A random variable $X$ has probability density function given by $$ \mathrm{f}(x) = \begin{cases} k(2x^2 - x^3) & 0 \leqslant x \leqslant 2, \\ 0 & \text{otherwise.} \end{cases} $$ (a) Show that $k = \frac{3}{4}$. (b) The median of $X$ is denoted by $m$. (i) Write down the value of $\mathrm{P}(X \leqslant m)$. (ii) Hence find $\mathrm{P}(\mathrm{E}(X) \leqslant X \leqslant m)$.

(a) k = 3/4 · (b)(i) 0.5 · (b)(ii) 0.0248 or 31/1250

9709/62 Oct/Nov 2025 — Mathematics Solved Past Paper

Q1MediumThe Poisson Distribution

The number, $X$ , of used computers donated to a charity has a constant average rate of 2.4 computers per week.

(a)

State a necessary condition for $X$ to have a Poisson distribution.

1M Similar

(b)

Now assume that $X$ has a Poisson distribution.

Calculate the probability that the number of computers donated during a 4-week period is more than 6 and less than 9.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Over 4 weeks, the Poisson mean is $\lambda = 2.4 \times 4 = 9.6$ . The event 'more than 6 and less than 9' means $X = 7$ or $X = 8$ , so add the two Poisson probabilities.

Working

\lambda = 2.4 \times 4 = 9.6

P(X=7) = \frac{\mathrm{e}^{-9.6}9.6^7}{7!} = 0.10098

P(X=8) = \frac{\mathrm{e}^{-9.6}9.6^8}{8!} = 0.12118

P(6 < X < 9) = P(X=7) + P(X=8) = 0.10098 + 0.12118 = 0.22216

Answer

0.222 \ (3 \text{ sf})

Final answer

0.222 (3 sf)

Detailed explanation

Walkthrough

The average rate is 2.4 computers per week. Over a 4-week period, the mean number of computers donated is

\lambda = 2.4 \times 4 = 9.6.

Since $X$ has a Poisson distribution, the probability that $X$ takes a particular value is

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

The event 'more than 6 and less than 9' means $X = 7$ or $X = 8$ , because $X$ can only take integer values. Therefore,

P(6 < X < 9) = P(X=7) + P(X=8).

Substitute $\lambda = 9.6$ :

P(X=7) = \frac{\mathrm{e}^{-9.6}9.6^7}{7!} = 0.10098

and

P(X=8) = \frac{\mathrm{e}^{-9.6}9.6^8}{8!} = 0.12118.

Adding these gives $0.22216$ , so the required probability is $0.222$ to 3 significant figures.

Key Takeaways

When the time period changes, multiply the rate by the number of periods to get the new Poisson mean.
For a Poisson variable, probabilities of ranges of integer values are found by summing individual point probabilities.

Common Mistakes

Using $\lambda = 2.4$ instead of $\lambda = 9.6$ for the 4-week period.
Including $X=6$ or $X=9$ in the sum; 'more than 6 and less than 9' means only $X=7$ and $X=8$ .
Forgetting to multiply by $\mathrm{e}^{-9.6}$ .
Using the wrong factorial in the Poisson formula.

Things to Be Careful About

The final answer must be given to 3 significant figures. The mark scheme allows one end error, but the correct method uses exactly the two probabilities for 7 and 8. An unsupported answer of 0.222 would not receive full marks because the method must be shown.

Techniques used

scale the Poisson mean to the time periodapply the Poisson probability formulasum probabilities for valid integer values

(c)

Use a suitable approximating distribution to calculate the probability that more than 50 computers are donated during a 20-week period.

4M Similar

DifficultyMedium

Worked solution

Approach

The number donated in 20 weeks is Poisson with $\lambda = 2.4 \times 20 = 48$ . Since $\lambda$ is large, approximate by a normal distribution with mean and variance both $48$ : $X \approx \mathrm{N}(48, 48)$ . Apply a continuity correction because $X$ is discrete: 'more than 50' becomes $X > 50.5$ . Then standardise and find the upper-tail probability.

Working

\lambda = 2.4 \times 20 = 48

X \approx \mathrm{N}(48, 48)

Apply the continuity correction:

P(X > 50) \approx P(X > 50.5)

Standardise:

z = \frac{50.5 - 48}{\sqrt{48}} = \frac{2.5}{6.928} = 0.361

P(X > 50) \approx 1 - \Phi(0.361) = 1 - 0.6409 = 0.3591

Answer

0.359 \ (3 \text{ sf})

Final answer

0.359 (3 sf)

Detailed explanation

Walkthrough

Over a 20-week period, the mean number of computers donated is

\lambda = 2.4 \times 20 = 48.

Because $\lambda = 48$ is large, the Poisson distribution can be approximated by a normal distribution with the same mean and variance:

X \approx \mathrm{N}(48, 48).

Since $X$ is discrete, a continuity correction is needed. The event 'more than 50' means $X \geq 51$ , so we use the boundary $50.5$ :

P(X > 50) \approx P(X > 50.5).

