use the standard normal table.

Working

The approximating normal distribution is

Y \sim \text{N}(145, 145)

Apply the continuity correction:

P(X \leq 150) \approx P(Y \leq 150.5)

Standardise:

z = \frac{150.5 - 145}{\sqrt{145}} = 0.457 \quad (3\text{ sf})

Therefore

P(X \leq 150) \approx \Phi(0.457) = 0.676 \quad (3\text{ sf})

Answer

P(X \leq 150) \approx 0.676

Final answer

P(X ≤ 150) ≈ 0.676

Detailed explanation

Walkthrough

The random variable $X$ follows a Poisson distribution with parameter $\lambda = 145$ . For such a large value of $\lambda$ , calculating the Poisson probability $P(X \leq 150)$ directly would require summing many terms, so we use a normal approximation. The rule is that $\text{Po}(\lambda)$ can be approximated by $\text{N}(\lambda, \lambda)$ , because a Poisson distribution has mean and variance both equal to $\lambda$ . Here, $\lambda = 145$ , so the approximating distribution is $\text{N}(145,145)$ .

Since the Poisson distribution is discrete and the normal distribution is continuous, we need a continuity correction. The event $X \leq 150$ uses the upper limit $150$ , so in the continuous approximation we use $150.5$ . This accounts for the fact that integer values up to $150$ correspond to a continuous interval ending halfway between $150$ and $151$ .

Now we convert the boundary to a standard normal $z$ -score:

z = \frac{150.5 - 145}{\sqrt{145}} \approx 0.457

The required probability is the area under the standard normal curve to the left of $0.457$ , written $\Phi(0.457)$ . Using normal distribution tables, this area is approximately $0.676$ .

Key Takeaways

The normal approximation to a Poisson distribution is used when $\lambda$ is large. It replaces $\text{Po}(\lambda)$ with $\text{N}(\lambda,\lambda)$ .

A continuity correction is essential when approximating a discrete distribution by a continuous one. For $P(X \leq k)$ , use $k + 0.5$ .

After standardising, the standard normal table gives the final probability.

Common Mistakes

Omitting the continuity correction and using $150$ instead of $150.5$ . This gives an approximation that is not accepted, and the mark scheme only condones an incorrect or omitted continuity correction for the method mark, not for the final answer.
Using $145$ as the standard deviation instead of $\sqrt{145}$ . The variance is $145$ , so the standard deviation is the square root.
Finding the wrong tail. For $X \leq 150$ , we use the area to the left of the $z$ -value.
Writing the unsupported answer $0.676$ . The mark scheme awards only partial credit if no working is shown.

Things to Be Careful About

The continuity correction boundary depends on the inequality sign: $\leq k$ becomes $k+0.5$ , $< k$ becomes $k-0.5$ , and $\geq k$ becomes $k-0.5$ .

Keep the intermediate $z$ -value accurate enough before reading the table.

State the approximating distribution $\text{N}(145,145)$ explicitly, as this earns a mark.

Use the correct formula $z = \frac{x - \mu}{\sqrt{\sigma^2}}$ ; here $\mu = 145$ and $\sigma^2 = 145$ .

Techniques used

approximate Poisson by a normal distributionapply continuity correction for a discrete-to-continuous approximationstandardise using the mean and varianceread a probability from the standard normal distribution

(b)

Justify the use of your approximating distribution in this case.

1M Similar

DifficultyEasy

Worked solution

Approach

The normal approximation to a Poisson distribution is reliable when the Poisson mean $\lambda$ is sufficiently large. Justify it by writing down the value of $\lambda$ and comparing it with the standard condition $\lambda > 15$ .

Working

Here $X \sim \text{Po}(145)$ , so

\lambda = 145

The usual condition for using the normal approximation is $\lambda > 15$ . In this case

145 > 15

so the condition is satisfied.

Answer

Since $\lambda = 145 > 15$ , the normal approximation is appropriate.

Final answer

λ = 145 > 15, so the normal approximation is appropriate.

Detailed explanation

Walkthrough

To justify using the normal approximation, we need to confirm that the Poisson parameter is large enough that the distribution is approximately symmetric and bell-shaped. The standard rule used in this syllabus is $\lambda > 15$ . Since the given distribution is $\text{Po}(145)$ , we have $\lambda = 145$ , which is much larger than $15$ . Therefore the normal approximation is appropriate. It is essential to state the actual value $\lambda = 145$ , not just write $\lambda > 15$ , because the condition must be checked with the specific parameter.

Key Takeaways

The normal approximation to the Poisson distribution requires $\lambda$ to be large; the A-Level convention is $\lambda > 15$ .

The mean and variance of $\text{Po}(\lambda)$ are both $\lambda$ , so the approximating normal distribution is $\text{N}(\lambda, \lambda)$ .

Common Mistakes

Writing only $\lambda > 15$ without identifying that $\lambda = 145$ . The mark scheme explicitly disallows this: you must state the value of $\lambda$ .
Comparing $\lambda$ with the wrong threshold, such as 10 or 30. Use the syllabus condition $\lambda > 15$ .
Using $n > 30$ or a sample-size rule, which is for the Central Limit Theorem, not for a Poisson approximation.

Things to Be Careful About

The condition is often accepted as $\lambda \geq 15$ as well, but the value $145$ must still be stated.
A shorter sentence such as 'the mean is large' may be accepted only if it is clear that the mean is $145$ and is greater than $15$ .
This justification is separate from the calculation and carries its own mark, so it should not be skipped.

Techniques used

state the Poisson parametercompare the Poisson parameter with the large-sample threshold for normal approximation

Q2Medium-EasySampling and Estimation

Henri wants to choose a random sample from the 804 students at his college. He numbers the students from 1 to 804 and then uses random numbers generated by his calculator. The first 20 random digits produced by his calculator are as follows.

5 6 7 1 0 9 8 4 3 1 0 9 6 6 5 0 2 1 7 6

Henri’s first two student numbers are 567 and 109.

(a)

Use Henri’s digits to find the numbers of the next two students in the sample.

