The random variable $X$ has the distribution $\text{Po}(15)$. (a) Write down an expression in terms of $e$ for $\text{P}(X = 12)$. (b) It is given that $\text{P}(X = n) = \text{P}(X = n + 1)$. Write down an equation in $n$, and hence find the value of $n$.

(a) P(X = 12) = 15^12 e^-15 / 12! · (b) n = 14

9709/62 May/June 2025 — Mathematics Solved Past Paper

Q1MediumSampling and EstimationLinear Combinations of Random Variables

(a)

One of a group of three students is to be chosen at random.

Explain how a single throw of a fair six-sided dice could be used to make the choice.

1M Similar

DifficultyEasy

Worked solution

Approach

Use the six equally likely outcomes of a fair dice and assign exactly two of them to each of the three students, so that each student has the same probability of being chosen.

Working

For example, let the three students be A, B and C. On a single throw of the dice:

\begin{aligned} 1, 2 &\rightarrow \text{choose A} \\ 3, 4 &\rightarrow \text{choose B} \\ 5, 6 &\rightarrow \text{choose C} \end{aligned}

Each student is chosen with probability

\frac{2}{6} = \frac{1}{3}

so the choice is random and fair. All six numbers on the dice are used, so the rule is unambiguous.

Answer

Assign two different dice numbers to each student, for example 1–2 for student 1, 3–4 for student 2 and 5–6 for student 3, and choose the student whose pair is thrown.

Final answer

Assign two dice numbers to each student, e.g. 1-2, 3-4, 5-6, and choose the student whose pair is thrown.

Detailed explanation

Walkthrough

The dice has six equally likely outcomes. To choose fairly between three students, each student must have the same chance. Give each student two of the six outcomes. Then each has probability $2/6 = 1/3$ of being chosen. It is important that the mapping covers all six outcomes so that every throw selects exactly one student and the rule is unambiguous.

Key Takeaways

A fair random choice can be made from a fair dice by partitioning the outcomes into equal-sized groups. The probability of each choice is the number of favourable outcomes divided by the total number of outcomes.

Common Mistakes

Giving one student fewer outcomes than another, which makes the choice unfair.
Not using all six numbers, so some throws would not correspond to a student.
Using more than one throw when the question explicitly requires a single throw.

Things to Be Careful About

The rule must be unambiguous: every possible outcome of the single throw must select exactly one of the three students. The mark scheme requires that two different numbers are assigned to each person and that these are different from the other people's numbers, so all six numbers on the dice are used.

Techniques used

partition six equally likely outcomes into three equal groupsensure each student has equal probabilityverify all outcomes are used

(b)

The times, in minutes, taken by students to complete a test are normally distributed with mean 125 and variance 50. Two students are chosen at random.

Find the probability that the difference between the times taken by these two students to complete the test is more than 12 minutes.

5M Similar

DifficultyMedium

Worked solution

Approach

Let $X_1$ and $X_2$ be the times taken by the two students. Since they are independent normal variables, their difference $D = X_1 - X_2$ is also normal. Find the mean and variance of $D$ , standardise to find $P(D > 12)$ , then double it to include the case where the second student is more than 12 minutes slower.

Working

Given:

X_1 \sim N(125, 50), \quad X_2 \sim N(125, 50)

with $X_1$ and $X_2$ independent. Let $D = X_1 - X_2$ . Then

\begin{aligned} E(D) &= E(X_1) - E(X_2) = 125 - 125 = 0 \\ \mathrm{Var}(D) &= \mathrm{Var}(X_1) + \mathrm{Var}(X_2) = 50 + 50 = 100 \end{aligned}

So $D \sim N(0, 100)$ . Standardising:

z = \frac{12 - 0}{\sqrt{100}} = \frac{12}{10} = 1.2

Therefore

P(D > 12) = 1 - \Phi(1.2) = 0.1151

The event "the difference is more than 12 minutes" means $|D| > 12$ , so by symmetry:

P(|D| > 12) = 2 \times P(D > 12) = 2 \times 0.1151 = 0.230

Answer

P(\text{difference} > 12) = 0.230 \ (3\text{ sf})

Final answer

0.230

Detailed explanation

Walkthrough

Let $X_1$ be the time of the first student and $X_2$ the time of the second student. Both are normally distributed with mean 125 and variance 50. Because the two students are chosen independently, $X_1$ and $X_2$ are independent.

Define $D = X_1 - X_2$ . Since a linear combination of independent normal variables is normal, $D$ is also normal. The mean of a difference is the difference of the means, so $E(D) = 125 - 125 = 0$ . For independent variables, the variance of a difference is the sum of the variances, because the coefficient $-1$ is squared: $\mathrm{Var}(D) = 50 + 50 = 100$ . Thus $D \sim N(0, 100)$ , so its standard deviation is $\sqrt{100} = 10$ .