Standardise using the normal mean $48$ and standard deviation $\sqrt{48}$ :

z = \frac{50.5 - 48}{\sqrt{48}} = \frac{2.5}{6.928} = 0.361.

The required probability is the upper tail:

P(X > 50) \approx 1 - \Phi(0.361) = 1 - 0.6409 = 0.3591.

So the probability is $0.359$ to 3 significant figures.

Key Takeaways

A Poisson distribution with a large mean can be approximated by a normal distribution with mean $\lambda$ and variance $\lambda$ .
Always apply a continuity correction when approximating a discrete distribution by a continuous one.
For 'more than $k$ ', use the boundary $k + 0.5$ ; for 'at least $k$ ', use $k - 0.5$ .

Common Mistakes

Forgetting the continuity correction.
Using $50$ instead of $50.5$ in the standardisation.
Using $\sqrt{48}$ incorrectly as the variance rather than the standard deviation.
Finding $\Phi(0.361)$ instead of the upper-tail probability $1 - \Phi(0.361)$ .

Things to Be Careful About

The normal approximation is appropriate here because $\lambda = 48$ is large. The mark scheme gives the method mark for standardising with the chosen normal distribution, even if the continuity correction is missing, but the final answer must be correct to 3 significant figures.

Techniques used

scale the Poisson mean to the time periodchoose the normal approximationapply continuity correctionstandardise and find upper-tail probability

Q2Medium-EasyThe Poisson DistributionLinear Combinations of Random Variables

The random variable $X$ has a normal distribution with mean 10 and standard deviation 3. The independent random variable $Y$ has a Poisson distribution with mean 4.

(a)

Find the standard deviation of $X + Y$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $X$ and $Y$ are independent, the variance of their sum is the sum of their variances. We first find $\operatorname{Var}(X)$ from the given standard deviation, and $\operatorname{Var}(Y)$ from the Poisson mean, then add and take the square root.

Working

\operatorname{Var}(X) = 3^2 = 9

For a Poisson distribution, the variance equals the mean, so

\operatorname{Var}(Y) = 4

Since $X$ and $Y$ are independent,

\operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) = 9 + 4 = 13

Therefore the standard deviation is

\sqrt{13} = 3.6055\ldots = 3.61 \text{ (3 sf)}

Answer

\sqrt{13} \approx 3.61

Final answer

sqrt(13) = 3.61

Detailed explanation

Walkthrough

We are told $X$ is normal with mean 10 and standard deviation 3, and $Y$ is Poisson with mean 4. The question asks for the standard deviation of $X+Y$ .

First, convert the standard deviation of $X$ into a variance: $\operatorname{Var}(X)=3^2=9$ . Variances are needed because they add for independent random variables, whereas standard deviations do not.

Next, use the Poisson property that the variance equals the mean. Since $Y$ has mean 4, $\operatorname{Var}(Y)=4$ .

Because $X$ and $Y$ are independent, $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)=9+4=13$ .

Finally, take the square root to return to standard deviation units: $\sqrt{13}\approx 3.61$ .

Key Takeaways

For independent random variables, variances add: $\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)$ .
For a Poisson distribution, the variance is equal to the mean.
Always work in variances when combining independent random variables, then convert back to standard deviation at the end.

Common Mistakes

Adding standard deviations directly instead of variances.
Forgetting that a Poisson distribution has variance equal to its mean.
Stopping at variance 13 instead of giving the standard deviation $\sqrt{13}$ .

Things to Be Careful About

The independence of $X$ and $Y$ is essential for the variance addition rule.
The mean of $X$ (10) is not needed for this calculation; only the standard deviation matters.
Give the final answer to 3 significant figures as requested, or leave it as $\sqrt{13}$ .

Techniques used

use variance of a Poisson distributionadd variances of independent random variablestake the square root of a variance

(b)

Find the standard deviation of $5X - Y$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For independent variables, the variance of a linear combination is obtained by adding the variances of each scaled term. The coefficient 5 multiplies the standard deviation, so it squares when applied to the variance; the subtraction does not change the fact that variances are added.

Working

\operatorname{Var}(5X) = 5^2 \operatorname{Var}(X) = 25 \times 3^2 = 25 \times 9 = 225

From part (a), $\operatorname{Var}(Y) = 4$ . Since $X$ and $Y$ are independent,

\operatorname{Var}(5X - Y) = \operatorname{Var}(5X) + \operatorname{Var}(Y) = 225 + 4 = 229

Therefore the standard deviation is

\sqrt{229} = 15.1327\ldots = 15.1 \text{ (3 sf)}

Answer

\sqrt{229} \approx 15.1

Final answer

sqrt(229) = 15.1

Detailed explanation

Walkthrough

We need the standard deviation of $5X-Y$ . Start by finding the variance of the scaled variable $5X$ . Scaling a random variable by a constant multiplies its variance by the square of the constant: $\operatorname{Var}(5X)=5^2\operatorname{Var}(X)=25\times 9=225$ .

From part (a), $\operatorname{Var}(Y)=4$ .