2M Similar

(b)

There were 30 students in Henri’s sample. He asked each of them how much time, $X$ hours, they spent on social media each week, on average. He summarised the results as follows.

n = 30 \quad \sum x = 610 \quad \sum x^2 = 12405

Use this information to calculate an unbiased estimate of the mean of $X$ and show that an unbiased estimate of the variance of $X$ is less than 0.1 .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

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

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

Substitute $n=30$ , $\sum x = 610$ , $\sum x^2 = 12405$ .

Working

\bar{x} = \frac{610}{30} = \frac{61}{3} \approx 20.3

\begin{aligned} s^2 &= \frac{1}{29}\left(12405 - \frac{610^2}{30}\right) \\ &= \frac{1}{29}\left(12405 - \frac{372100}{30}\right) \\ &= \frac{1}{29}\left(\frac{50}{30}\right) \\ &= \frac{5}{87} \approx 0.0575 \end{aligned}

Since $0.0575 < 0.1$ , the unbiased estimate of the variance is less than 0.1.

Answer

\text{Unbiased estimate of mean} = \frac{61}{3} \approx 20.3

\text{Unbiased estimate of variance} = \frac{5}{87} \approx 0.0575

Final answer

Mean = 61/3 ≈ 20.3; variance = 5/87 ≈ 0.0575

Detailed explanation

Walkthrough

The unbiased estimate of the population mean is the sample mean, so substitute directly: $\bar{x} = 610/30 = 61/3$ . For the variance, the unbiased estimate uses denominator $n-1$ , not $n$ , because dividing by $n$ would underestimate the population variance. Substitute into $s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ . Compute $610^2/30 = 372100/30 = 12403\frac{1}{3}$ , so $12405 - 12403\frac{1}{3} = \frac{5}{3}$ . Dividing by 29 gives $\frac{5}{87} \approx 0.0575$ , which is less than 0.1.

Key Takeaways

Unbiased estimates use $n-1$ for variance. The mean estimate is just the sample mean. Always show the formula and substitution.

Common Mistakes

Using $n$ instead of $n-1$ in the variance denominator.
Using $\frac{\sum x^2}{n} - \bar{x}^2$ without multiplying by $\frac{n}{n-1}$ .
Arithmetic errors in $610^2/30$ .
Forgetting to state that $0.0575 < 0.1$ .

Things to Be Careful About

The two forms of the unbiased variance formula are equivalent:

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

Use whichever is easier, but keep the $n-1$ denominator. The exact fraction is $\frac{5}{87}$ .

Techniques used

apply unbiased estimate formula for the meanapply unbiased estimate formula for the variancesimplify fraction and compare with 0.1

(c)

Henri’s friend claims that Henri has probably made a mistake in his calculation of $\sum x$ or $\sum x^2$ .

Use your answer to part (b) to comment on this claim.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Compare the size of the estimated variance with what would be plausible for weekly social media use in hours. A very small variance suggests the data summary is unreliable.

Working

The estimated variance is about $0.0575$ , so the estimated standard deviation is:

\sqrt{0.0575} \approx 0.240

This means almost all students would be within about $0.24$ hours of the mean, which is unrealistically small for weekly social media times.

Answer

The variance is unrealistically small, so Henri has probably made a mistake in $\sum x$ or $\sum x^2$ ; the friend's claim is probably correct.

Final answer

The variance is unrealistically small, so Henri has probably made a mistake; the claim is probably correct.

Detailed explanation

Walkthrough

Part (b) gave variance estimate about 0.0575, so standard deviation about 0.240 hours. For weekly social media use across 30 students, this is unrealistically small: it would mean almost all students spend within about a quarter of an hour of the same time. Therefore the data summary is suspicious, so Henri has probably made an arithmetic mistake in $\sum x$ or $\sum x^2$ . The friend's claim is probably correct.

Key Takeaways

An estimate can be mathematically correct but practically implausible. A very small variance for real-world data suggests a recording or calculation error.

Common Mistakes

Only saying 'less than 0.1' without saying 'small'.
Commenting on the mean being large or small instead of the variance.
Not linking the small variance to the possibility of a mistake.

Things to Be Careful About

The mark scheme requires both ideas: the variance is small/unrealistic and Henri has probably made a mistake. Saying 'mean is large' alone is not enough, but 'mean is large compared with the variance' is acceptable.

Techniques used

interpret the size of the estimated variancerelate a very small variance to a likely data error

Q34MMediumSampling and Estimation

A student wishes to estimate the proportion, $p$ , of students at her college who have exactly one brother. She surveys a random sample of 50 students at her college and finds that 18 of them have exactly one brother. She calculates an approximate $\alpha\%$ confidence interval for $p$ and finds that the lower limit of the confidence interval is 0.244 correct to 3 significant figures.

Find $\alpha$ correct to the nearest integer.

DifficultyMedium

Worked solution

Approach

The lower limit of a confidence interval for a proportion $p$ has the form

\hat{p} - z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

We set the lower limit equal to the given value $0.244$ , solve for the critical value $z$ , then convert this $z$ to the total central probability of the standard normal distribution. That central probability equals $\alpha$ (as a percentage).

Working

The sample proportion is

\hat{p} = \frac{18}{50} = 0.36

The standard error of the sample proportion is

\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.36 \times 0.64}{50}} = \sqrt{0.004608}

Setting the lower limit equal to $0.244$ :

0.36 - z\sqrt{0.004608} = 0.244

So

z\sqrt{0.004608} = 0.36 - 0.244 = 0.116

Since $\sqrt{0.004608} = 0.067883$ , we get

z = \frac{0.116}{0.067883} \approx 1.709

The confidence level $\alpha$ corresponds to the central region of the standard normal distribution:

\alpha = 100 \times P(-1.709 < Z < 1.709)

From the standard normal table, $\Phi(1.709) = 0.9564$ . Hence

P(-1.709 < Z < 1.709) = 2\Phi(1.709) - 1 = 2(0.9564) - 1 = 0.9128

So

\alpha = 100 \times 0.9128 = 91.28

Answer

\alpha = 91

Final answer

α = 91

Detailed explanation

Walkthrough

This is a two-sided confidence interval problem, so the interval has the form: point estimate $\pm$ margin of error. The lower limit is the point estimate minus the margin. We know the point estimate $\hat{p} = 0.36$ (18 out of 50 students), we know the sample size $n = 50$ , and we are told the lower limit is $0.244$ . So the only unknown is the critical value $z$ — the quantity that depends on the confidence level.

We substitute into

0.36 - z\sqrt{\frac{0.36 \times 0.64}{50}} = 0.244

Step by step:

Compute the standard error: $\sqrt{0.004608} \approx 0.067883$ .
Rearrange the equation: the margin $z \times 0.067883 = 0.116$ , so $z = 0.116 / 0.067883 \approx 1.709$ .
Convert $z$ to a probability. Because this is a two-sided interval, the confidence level is the total area under the normal curve between $-1.709$ and $1.709$ . That is $2\Phi(1.709) - 1 = 2(0.9564) - 1 = 0.9128$ (equivalently the mark scheme writes $1 - 2(1 - 0.956) = 0.912$ ).
Multiply by 100 to express as a percentage: $\alpha = 91.28\%$ , which rounds to $91\%$ .

Key Takeaways

The confidence interval for a proportion is always of the form
$\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
When only one limit is given, substitute it back into the formula and solve for the critical value $z$ .
The confidence coefficient that pairs with $z$ is the central area $2\Phi(z) - 1$ , expressed as a percentage.
The words 'lower limit' or 'upper limit' tell you whether to use the $-$ of the $\pm$ sign.

Common Mistakes

Quoting $0.91$ instead of $91$ ; the central probability is a decimal, but the question asks for a percentage.
Using only one tail, such as $\Phi(z)$ or $1 - \Phi(z)$ , instead of the whole middle region, which gives an incorrect level.
Forgetting to divide by $n$ inside the square root when forming the standard error.
Rounding $z$ too early; keeping $z \approx 1.709$ preserves accuracy for the final conversion.

Things to Be Careful About

The question asks for the answer correct to the nearest integer, so $\alpha = 91.28$ rounds to $91$ .
The mark scheme accepts $z \approx 1.71$ if it is stated, but to retain accuracy use $1.709$ until the final step.
When reading the standard normal table for $\Phi(1.709)$ , a small misread can shift the last digit; only round at the very end.

Techniques used

set up the confidence interval equation for a proportionsolve for the critical z-value from the lower limitconvert the z-value into the central confidence probability

Approach

Since $X_1$ and $X_2$ are independent and identically distributed normals, the difference $X_1 - X_2$ is a normal random variable with mean $0$ . The value $0$ is the centre of this symmetric distribution, so the probability that it is exceeded is exactly $0.5$ .

Working

\begin{aligned} X_1 - X_2 &\sim N(10 - 10,\; 12 + 12) = N(0, 24) \end{aligned}

Therefore

\begin{aligned} P(X_1 > X_2) &= P(X_1 - X_2 > 0) \\ &= P(Z > 0) \\ &= 0.5 \end{aligned}

Answer

P(X_1 > X_2) = 0.5

Final answer

0.5

Detailed explanation

Walkthrough

Both $X_1$ and $X_2$ come from the same normal distribution $N(10, 12)$ and are chosen independently. The event $X_1 > X_2$ is exactly the event $X_1 - X_2 > 0$ . Because the two variables are drawn independently from the same distribution, there is complete symmetry: neither value is more likely to be the larger. Forming the difference shows this explicitly — $X_1 - X_2$ is normal with mean $10 - 10 = 0$ . The central value $0$ of a normal distribution is its mean, and a normal distribution is symmetric about its mean, so the probability of being above $0$ is exactly half: $P(X_1 - X_2 > 0) = 0.5$ . Note that the variance does not matter at all for this part: even though $\mathrm{Var}(X_1 - X_2) = 12 + 12 = 24$ , the relevant boundary is the mean, so the tail probability is exactly $0.5$ .

A quicker conceptual route: since $P(X_1 > X_2) = P(X_2 > X_1)$ by symmetry and these two events together cover everything (the probability of equality is $0$ for continuous random variables), each equals $0.5$ .

Key Takeaways

For two independent variables from the same continuous distribution, the probability that one exceeds the other is always $0.5$ , by symmetry.
The difference of two independent normal random variables is again normal, a fact that follows from the linear-combination results for normals.
When the boundary of interest is the mean of a symmetric distribution, no standardising is needed — the answer is immediate.

Common Mistakes

Trying to compute the probability by integrating the joint density instead of using symmetry.
Forgetting that $X_1 - X_2$ has mean $0$ , or incorrectly thinking the mean is $10$ .
Adding variances incorrectly as $12 - 12 = 0$ and then concluding the difference is degenerate; the variance is actually $24$ .

Things to Be Careful About

For continuous distributions the probability of a tie is $0$ , so the events $X_1 > X_2$ and $X_2 > X_1$ are exhaustive and disjoint (apart from a zero-probability tie).
The symmetry argument requires the two variables to be independent and from the same distribution.
No calculator work or normal tables are needed for this part.

Techniques used

form the difference of independent normalsrecognise the symmetry of the symmetric normal distributionevaluate the upper-tail probability at the mean

(b)

Find $P(X_1 > 2X_2 - 3)$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Bring all variable terms to one side to obtain the single linear combination $X_1 - 2X_2 + 3$ . Since $X_1$ and $X_2$ are independent normals, this combination is also normal, so its mean and variance can be computed and the required probability found by standardising.