For independent variables, the variance of a difference is still the sum of the variances: $\operatorname{Var}(5X-Y)=\operatorname{Var}(5X)+\operatorname{Var}(Y)=225+4=229$ .

Take the square root to get the standard deviation: $\sqrt{229}\approx 15.1$ .

Key Takeaways

$\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$ .
For independent $X$ and $Y$ , $\operatorname{Var}(aX \pm bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$ .
The minus sign in $5X-Y$ does not make the variance negative; variances are always added.

Common Mistakes

Writing $\operatorname{Var}(5X-Y)=\operatorname{Var}(5X)-\operatorname{Var}(Y)$ .
Forgetting to square the 5 when scaling the variance.
Using the standard deviation 3 directly in $\operatorname{Var}(5X)$ instead of $3^2$ .
Forgetting to take the square root at the end.

Things to Be Careful About

The coefficient 5 applies to $X$ only, not to $Y$ .
The variance of $Y$ from part (a) is carried forward; it is 4.
The final answer should be $\sqrt{229}$ or 15.1 to 3 significant figures.

Techniques used

square the coefficient when scaling a varianceadd variances of independent random variablestake the square root of a variance

Q3MediumSampling and Estimation

The times, in minutes, taken by students to complete a test have mean $\mu$ and standard deviation $\sigma$ . The times taken by a random sample of 100 students are noted and are used to calculate a 95% confidence interval for $\mu$ .

(a)

Given that the end points of the 95% confidence interval are 31.02 and 33.98, correct to 4 significant figures, calculate the value of $\sigma$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a 95% confidence interval for the population mean, the sample mean is the midpoint of the endpoints and the margin of error is half the width. Use

\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}

with $n=100$ . Equate the upper endpoint to $\bar{x} + 1.96\sigma/10$ and solve for $\sigma$ .

Working

The sample mean is

\bar{x} = \frac{31.02+33.98}{2} = 32.5.

The width of the interval is

33.98 - 31.02 = 2.96,

so the margin of error is $2.96/2 = 1.48$ .

For a 95% confidence interval,

\bar{x} + 1.96\frac{\sigma}{\sqrt{100}} = 33.98.

Substitute $\bar{x}=32.5$ :

32.5 + 1.96\frac{\sigma}{10} = 33.98.

Thus

0.196\sigma = 1.48,

so

\sigma = \frac{1.48}{0.196} = 7.551\ldots

Answer

\sigma = 7.55 \quad (3\text{ s.f.})

Final answer

σ = 7.55 (3 s.f.)

Detailed explanation

Walkthrough

A 95% confidence interval for the population mean is centred at the sample mean. Because the interval is symmetric, the sample mean is exactly the midpoint of the two endpoints. The distance from the sample mean to either endpoint is the margin of error, which is half the total width.

For a 95% confidence interval the critical value is $z=1.96$ . With $n=100$ , $\sqrt{n}=10$ , so the margin is $1.96\sigma/10 = 0.196\sigma$ . Setting the upper endpoint equal to the sample mean plus this margin gives $32.5 + 0.196\sigma = 33.98$ , so $0.196\sigma = 1.48$ . Dividing gives $\sigma = 1.48/0.196 = 7.55$ to 3 significant figures.

Key Takeaways

The midpoint of a confidence interval is the sample mean.
Half the width is the margin of error.
The 95% confidence interval formula is $\bar{x} \pm 1.96\sigma/\sqrt{n}$ .
Rearranging the interval formula lets us recover the population standard deviation.

Common Mistakes

Using the full width $2.96$ as the margin of error instead of half of it. This gives $\sigma = 15.1$ and loses the final accuracy mark.
Using $z=1.645$ instead of $1.96$ ; this may earn method credit but not the correct final answer.
Rounding too early; keep $7.551\ldots$ before giving $7.55$ .

Things to Be Careful About

The mark scheme allows either the sample mean or the width to be seen as the first step. It then requires a correct equation relating the endpoints to $\sigma$ ; a factor-of-2 error can still receive method credit, but the final answer must be $7.55$ . The final answer must be given to 3 significant figures.

Techniques used

find the sample mean as the midpoint of the intervaluse the 95% confidence interval formulasolve for the population standard deviation

(b)

The calculation of the confidence interval required the use of the Central Limit theorem.

Explain why it is valid to use the Central Limit theorem in this case.

1M Similar

DifficultyEasy

Worked solution

Approach

The Central Limit Theorem states that for a sufficiently large sample, the sample mean is approximately normally distributed, regardless of the population distribution. Here the sample size is 100, which is large.

Working

The sample consists of 100 students, so $n=100$ . Since $n$ is large, the Central Limit Theorem justifies treating

\bar{X} \sim N\left(\mu, \frac{\sigma^2}{100}\right)

as an approximate normal distribution, even though the distribution of individual completion times is unknown.

Answer

It is valid because the sample size is large ( $n=100$ ).

Final answer

It is valid because the sample size is large (n = 100).