Working

Compute the mean:

\begin{aligned} E(X_1 - 2X_2 + 3) &= E(X_1) - 2E(X_2) + 3 \\ &= 10 - 2(10) + 3 \\ &= -7 \end{aligned}

Compute the variance. The variables are independent, so the variances add with each coefficient squared; the constant $+3$ contributes nothing:

\begin{aligned} \mathrm{Var}(X_1 - 2X_2 + 3) &= \mathrm{Var}(X_1) + (-2)^2\mathrm{Var}(X_2) + 0 \\ &= 12 + 4(12) \\ &= 60 \end{aligned}

Hence $X_1 - 2X_2 + 3 \sim N(-7, 60)$ . Now standardise:

\begin{aligned} P(X_1 > 2X_2 - 3) &= P(X_1 - 2X_2 + 3 > 0) \\ &= P\left(Z > \frac{0 - (-7)}{\sqrt{60}}\right) \\ &= P\left(Z > \frac{7}{\sqrt{60}}\right) \\ &= P(Z > 0.904) \\ &= 1 - \Phi(0.904) \\ &= 0.183 \end{aligned}

Answer

P(X_1 > 2X_2 - 3) = 0.183

Final answer

0.183

Detailed explanation

Walkthrough

The inequality $X_1 > 2X_2 - 3$ is rearranged by bringing every variable term to the left and the constant to the right:

X_1 - 2X_2 + 3 > 0

The left-hand side is a linear combination of the two independent normal variables $X_1$ and $X_2$ , so it is itself normal. Its mean is found using linearity of expectation:

E(X_1 - 2X_2 + 3) = E(X_1) - 2E(X_2) + 3 = 10 - 20 + 3 = -7

Note the $+3$ : the $-3$ on the right of the original inequality becomes $+3$ when moved to the left side. Its variance uses the rule that for independent variables the variances add, with each coefficient squared, and a constant adds nothing:

\mathrm{Var}(X_1 - 2X_2 + 3) = \mathrm{Var}(X_1) + 4\mathrm{Var}(X_2) = 12 + 48 = 60

So $X_1 - 2X_2 + 3 \sim N(-7, 60)$ . The probability we want is therefore

P\left(Z > \frac{0 - (-7)}{\sqrt{60}}\right) = P(Z > 0.904)

Finally this is an upper-tail probability:

P(Z > 0.904) = 1 - \Phi(0.904) = 0.183

Each step matches a mark-scheme requirement: the mean (B1), the variance (B1), the standardisation (M1), the use of the upper tail (M1), and the final value (A1).

Key Takeaways

Inequalities between linear expressions of independent normal variables are solved by first forming one linear combination and standardising.
$E(aX + bY + c) = aE(X) + bE(Y) + c$ for any constants.
$\mathrm{Var}(aX + bY + c) = a^2\mathrm{Var}(X) + b^2\mathrm{Var}(Y)$ when $X$ and $Y$ are independent; the constant term contributes nothing.
The resulting combination is normal, so standard-normal tables give the answer.

Common Mistakes

Using the wrong sign on the constant: $10 - 20 - 3 = -13$ instead of $10 - 20 + 3 = -7$ .
Forgetting to square the coefficient: using $\mathrm{Var}(2X_2) = 2 \times 12 = 24$ instead of $4 \times 12 = 48$ .
Writing the numerator as $-3 - 10 = -13$ ; the numerator must be $7$ or $-7$ (i.e. $0 - \text{mean}$ ), not $-13$ .
Quoting $\Phi(0.904)$ directly as the answer instead of $1 - \Phi(0.904)$ , since the required region is the upper tail.

Things to Be Careful About

The variance is in squared units; the standard deviation, $\sqrt{60}$ , is what is used in the denominator when standardising.
The standardised value is $(0 - (-7))/\sqrt{60} = 7/\sqrt{60} \approx 0.904$ ; keep the sign consistent.
Read $\Phi(0.904) \approx 0.817$ from tables, then subtract from 1 to give $0.183$ .
The final answer should be given to a sensible degree of accuracy, typically 3 significant figures or 3 decimal places.

Techniques used

rearrange the inequality into a single linear combination of the variablescompute the mean of a linear combinationcompute the variance of a linear combination of independent variablesstandardise a normal random variableapply the complement rule for the upper-tail probability

Q5Medium-HardThe Poisson DistributionLinear Combinations of Random Variables

The number of goals scored by a sports team in the first half of any match has the distribution $X \sim \text{Po}(3.1)$ . The number of goals scored by the same team in the second half of any match has the distribution $Y \sim \text{Po}(2.4)$ . You may assume that the distributions of $X$ and $Y$ are independent.

(a)

Find $P(X < 4)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a Poisson random variable $X \sim \text{Po}(3.1)$ , find $P(X < 4)$ by summing the probabilities of all values less than 4: $X = 0, 1, 2, 3$ , using the Poisson probability formula.

Working

The Poisson probability mass function is

P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}.

Here $\lambda = 3.1$ . Since $X < 4$ means $x = 0, 1, 2, 3$ ,

P(X < 4) = e^{-3.1}\left(1 + 3.1 + \frac{3.1^2}{2!} + \frac{3.1^3}{3!}\right).

Evaluating term by term,

P(X < 4) = 0.0450 + 0.1397 + 0.2165 + 0.22368 = 0.6249.

Answer

The probability is $P(X < 4) = 0.625$ to 3 significant figures.

Final answer

0.625 (3 s.f.)

Detailed explanation

Walkthrough

Start by recognising that $X$ is Poisson with mean $3.1$ . The inequality $X<4$ includes the whole-number values $0,1,2,3$ and excludes $4$ . For each such value we use the Poisson mass function $P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$ . Substituting $\lambda=3.1$ gives the four terms inside the brackets. Adding them gives the cumulative probability $0.6249$ , which rounds to $0.625$ .

Key Takeaways

A Poisson random variable takes only non-negative integer values, so inequalities must be converted into sums over those integers. The probability mass function $\frac{e^{-\lambda}\lambda^x}{x!}$ is the key tool.

Common Mistakes

The most common mistake is to confuse $X<4$ with $X\le 4$ , which would add the $X=4$ term. Another is to omit the $X=0$ term. The mark scheme also says a correct answer with no working scores only special case B1, so the expression must be shown.

Things to Be Careful About

Check the endpoint: $X<4$ excludes $4$ . Use the factorial denominators correctly. Use the given value $\lambda=3.1$ throughout, and round only at the end.

Techniques used

sum Poisson probability mass function termsevaluate exponential and factorial terms

(b)

Find the probability that, in a randomly chosen match, the team scores at least 5 goals.

3M Similar

DifficultyMedium

Worked solution

Approach

Let $T = X + Y$ be the total number of goals in a match. Because $X$ and $Y$ are independent Poisson variables, $T$ is also Poisson with parameter $3.1 + 2.4 = 5.5$ . The probability of at least 5 goals is $P(T \geq 5)$ , which we find as $1 - P(T \leq 4)$ .

Working

Since $T \sim \text{Po}(5.5)$ ,

P(T \leq 4) = e^{-5.5}\left(1 + 5.5 + \frac{5.5^2}{2!} + \frac{5.5^3}{3!} + \frac{5.5^4}{4!}\right).

Evaluating term by term,

P(T \leq 4) = 0.00408677 + 0.02247723 + 0.06181239 + 0.11332271 + 0.15581873 = 0.357518.

Therefore

P(T \geq 5) = 1 - 0.357518 = 0.642482.

Answer

The probability of at least 5 goals is $0.642$ or $0.643$ to 3 significant figures.

Final answer

0.642 or 0.643 (3 s.f.)

Detailed explanation

Walkthrough

The question asks about the total goals in a match. The first half and second half are independent and each follows a Poisson distribution, so their sum also follows a Poisson distribution with mean the sum of the means. This is the crucial step: $T\sim\text{Po}(5.5)$ . We then want $P(T\geq 5)$ . Since it is easier to calculate up to a value, we use the complement rule: $P(T\geq 5)=1-P(T\leq 4)$ . Substitute $\lambda=5.5$ into the Poisson formula and sum the probabilities for $T=0,1,2,3,4$ . Subtract from 1.

Key Takeaways

The sum of independent Poisson random variables is Poisson, with parameter equal to the sum of the parameters. For 'at least' probabilities with discrete distributions, consider using the complement $1-P(T\le k)$ .

Common Mistakes

Using only $\lambda=3.1$ or $\lambda=2.4$ instead of $5.5$ . Forgetting to subtract the cumulative sum from 1. Including $P(T=5)$ in the subtracted sum instead of stopping at $P(T=4)$ . Omitting the factorial denominators in the Poisson terms.

Things to Be Careful About

'At least 5' means $5,6,7,\ldots$ , so the complement is $0,1,2,3,4$ . The addition property requires independence, which is stated in the question. Show the expression for the cumulative probability because an unsupported numerical answer would not receive full marks.

Techniques used

add parameters of independent Poisson variablescompute cumulative Poisson probabilitiesapply complement rule for at least condition

(c)

Given that the team scores a total of 5 goals in a randomly chosen match, find the probability that they score exactly 3 goals in the first half.

4M Similar

DifficultyMedium-Hard

Worked solution

Approach

We need $P(X = 3 \mid X + Y = 5)$ . Use the conditional probability formula:

P(X=3 \mid X+Y=5) = \frac{P(X=3 \text{ and } X+Y=5)}{P(X+Y=5)}.

If $X=3$ and the total is 5, then $Y=2$ . Because $X$ and $Y$ are independent, the numerator is $P(X=3)\,P(Y=2)$ . The denominator uses the sum $X+Y \sim \text{Po}(5.5)$ .

Working

First find the numerator:

P(X=3) = e^{-3.1}\frac{3.1^3}{3!} = 0.223676,

P(Y=2) = e^{-2.4}\frac{2.4^2}{2!} = 0.261267.

Thus, by independence,

P(X=3 \text{ and } Y=2) = 0.223676 \times 0.261267 = 0.05844.

The total number of goals satisfies $X + Y \sim \text{Po}(5.5)$ , so

P(X+Y=5) = e^{-5.5}\frac{5.5^5}{5!} = 0.17140.

Therefore

P(X=3 \mid X+Y=5) = \frac{0.05844}{0.17140} = 0.341.

Answer

The probability is $0.341$ to 3 significant figures.

Final answer

0.341 (3 s.f.)

Detailed explanation

Walkthrough

This is a conditional probability question. We are told the total is 5 and want the probability that 3 goals came in the first half. By definition, $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ . Here $A$ is 'first half has exactly 3' and $B$ is 'total is 5'. If the first half has 3 and the total is 5, then the second half has exactly 2. Because $X$ and $Y$ are independent, the intersection probability is the product of the two individual Poisson probabilities. The denominator is the probability that $X+Y=5$ . The sum of independent Poisson variables is Poisson with mean $3.1+2.4=5.5$ , so we compute $P(X+Y=5)$ from $\text{Po}(5.5)$ . Dividing gives the conditional probability.

Key Takeaways

Conditional probability is the ratio of an intersection probability to the probability of the conditioning event. Independence lets us multiply $P(X=3)$ and $P(Y=2)$ . The sum of independent Poisson variables remains Poisson with added parameters.

Common Mistakes

Forgetting that the second half must have exactly 2 goals when the first half has 3 and the total is 5. Using the unconditional probability $P(X=3)$ as the answer instead of dividing by $P(X+Y=5)$ . Forgetting to use $\lambda=5.5$ for the total. Not showing the quotient formula when the mark scheme requires it.

Things to Be Careful About

Use the conditional probability formula explicitly so that the result is interpreted correctly. The independence assumption is essential for multiplying the two half probabilities. Make sure the denominator is $P(X+Y=5)$ , not $P(X=5)$ or $P(Y=5)$ . Round only at the end; the final answer is $0.341$ .

Techniques used

add parameters of independent Poisson variablesuse conditional probability formulamultiply independent Poisson probabilitiesevaluate Poisson probability mass functions

Q6Medium-HardHypothesis Tests

The masses of cereal boxes filled by a certain machine have mean 510 grams. An adjustment is made to the machine and an inspector wishes to test whether the mean mass of cereal boxes filled by the machine has decreased.

After the adjustment is made, he chooses a random sample of 120 cereal boxes. The mean mass of these boxes is found to be 508 grams.

Assume that the standard deviation of the masses is 10 grams.