Detailed explanation

Walkthrough

The Central Limit Theorem says that when the sample size is large, the distribution of the sample mean is approximately normal, whatever the distribution of the individual observations. Here the sample size is 100, which is comfortably large, so the sample mean can be treated as approximately normal. This is why a normal-based confidence interval can be used even though we are not told that the completion times themselves are normally distributed.

Key Takeaways

The Central Limit Theorem applies to the sample mean, not to the individual data values.
A large sample size, such as $n=100$ , is enough to justify using the normal distribution for the sample mean.
No assumption about the population distribution is needed.

Common Mistakes

Saying only that 'the number of students is large' without referring to the sample. The mark scheme requires a reference to the sample size or to $n=100$ .
Claiming the population times are normally distributed; this is not given and is not needed because of the Central Limit Theorem.

Things to Be Careful About

Use the phrase 'the sample size is large' or ' $n=100$ is large'. The mark scheme accepts $n>30$ (and condones $n>50$ ), but a statement about the population being large is not enough.

Techniques used

state that the sample size is largeapply the Central Limit Theorem to the sample mean

(c)

A researcher calculates a number, $r$ , of 95% confidence interval for $\mu$ .

Find the largest value of $r$ such that the probability that all $r$ confidence intervals contain the true value of $\mu$ is greater than 0.5.

4M Similar

DifficultyMedium

Worked solution

Approach

Each 95% confidence interval has probability 0.95 of containing $\mu$ . Assuming the intervals are independent, the probability that all $r$ intervals contain $\mu$ is $0.95^r$ . We require $0.95^r > 0.5$ , take logs and solve for $r$ , then choose the largest integer.

Working

P(\text{all } r \text{ intervals contain } \mu) = 0.95^r.

We need

0.95^r > 0.5.

Take natural logarithms:

r\ln 0.95 > \ln 0.5.

Since $\ln 0.95 < 0$ , dividing reverses the inequality:

r < \frac{\ln 0.5}{\ln 0.95} = \frac{-0.693147\ldots}{-0.051293\ldots} = 13.513\ldots

Therefore the largest integer $r$ is 13.

Check:

0.95^{13} = 0.5133\ldots > 0.5,

0.95^{14} = 0.4877\ldots < 0.5.

Answer

r = 13

Final answer

r = 13

Detailed explanation

Walkthrough

Each 95% confidence interval has probability $0.95$ of containing the true mean $\mu$ . If the intervals are independent, the probability that all $r$ intervals contain $\mu$ is the product of the individual probabilities, $0.95^r$ .

We need this probability to be greater than $0.5$ , so we solve $0.95^r > 0.5$ . Taking natural logarithms gives $r\ln 0.95 > \ln 0.5$ . Since $\ln 0.95$ is negative, dividing by it reverses the inequality, giving $r < \ln 0.5/\ln 0.95 \approx 13.513$ . Because $r$ must be an integer and must be less than $13.513$ , the largest possible value is $13$ . Checking, $0.95^{13} \approx 0.513 > 0.5$ and $0.95^{14} \approx 0.488 < 0.5$ .

Key Takeaways

The probability that all independent events occur is the product of their probabilities.
When solving an inequality involving a logarithm of a number less than 1, the inequality sign reverses on division.
After finding a decimal bound, choose the appropriate integer (less than the bound, not the nearest integer).

Common Mistakes

Forgetting to reverse the inequality when dividing by $\ln 0.95$ , which is negative.
Rounding $13.513$ to $14$ instead of choosing the largest integer less than it, which is $13$ .
Giving $13.5$ or $13.51$ as the final answer instead of the integer $13$ .
Not stating that the intervals are independent when forming $0.95^r$ .

Things to Be Careful About

The mark scheme allows the use of ' $=$ ' in the inequality and condones incorrect inequality signs throughout, but the final largest value must be $13$ . If you evaluate powers instead of using logs, you should still check both $0.95^{13}$ and $0.95^{14}$ to justify the answer.

Techniques used

model each interval as an independent event with probability 0.95set up the probability inequality 0.95^r > 0.5solve the logarithmic inequalitychoose the largest integer satisfying the inequality

Q4MediumHypothesis TestsThe Poisson Distribution

An inspector believes that 18% of cups made at a certain factory contain flaws. The factory owner claims that the true percentage is less than 18%. The inspector examines a random sample of 40 cups and finds that 3 of them contain flaws.

(a)

Stating a necessary assumption, use a binomial distribution to test the factory owner's claim at the 5% significance level.

6M Similar

DifficultyMedium

Worked solution

Approach

Let $X$ be the number of flawed cups in a sample of 40. Assume that flaws occur independently and that the probability of a cup containing a flaw is the same for every cup. Then $X \sim \mathrm{B}(40,0.18)$ . Since the owner claims the true percentage is less than 18%, use a one-tailed lower-tail test at the 5% significance level.

Working

State the hypotheses:

\mathrm{H}_0: p = 0.18, \qquad \mathrm{H}_1: p < 0.18

where $p$ is the probability that a cup contains a flaw.