(a)

Test at the 2.5% significance level whether the mean mass of cereal boxes filled by the machine has decreased.

5M Similar

DifficultyMedium

Worked solution

Approach

This is a one-tailed hypothesis test for a population mean. Since the sample size is large ( $n = 120$ ), the sample mean $\overline{X}$ is approximately normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$ . State the hypotheses, standardise the observed sample mean, and compare with the critical value at the 2.5% significance level.

Working

State the hypotheses:

H_0: \mu = 510 \text{ g}, \quad H_1: \mu < 510 \text{ g}

The test statistic is:

z = \frac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{508 - 510}{10/\sqrt{120}} = \frac{-2}{0.9129} = -2.191

For a one-tailed test at the 2.5% significance level, the critical value is $z = -1.96$ .

Since $-2.191 < -1.96$ , the test statistic lies in the critical region, so we reject $H_0$ .

Alternatively, the $p$ -value is $P(Z < -2.191) = 0.0142$ , which is less than $0.025$ .

Answer

Reject $H_0$ . There is sufficient evidence to suggest that the mean mass of cereal boxes filled by the machine has decreased.

Final answer

Reject H0. There is sufficient evidence that the mean mass has decreased.

Detailed explanation

Walkthrough

This is a hypothesis test about a population mean. The machine previously filled boxes with mean mass 510 g, and after an adjustment we want to know whether the mean has decreased. We set up the null hypothesis $H_0: \mu = 510$ (no change) and the alternative $H_1: \mu < 510$ (decreased). Because the sample size is 120, which is large, the Central Limit Theorem tells us the sample mean is approximately normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{n} = 10/\sqrt{120}$ . We standardise the observed sample mean of 508 g to get a z-score. The z-score measures how many standard errors 508 is below 510. We then compare this z-score with the critical value for a one-tailed test at 2.5% significance, which is -1.96. Since -2.191 is more extreme (further left) than -1.96, we reject the null hypothesis. The conclusion must be stated in context: there is sufficient evidence that the mean mass has decreased.

Key Takeaways

How to set up null and alternative hypotheses for a one-tailed test.
Using the Central Limit Theorem to justify the normal distribution of the sample mean for large samples.
Standardising the sample mean: $z = (\bar{x} - \mu)/(\sigma/\sqrt{n})$ .
Comparing the test statistic with the critical value, or the p-value with the significance level.
Writing a conclusion in context without overstating certainty.

Common Mistakes

Using a two-tailed test when the question asks whether the mean has decreased (one-tailed). The mark scheme says a two-tail test scores maximum B0 M1 A1 M1 A0.
Forgetting to divide by $\sqrt{120}$ when standardising.
Getting the inequality sign the wrong way round — the mark scheme says this scores M1 A0.
Writing "changed" instead of "decreased" in the conclusion.
Not stating hypotheses in terms of the population mean $\mu$ .

Things to Be Careful About

The hypotheses must refer to the population mean, not the sample mean.
The significance level is 2.5% for a one-tailed test, so the critical z-value is -1.96.
The conclusion must not be definite ("sufficient evidence", not "proves").
If using the p-value approach, compare 0.0142 with 0.025.

Techniques used

state null and alternative hypothesesstandardise the sample meancompare test statistic with critical valueconclude in context

(b)

Later the inspector carries out a similar test at the 2.5% significance level, using the same hypotheses and another 120 randomly chosen cereal boxes.

Given that the mean mass is now actually 506 grams, find the probability of a Type II error.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

A Type II error occurs when we fail to reject $H_0$ even though the alternative hypothesis is true. First find the critical value of the sample mean that separates the acceptance and rejection regions under $H_0$ . Then, assuming the true mean is 506 g, compute the probability that the sample mean falls in the acceptance region (i.e., above the critical value).

Working

Find the critical value $c$ such that $P(\overline{X} < c \mid \mu = 510) = 0.025$ :

\frac{c - 510}{10/\sqrt{120}} = -1.96

c = 510 - 1.96 \times \frac{10}{\sqrt{120}} = 510 - 1.96 \times 0.9129 = 510 - 1.789 = 508.21

Now compute the probability of a Type II error when $\mu = 506$ :

P(\text{Type II error}) = P(\overline{X} > 508.21 \mid \mu = 506)

= P\left(Z > \frac{508.21 - 506}{10/\sqrt{120}}\right) = P(Z > 2.421)

= 1 - \Phi(2.421) = 1 - 0.9922 = 0.0078

Answer

The probability of a Type II error is 0.0078 (2 s.f.).

Final answer

0.0078 (2 s.f.)

Detailed explanation

Walkthrough

A Type II error is failing to reject the null hypothesis when it is false. Here, the true mean is 506 g, so we want the probability that the test fails to reject $H_0$ (i.e., concludes the mean has not decreased) even though it actually has decreased to 506 g. First we need the critical value of the sample mean: the value below which we would reject $H_0$ . Under $H_0$ ( $\mu = 510$ ), the critical value $c$ satisfies $P(\bar{X} < c) = 0.025$ , so $(c - 510)/(10/\sqrt{120}) = -1.96$ , giving $c = 508.21$ . We reject $H_0$ when $\bar{X} < 508.21$ and fail to reject when $\bar{X} \geq 508.21$ . Now, under the true mean 506, the probability of a Type II error is $P(\bar{X} > 508.21 \mid \mu = 506)$ . We standardise: $z = (508.21 - 506)/(10/\sqrt{120}) = 2.421$ . Then $P(Z > 2.421) = 1 - \Phi(2.421) = 0.0078$ .

Key Takeaways

The definition of a Type II error: failing to reject $H_0$ when $H_1$ is true.
Finding the critical value of the test statistic (in original units) using the null distribution.
Recomputing the probability under the alternative distribution to find the Type II error.
The Type II error probability depends on the actual value of the parameter under the alternative.

Common Mistakes

Using the wrong mean in the standardisation: the critical value uses 510 (under $H_0$ ), but the Type II error probability uses 506 (the true mean).
Using $\frac{510-506}{10/\sqrt{120}}$ directly — the mark scheme says this scores max M0 A0 M1 M1 A0 because it misses the critical value step.
Forgetting to subtract from 1 when finding the upper-tail probability.
Confusing Type I and Type II errors.