Under $\mathrm{H}_0$ , $X \sim \mathrm{B}(40,0.18)$ . The observed result is $X = 3$ , so the $p$ -value is:

\mathrm{P}(X \le 3) = 0.82^{40} + 40(0.82)^{39}(0.18) + {}^{40}\mathrm{C}_2(0.82)^{38}(0.18)^2 + {}^{40}\mathrm{C}_3(0.82)^{37}(0.18)^3

Evaluating:

\mathrm{P}(X \le 3) = 0.0003569 + 0.0031338 + 0.013414 + 0.037298 = 0.0542 \quad (3\text{ sf})

Compare with the significance level:

0.0542 > 0.05

so the result is not significant. There is insufficient evidence to reject $\mathrm{H}_0$ .

Answer

There is insufficient evidence to support the factory owner's claim that the true percentage of flawed cups is less than 18%.

Final answer

There is insufficient evidence to support the factory owner's claim that the true percentage of flawed cups is less than 18%.

Detailed explanation

Walkthrough

Start by identifying the random variable: $X$ , the number of flawed cups in 40. For a binomial model we need independent trials and the same probability on each trial; this is the necessary assumption. Set up hypotheses: null $p=0.18$ , alternative $p<0.18$ because the owner claims less. The test is lower-tailed. Under the null, $X\sim \mathrm{B}(40,0.18)$ . Since 3 flaws were observed, compute the probability of 3 or fewer, i.e. the lower tail. This is the $p$ -value. Add the four binomial probabilities. The total is 0.0542. Because 0.0542 > 0.05, the observed result could reasonably occur by chance under the null, so we do not reject $\mathrm{H}_0$ . Conclude in context: insufficient evidence that the true percentage is below 18%.

Key Takeaways

This question tests one-tailed binomial hypothesis testing: stating hypotheses, computing a tail probability, comparing with the significance level, and writing a conclusion in context. The assumption of independence or equal probability is part of the binomial model.

Common Mistakes

Forgetting to state the independence or equal-probability assumption.
Using a two-tailed test; this would compare with 0.025 and lose marks.
Writing a conclusion that is too definite, such as 'the percentage is 18%' or 'not less than 18%'.
Quoting 0.0542 without showing the binomial probability calculation; the method mark requires the sum of the four terms.

Things to Be Careful About

The comparison must be a tail probability, not just any probability. If using a critical region, it is $X\le 2$ since $\mathrm{P}(X\le2)=0.0169$ and $\mathrm{P}(X\le3)=0.0542$ ; then compare the observed 3 with the critical value. Use $p<0.18$ for the one-tailed alternative, and make sure the final conclusion is in context and not definitive.

Techniques used

state the necessary independence assumption for a binomial modelset up null and alternative hypothesescompute a lower-tail binomial probabilitycompare the p-value with the significance leveldraw a conclusion in context

(b)

Explain why it would not be appropriate to use the Poisson approximation to the binomial distribution to carry out the test in part (a).

1M Similar

DifficultyEasy

Worked solution

Approach

For the Poisson approximation to the binomial distribution to be valid, we require $n$ large and $p$ small, typically $n \ge 50$ and $p \le 0.1$ , or equivalently $np \le 5$ . Check these conditions in context.

Working

Here $n = 40$ and $p = 0.18$ , so

np = 40 \times 0.18 = 7.2

Since $np = 7.2 > 5$ , the expected number of flawed cups is too large for the Poisson approximation to be appropriate. Alternatively, $n = 40$ is not greater than 50, and $p = 0.18$ is not small.

Answer

The Poisson approximation is not appropriate because $np = 7.2 > 5$ (or because $n = 40$ is not large enough, or $p = 0.18$ is not small enough).

Final answer

np = 7.2 > 5, so the Poisson approximation to the binomial is not appropriate.

Detailed explanation

Walkthrough

The Poisson approximation to the binomial is suitable when the number of trials $n$ is large and the probability $p$ is small. Here $n=40$ and $p=0.18$ , so $np=40\times0.18=7.2$ . Since $np>5$ , the expected count is too large for the Poisson approximation to be reliable. Equivalently, $n=40$ is not greater than 50, and $p=0.18$ is not small. One correct contextual condition is enough.

Key Takeaways

The conditions for using Poisson as an approximation to binomial are usually $n\ge50$ and $p\le0.1$ , or $np\le5$ . The answer must refer to the actual context, e.g. $np=7.2$ .

Common Mistakes

Giving a condition without context, such as just 'np is too large'.
Confusing the condition: saying $n$ is large when $n=40$ is not large enough.
Using $np=7.2$ but not explaining why this makes the approximation inappropriate.

Things to Be Careful About

Use the values from the question: $n=40$ , $p=0.18$ , $np=7.2$ . Any one of the equivalent reasons is acceptable, but it must be stated in context.

Techniques used

check the conditions for Poisson approximation to the binomialrelate np to the approximation condition

Q5MediumContinuous Random Variables

A random variable $X$ has probability density function given by

\mathrm{f}(x) = \begin{cases} k(2x^2 - x^3) & 0 \leqslant x \leqslant 2, \\ 0 & \text{otherwise.} \end{cases}

(a)

Show that $k = \frac{3}{4}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

A probability density function must have total area $1$ over the whole sample space. Integrate $\mathrm{f}(x)$ from $0$ to $2$ , set it equal to $1$ , and solve for $k$ .