Things to Be Careful About

The critical value must be computed with the null mean 510 and $\sigma/\sqrt{n}$ .
The probability is an upper-tail probability $P(Z > 2.421)$ , so it equals $1 - \Phi(2.421)$ .
The answer should be given to 2 significant figures (0.0078), and the mark scheme accepts 0.0077 to 0.0080 depending on rounding.

Techniques used

find the critical value of the teststandardise under the alternative meancompute the upper-tail probability

Q7MediumContinuous Random Variables

The probability density function, $f$ , of a random variable $X$ is given by

f(x) = \begin{cases} k(1 + \cos x) & 0 \leq x \leq \pi, \\ 0 & \text{otherwise,} \end{cases}

where $k$ is a constant.

(a)

Show that $k = \frac{1}{\pi}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For $f$ to be a valid probability density function, the total area under the curve over its entire domain must equal 1. Integrate $f(x) = k(1 + \cos x)$ from $0$ to $\pi$ and equate the result to 1, then solve for $k$ .

Working

Since $f$ is a PDF, the integral over the whole domain must equal 1:

\int_0^{\pi} k(1 + \cos x)\,\mathrm{d}x = 1

Integrating term by term:

k\left[x + \sin x\right]_0^{\pi} = 1

Substituting the limits:

k\left(\pi + \sin\pi - (0 + \sin 0)\right) = 1

Since $\sin\pi = 0$ and $\sin 0 = 0$ :

k\pi = 1

Therefore:

k = \frac{1}{\pi}

Final answer

k = 1/π

Detailed explanation

Walkthrough

For any function to be a valid probability density function, the total area under its curve across the entire domain must be exactly 1. This is the fundamental normalisation property of a PDF. Here the domain is $0 \le x \le \pi$ , and the PDF is $f(x) = k(1 + \cos x)$ there, and 0 elsewhere.

The antiderivative of $1 + \cos x$ is $x + \sin x$ . Evaluating this from $0$ to $\pi$ :

at $x = \pi$ : $\pi + \sin\pi = \pi + 0 = \pi$
at $x = 0$ : $0 + \sin 0 = 0$

So the definite integral is $\pi$ . Multiplying by $k$ and equating to 1 gives $k\pi = 1$ , hence $k = \frac{1}{\pi}$ .

Key Takeaways

The total integral of a PDF over its domain must equal 1 — this is the defining property used to find a normalising constant.
The antiderivative of $\cos x$ is $\sin x$ , and both $\sin\pi$ and $\sin 0$ equal 0, making the limit evaluation simple.

Common Mistakes

Forgetting that the integral of the PDF must equal 1 before solving for $k$ .
Dropping the factor $k$ while integrating, then solving incorrectly.
Writing $\sin\pi = 1$ or $\sin 0 = 1$ — both are 0 in radians.
Jumping straight to $k = \frac{1}{\pi}$ without showing the intermediate step $k\pi = 1$ ; the mark scheme explicitly requires evidence of substituting the limits.

Things to Be Careful About

The mark scheme awards the final A1 only if at least one interim step (e.g. $k\pi = 1$ ) is shown after integrating. A bare answer with no substitution of limits loses the mark.
Keep the constant $k$ outside the integral throughout — do not lose it when integrating.

Techniques used

integrate the PDF over its full domainequate the integral to 1solve for the normalising constant

(b)

Verify that the median of $X$ lies between 0.83 and 0.84 .

3M Similar

DifficultyMedium

Worked solution

Approach

The median $m$ of a continuous random variable satisfies $\int_0^m f(x)\,\mathrm{d}x = 0.5$ . Using the cumulative distribution function found in part (a), evaluate it at $m = 0.83$ and $m = 0.84$ , and show that $0.5$ lies between the two values.

Working

The cumulative distribution function is:

F(m) = \frac{1}{\pi}\int_0^m (1 + \cos x)\,\mathrm{d}x = \frac{1}{\pi}\left[x + \sin x\right]_0^m = \frac{m + \sin m}{\pi}

At $m = 0.83$ :

F(0.83) = \frac{0.83 + \sin(0.83)}{\pi} = \frac{0.83 + 0.73793}{\pi} = \frac{1.56793}{\pi} \approx 0.499

At $m = 0.84$ :

F(0.84) = \frac{0.84 + \sin(0.84)}{\pi} = \frac{0.84 + 0.74453}{\pi} = \frac{1.58453}{\pi} \approx 0.504

Since $0.499 < 0.5 < 0.504$ , the value of $m$ where $F(m) = 0.5$ must lie strictly between 0.83 and 0.84.

Answer

The median lies strictly between 0.83 and 0.84.

Final answer

0.83 < median < 0.84

Detailed explanation

Walkthrough

The median $m$ of a continuous random variable is defined as the value where the cumulative distribution function equals 0.5, i.e. $P(X \le m) = 0.5$ . Equivalently, $\int_0^m f(x)\,\mathrm{d}x = 0.5$ .

From part (a), we know $k = \frac{1}{\pi}$ , so the CDF is $F(m) = \frac{1}{\pi}\int_0^m (1 + \cos x)\,\mathrm{d}x = \frac{m + \sin m}{\pi}$ .

To verify the median lies between 0.83 and 0.84, we evaluate the CDF at both endpoints:

$F(0.83) = \frac{0.83 + \sin(0.83)}{\pi} \approx 0.499$ , which is just below 0.5.
$F(0.84) = \frac{0.84 + \sin(0.84)}{\pi} \approx 0.504$ , which is just above 0.5.

Since the CDF is continuous and increasing on the interval (the PDF is positive everywhere), the median must lie strictly between these two values.

Key Takeaways

The median of a continuous random variable is the value where the CDF equals 0.5.
To bracket the median between two values, evaluate the CDF at those values and check that 0.5 lies between the results.