Working

Since $\mathrm{f}(x)=0$ outside $0\le x\le 2$ , the normalisation condition is

\int_0^2 k(2x^2-x^3)\mathrm{d}x = 1.

As $k$ is a constant,

k\int_0^2 (2x^2-x^3)\mathrm{d}x = 1.

Integrating term by term:

k\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^2 = 1.

At $x=2$ :

\frac{2(2)^3}{3}-\frac{(2)^4}{4} = \frac{16}{3}-4 = \frac{4}{3}.

The lower limit $x=0$ gives $0$ , so

k\cdot\frac{4}{3}=1.

Therefore

k=\frac{3}{4}.

Final answer

k = 3/4

Detailed explanation

Walkthrough

A continuous probability density function has total area under the curve equal to $1$ . Since the PDF is non-zero only on $[0,2]$ , we integrate $k(2x^2-x^3)$ from $0$ to $2$ and set the result equal to $1$ . Factor the constant $k$ outside the integral, then integrate each term using the power rule. Substituting $x=2$ gives $\frac{16}{3}-4=\frac{4}{3}$ , and substituting $x=0$ gives $0$ . Thus $k\cdot\frac{4}{3}=1$ , so $k=\frac{3}{4}$ . This is a "show that" question, so the intermediate simplification must be shown; writing only the final value is not enough.

Key Takeaways

This question tests the defining property of a PDF: the total probability over the whole sample space is $1$ . It also practises integrating a polynomial and using the result to find an unknown normalising constant.

Common Mistakes

Forgetting to set the integral of $\mathrm{f}(x)$ equal to $1$ .
Omitting the lower and upper limits when evaluating the definite integral.
Making the final simplification $k=\frac{3}{4}$ without showing any working, which loses the "AG" mark in a prove/show question.
Arithmetic error in $\frac{16}{3}-4=\frac{4}{3}$ .

Things to Be Careful About

The lower limit $x=0$ contributes $0$ , but it must still be included in the evaluation.
Since the question says "Show that", the answer $k=\frac{3}{4}$ must be derived with at least one clear step and no errors.
The constant $k$ is positive, so it can be moved outside the integral without changing the sign.

Techniques used

integrate the probability density function over its supportset the total probability equal to 1solve for the normalising constant

(b)

The median of $X$ is denoted by $m$ .

6M

(i)

Write down the value of $\mathrm{P}(X \leqslant m)$ .

1M Similar

(ii)

Hence find $\mathrm{P}(\mathrm{E}(X) \leqslant X \leqslant m)$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the identity

\mathrm{P}(\mathrm{E}(X)\le X \le m) = \mathrm{P}(X\le m) - \mathrm{P}(X\le \mathrm{E}(X)).

From part (b)(i), $\mathrm{P}(X\le m)=0.5$ . First calculate $\mathrm{E}(X)$ using $\int x\mathrm{f}(x)\mathrm{d}x$ , then calculate $\mathrm{P}(X\le \mathrm{E}(X))$ by integrating the PDF up to $\mathrm{E}(X)$ .

Working

Using $\mathrm{f}(x)=\frac{3}{4}(2x^2-x^3)$ on $0\le x\le2$ ,

\mathrm{E}(X)=\int_0^2 x\mathrm{f}(x)\mathrm{d}x = \frac{3}{4}\int_0^2 (2x^3-x^4)\mathrm{d}x.

Integrating:

\mathrm{E}(X)=\frac{3}{4}\left[\frac{2x^4}{4}-\frac{x^5}{5}\right]_0^2.

At $x=2$ :

\frac{2(2)^4}{4}-\frac{(2)^5}{5} = \frac{32}{4}-\frac{32}{5} = 8-\frac{32}{5} = \frac{8}{5}.

So

\mathrm{E}(X)=\frac{3}{4}\cdot\frac{8}{5}=\frac{6}{5}=1.2.

Now calculate $\mathrm{P}(X\le \mathrm{E}(X))$ . Since $\mathrm{E}(X)=\frac{6}{5}$ ,

\mathrm{P}\left(X\le\frac{6}{5}\right)=\frac{3}{4}\int_0^{6/5}(2x^2-x^3)\mathrm{d}x.

Integrating:

\mathrm{P}\left(X\le\frac{6}{5}\right)=\frac{3}{4}\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^{6/5}.

At $x=\frac{6}{5}$ :

\frac{2\left(\frac{6}{5}\right)^3}{3}-\frac{\left(\frac{6}{5}\right)^4}{4} = \frac{144}{125}-\frac{324}{625}= \frac{396}{625}.

Therefore

\mathrm{P}\left(X\le\frac{6}{5}\right)=\frac{3}{4}\cdot\frac{396}{625}=\frac{297}{625}=0.4752.

Since $m>\mathrm{E}(X)$ here,

\mathrm{P}(\mathrm{E}(X)\le X\le m) = 0.5 - 0.4752 = 0.0248 = \frac{31}{1250}.

Final answer

0.0248 or 31/1250

Detailed explanation

Walkthrough

We want the probability that $X$ lies between the mean and the median. For a continuous random variable this equals the probability up to the median minus the probability up to the mean. Part (b)(i) gives the first of these as $0.5$ . To get the second, compute $\mathrm{E}(X)$ by integrating $x\mathrm{f}(x)$ from $0$ to $2$ . This gives $\frac{6}{5}=1.2$ . Then integrate $\mathrm{f}(x)$ from $0$ to $\frac{6}{5}$ to find $\mathrm{P}(X\le1.2)=0.4752$ . Finally subtract: $0.5-0.4752=0.0248$ . The median $m$ itself is never needed explicitly.

Key Takeaways

This question combines three important ideas: the mean formula $\mathrm{E}(X)=\int x\mathrm{f}(x)\mathrm{d}x$ , the cumulative probability integral $\mathrm{P}(X\le a)=\int_0^a\mathrm{f}(x)\mathrm{d}x$ , and the median condition $\mathrm{P}(X\le m)=0.5$ . It also shows how interval probabilities can be written as differences of cumulative probabilities.

Common Mistakes

Forgetting the factor $\frac{3}{4}$ when computing $\mathrm{E}(X)$ or $\mathrm{P}(X\le\mathrm{E}(X))$ .
Using $\mathrm{f}(x)$ instead of $x\mathrm{f}(x)$ when computing the mean.
Subtracting in the wrong order, e.g. $0.4752-0.5$ , which gives a negative probability.
Trying to solve for $m$ by numerical integration when the difference-of-cumulative-probabilities method is much simpler; this also risks rounding errors.
Arithmetic slips with fractions such as $\frac{396}{625}\times\frac{3}{4}$ .

Things to Be Careful About

For a continuous random variable, strict and non-strict inequalities give the same probability, so $\mathrm{P}(X\le a)=\mathrm{P}(X<a)$ .
Here $m>\mathrm{E}(X)$ , so the interval probability is $0.5-\mathrm{P}(X\le\mathrm{E}(X))$ ; check the order before subtracting.
The final answer can be given as a decimal $0.0248$ or an exact fraction $\frac{31}{1250}$ ; the mark is for the correct final value.
"CWO" in the mark scheme means correct working only, so unsupported or incorrect intermediate values may lose the final mark even if the answer is plausible.

Techniques used

calculate the mean by integrating x times the PDFevaluate a cumulative probability by integrating the PDFsubtract cumulative probabilities to obtain an interval probability

Q6Medium-HardHypothesis Tests

The weekly profit, in dollars, made by a certain firm has a normal distribution. In the past, the weekly profit had the distribution $\mathrm{N}(736, 26^2)$ . Following a change in management, the mean weekly profit for 35 randomly chosen weeks is $725.

(a)

Stating a necessary assumption, test at the 2% significance level whether the mean weekly profit has decreased.

6M Similar

DifficultyMedium

Worked solution

Approach

This is a one-tailed hypothesis test for the mean of a normal distribution with a known population standard deviation. We first state the necessary assumption, set up the null and alternative hypotheses, calculate the test statistic, and compare it with the lower 2% critical value (or compare the $p$ -value with 0.02).

Working

A necessary assumption is that the population standard deviation is still $\sigma = 26$ ; the change in management has not changed the variability.

Let $\mu$ be the population mean weekly profit in dollars.

\mathrm{H}_0: \mu = 736

\mathrm{H}_1: \mu < 736

The sample size is $n = 35$ , the sample mean is $\bar{x} = 725$ , and $\sigma = 26$ . The test statistic is

z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{725 - 736}{26/\sqrt{35}}

z = \frac{-11}{26/\sqrt{35}} = -2.503

The lower 2% critical value is $z = -2.054$ . Since $-2.503 < -2.054$ , or equivalently since $p = 0.0062 < 0.02$ , the result is significant at the 2% level.

Reject $\mathrm{H}_0$ .

Answer

There is sufficient evidence that the mean weekly profit has decreased.

Final answer

Reject H0; sufficient evidence that the mean weekly profit has decreased

Detailed explanation

Walkthrough

This question is a one-tailed hypothesis test about a normal population mean. The past distribution is $\mathrm{N}(736,26^2)$ , so under the null hypothesis the population mean is 736 and the standard deviation is 26. The management changed and we want to know whether the mean weekly profit has decreased, so the alternative hypothesis is one-sided: $\mu < 736$ . The necessary assumption is that the standard deviation is still 26; without this assumption we could not standardise the sample mean.

The sample mean is $\bar{x} = 725$ with $n = 35$ . Since the population is normal and $\sigma$ is known, we use

z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}.

Substituting gives $z = -2.503$ . At the 2% significance level for a lower-tailed test, the critical value is $z = -2.054$ . Our test statistic is less than this critical value, so it lies in the rejection region. Equivalently, the $p$ -value is $0.0062$ , which is less than $0.02$ . We therefore reject the null hypothesis. In context, there is sufficient evidence that the mean weekly profit has decreased, but the test does not prove the decrease definitely.