Common Mistakes

Forgetting the factor $\frac{1}{\pi}$ when evaluating the CDF — this changes the comparison entirely.
Using degrees instead of radians for $\sin(0.83)$ and $\sin(0.84)$ . The angles are in radians.
Only showing that $F(0.83) < 0.5$ without also showing $F(0.84) > 0.5$ . Both inequalities are required.
Confusing the median with the mean — the median uses the CDF, not the expectation integral.

Things to Be Careful About

The mark scheme allows several equivalent approaches (integrating from 0 to 0.83 and 0.84 to $\pi$ ; or using $g(m) = m + \sin m - \frac{\pi}{2}$ ; or using $h(m) = m + \sin m$ ). Whichever you use, both evaluations must be shown.
The final conclusion must state BOTH that $F(0.83) < 0.5$ and $F(0.84) > 0.5$ before concluding $0.83 < m < 0.84$ .
If you get no marks, the mark scheme offers a special case: setting $\frac{1}{\pi}(m + \sin m) = 0.5$ and solving to $m \approx 0.831$ earns 2 marks.

Techniques used

form the cumulative distribution functionevaluate the CDF at two candidate valuescompare the CDF values against 0.5 to bracket the median

(c)

Find the exact value of $E(X)$ .

4M Similar

DifficultyMedium

Worked solution

Approach

The expectation of a continuous random variable is $E(X) = \int_0^{\pi} x f(x)\,\mathrm{d}x$ . Substitute $f(x) = \frac{1}{\pi}(1 + \cos x)$ , split the integral into two parts, and use integration by parts for the $x\cos x$ term.

Working

E(X) = \frac{1}{\pi}\int_0^{\pi} x(1 + \cos x)\,\mathrm{d}x = \frac{1}{\pi}\int_0^{\pi} (x + x\cos x)\,\mathrm{d}x

Split into two integrals:

E(X) = \frac{1}{\pi}\left[\int_0^{\pi} x\,\mathrm{d}x + \int_0^{\pi} x\cos x\,\mathrm{d}x\right]

First integral:

\int_0^{\pi} x\,\mathrm{d}x = \left[\frac{x^2}{2}\right]_0^{\pi} = \frac{\pi^2}{2}

Second integral — integrate by parts with $u = x$ , $\mathrm{d}v = \cos x\,\mathrm{d}x$ , so $\mathrm{d}u = \mathrm{d}x$ and $v = \sin x$ :

\int_0^{\pi} x\cos x\,\mathrm{d}x = \left[x\sin x\right]_0^{\pi} - \int_0^{\pi} \sin x\,\mathrm{d}x

= (\pi\sin\pi - 0) - \left[-\cos x\right]_0^{\pi} = 0 - \left(-\cos\pi + \cos 0\right) = -(-1 + 1) = -2

Combining the two results:

E(X) = \frac{1}{\pi}\left(\frac{\pi^2}{2} - 2\right) = \frac{\pi}{2} - \frac{2}{\pi}

Final answer

E(X) = π/2 - 2/π

Detailed explanation

Walkthrough

The expected value of a continuous random variable with PDF $f$ is defined as $E(X) = \int x f(x)\,\mathrm{d}x$ over the domain. Substituting $f(x) = \frac{1}{\pi}(1 + \cos x)$ gives:

E(X) = \frac{1}{\pi}\int_0^{\pi} x(1 + \cos x)\,\mathrm{d}x = \frac{1}{\pi}\int_0^{\pi} (x + x\cos x)\,\mathrm{d}x

The integral splits into two parts:

$\int_0^{\pi} x\,\mathrm{d}x = \left[\frac{x^2}{2}\right]_0^{\pi} = \frac{\pi^2}{2}$ — a straightforward power rule.
$\int_0^{\pi} x\cos x\,\mathrm{d}x$ — this needs integration by parts. We choose $u = x$ (so $\mathrm{d}u = \mathrm{d}x$ ) and $\mathrm{d}v = \cos x\,\mathrm{d}x$ (so $v = \sin x$ ). The formula $\int u\,\mathrm{d}v = uv - \int v\,\mathrm{d}u$ gives:

\int x\cos x\,\mathrm{d}x = x\sin x - \int \sin x\,\mathrm{d}x = x\sin x + \cos x

Evaluating from 0 to $\pi$ : at $\pi$ we get $\pi\sin\pi + \cos\pi = 0 - 1 = -1$ ; at 0 we get $0 + \cos 0 = 1$ . The difference is $-1 - 1 = -2$ .

Combining: $E(X) = \frac{1}{\pi}\left(\frac{\pi^2}{2} - 2\right) = \frac{\pi}{2} - \frac{2}{\pi}$ .

Key Takeaways

The defining formula for expectation of a continuous random variable is $E(X) = \int x f(x)\,\mathrm{d}x$ .
Integration by parts is the key tool for integrals of the form $\int x \cos x\,\mathrm{d}x$ . The choice $u = x$ is standard because its derivative is 1, simplifying the integral.
The exact value is preferred; a decimal is only an approximation.

Common Mistakes

Forgetting the factor $\frac{1}{\pi}$ when substituting $f(x)$ .
Sign errors in the integration by parts — particularly when evaluating $[-\cos x]_0^{\pi}$ . Remember $\cos\pi = -1$ and $\cos 0 = 1$ .
Writing $\sin\pi = 1$ — it is 0.
Mixing up the two integrals and combining them incorrectly.

Things to Be Careful About

The mark scheme allows the integration by parts to be done either as shown here (splitting the integral) or as one combined integral: $\frac{1}{\pi}\left(x^2 + x\sin x - \int(x + \sin x)\,\mathrm{d}x\right)$ . Both are acceptable.
An unsupported answer of $\frac{\pi}{2} - \frac{2}{\pi}$ scores only 3 marks (B3), and an unsupported decimal $0.934$ scores 2 marks (B2). Full working is required for all 4 marks.
The final answer must be exact — write $\frac{\pi}{2} - \frac{2}{\pi}$ , not a rounded decimal.

Techniques used

set up the expectation integralapply integration by partsevaluate the resulting definite integrals

Mathematics 9709/62 — May/June 2024

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