Key Takeaways

A hypothesis test for a normal population mean uses the standardised sample mean. We must state assumptions such as the standard deviation remaining unchanged, define hypotheses, choose the correct one-tailed direction, and compare either a test statistic with a critical value or a $p$ -value with the significance level. The conclusion must be written in the context of the problem.

Common Mistakes

Omitting the assumption that the standard deviation is still 26.
Using a two-tailed alternative; this changes the critical values and is not the correct test for 'decreased'.
Writing the alternative hypothesis as $\mu > 736$ instead of $\mu < 736$ .
Stating the conclusion without context, e.g. just 'reject $\mathrm{H}_0$ ' without saying what has decreased.
Saying the mean weekly profit 'has decreased' as a definite fact rather than saying there is sufficient evidence.

Things to Be Careful About

The critical value for a 2% lower-tailed test is $-2.054$ ; the two-tailed critical values would be $\pm 2.326$ , which would not test the stated one-sided claim.
The sign of the test statistic matters: 725 is below 736, so $z$ is negative.
The mark scheme accepts comparing $-2.503$ with $-2.054$ or comparing $0.0062$ with $0.02$ ; both are valid.
Do not use the sample standard deviation; the population standard deviation is 26.

Techniques used

state the required distributional assumptionformulate null and alternative hypothesesstandardise the sample meancompare test statistic with critical value or p-value with significance levelwrite a conclusion in context

(b)

The mean weekly profit for another random sample of 35 weeks is found and a similar test is carried out at the 2% significance level.

State the probability of a Type I error.

1M Similar

(c)

Given that the mean weekly profit is now in fact $718, find the probability of a Type II error.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

To find the probability of a Type II error, we need the probability of failing to reject $\mathrm{H}_0$ when the true mean is $718. First we find the critical value of the sample mean at the 2% significance level using the null mean 736. Then we standardise that critical value using the true mean 718 and find the corresponding upper-tail normal probability.

Working

Under $\mathrm{H}_0$ , the critical value $a$ of $\bar{X}$ is found from

\frac{a - 736}{26/\sqrt{35}} = -2.054

so

a = 736 - 2.054 \times \frac{26}{\sqrt{35}} = 726.973

A Type II error occurs if we do not reject $\mathrm{H}_0$ , i.e. if the observed sample mean is greater than $a$ , when in fact the true mean is 718.

Standardise $a$ using $\mu = 718$ :

z = \frac{726.973 - 718}{26/\sqrt{35}} = 2.042

Therefore

P(\text{Type II error}) = P\left(\bar{X} > 726.973 \mid \mu = 718\right) = 1 - \Phi(2.042)

= 0.0205

Answer

0.0205 (accepted range 0.0203 to 0.0207)

Final answer

0.0205 (accepted range 0.0203 to 0.0207)

Detailed explanation

Walkthrough

A Type II error is the probability of failing to reject the null hypothesis when the alternative is true. Here, if the true mean has become 718, we want the chance that the test still does not reject $\mathrm{H}_0$ .

First locate the boundary of the acceptance region in terms of the sample mean. For a lower-tailed 2% test under $\mu = 736$ , the critical sample mean $a$ satisfies $P(\bar{X} \le a) = 0.02$ . Standardising gives $(a - 736)/(26/\sqrt{35}) = -2.054$ , so $a = 726.973$ . This means: if the sample mean is below about 726.973, we reject $\mathrm{H}_0$ ; if it is above that value, we do not reject.

Now suppose the true mean is 718. The probability of a Type II error is the probability of not rejecting, i.e. $\bar{X} > 726.973$ . Standardising 726.973 with mean 718 gives $z = 2.042$ , so the probability is $1 - \Phi(2.042) = 0.0205$ .

Key Takeaways

A Type II error is not the significance level; it depends on the true parameter value. To compute it, find the critical value of the sample mean under the null hypothesis and then standardise it under the assumed true mean. The tail direction is the opposite of the rejection region.

Common Mistakes

Using 725 or 736 instead of 718 when calculating the Type II error probability. The true mean is now 718.
Forgetting to subtract from 1; we need the probability above the critical value, not below.
Using 0.02 as the Type II probability.
Omitting the critical-value calculation and directly standardising 736 or 725 against 718; this does not give the correct boundary.

Things to Be Careful About

Accept a critical value of 726.973 or 727.
The accepted final probability range is 0.0203 to 0.0207 depending on rounding of $a$ and $z$ .
If part (a) had been a two-tailed test, the critical value would be $-2.326$ ; but here we are using the correct one-tailed lower critical value $-2.054$ .
The probability of a Type II error is the chance of the sample mean lying in the acceptance region, which is above the critical value.

Techniques used

find the critical value of the sample meanstandardise the critical value under the true meancompute the upper-tail normal probabilityinterpret the result as a Type II error

Mathematics 9709/62 — October/November 2025

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