To find $P(D > 12)$ , standardise: $z = (12 - 0)/10 = 1.2$ . Then $P(D > 12) = 1 - \Phi(1.2) = 0.1151$ .

The word "difference" in the question means the absolute difference $|X_1 - X_2|$ . So we need both $D > 12$ and $D < -12$ . By symmetry these two probabilities are equal, so the required probability is $2 \times 0.1151 = 0.230$ .

Key Takeaways

The difference of two independent normal random variables is normal.
$E(X_1 - X_2) = E(X_1) - E(X_2)$ .
For independent variables, $\mathrm{Var}(X_1 - X_2) = \mathrm{Var}(X_1) + \mathrm{Var}(X_2)$ .
When a question asks for the probability that a difference exceeds a value, it usually means the absolute difference, so both tails must be included.

Common Mistakes

Writing $\mathrm{Var}(X_1 - X_2) = \mathrm{Var}(X_1) - \mathrm{Var}(X_2)$ , which is incorrect; variances always add for independent variables.
Forgetting to double the one-tailed probability, giving 0.115 instead of 0.230.
Using a continuity correction; this is not needed because the normal distribution is already continuous.
Standardising with the variance 100 instead of the standard deviation 10.

Things to Be Careful About

The mark scheme gives $E(D) = 0$ and $\mathrm{Var}(D) = 100$ for the first mark. The standardising mark requires using their $E(D)$ and $\mathrm{Var}(D)$ from a combination attempt. The final answer must be doubled only if the one-tailed probability is less than 0.5, so that the final probability is not greater than 1.

Techniques used

define difference of independent normal variablescompute mean and variance of differencestandardise to standard normalfind upper-tail probabilitydouble for absolute difference

Q23MMedium-EasySampling and Estimation

The height of a certain species of plant is denoted by $H$ cm. The heights of a random sample of 100 plants were measured, and the following results were found.

\begin{itemize}
\item The mean, $\bar{h}$ , for the sample was 80.2.
\item An unbiased estimate of the population variance of $H$ was 15.6.
\end{itemize}

Calculate the value of $\sum h^2$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the formula for the unbiased estimate of the population variance, which expresses the raw sum of squares $\sum h^2$ in terms of the sample size $n$ and the sample mean $\bar{h}$ . Substitute the known values and solve for $\sum h^2$ .

Working

For a sample of size $n$ , the unbiased estimate of the population variance is

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

Here $n = 100$ , $\bar{h} = 80.2$ and $s^2 = 15.6$ , so

15.6 = \frac{1}{99}\left(\sum h^2 - 100(80.2)^2\right).

Compute the correction term:

100(80.2)^2 = 100 \times 6432.04 = 643204.

Therefore

15.6 = \frac{\sum h^2 - 643204}{99}.

Multiply both sides by 99:

1544.4 = \sum h^2 - 643204.

Add 643204 to both sides:

\sum h^2 = 644748.4.

Answer

\sum h^2 = 644748.4 \quad (\text{or } 645000 \text{ to 3 significant figures}).

Final answer

644748.4 (or 645000 to 3 significant figures)

Detailed explanation

Walkthrough

We are told the sample mean is $\bar{h} = 80.2$ , the sample size is $n = 100$ , and the unbiased estimate of the population variance is $15.6$ . The unbiased variance estimate is

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

This formula comes from expanding $\sum (h - \bar{h})^2 = \sum h^2 - n\bar{h}^2$ and dividing by $n-1$ because the estimate is unbiased. Substituting the known values gives

15.6 = \frac{1}{99}\left(\sum h^2 - 100(80.2)^2\right).

First compute $100(80.2)^2 = 100 \times 6432.04 = 643204$ . The equation becomes

15.6 = \frac{\sum h^2 - 643204}{99}.

Multiplying both sides by 99 gives $1544.4 = \sum h^2 - 643204$ , so $\sum h^2 = 644748.4$ . This is the required raw sum of squares.

Key Takeaways

The unbiased estimate of the population variance is $s^2 = \frac{1}{n-1}\left(\sum h^2 - n\bar{h}^2\right)$ .
The sample mean and sample size allow the sum of squared deviations to be rewritten in terms of the raw sum of squares $\sum h^2$ .
When a question says the estimate is unbiased, use the denominator $n-1$ , not $n$ .

Common Mistakes

Using $n$ instead of $n-1$ in the denominator, which gives the biased estimate and changes the answer.
Forgetting to multiply $\bar{h}^2$ by $n$ before subtracting it from $\sum h^2$ .
Confusing the unbiased estimate with the biased sample variance.

Things to Be Careful About

Keep the order of operations correct: $\sum h^2 - n\bar{h}^2$ , not $\sum h^2 - \bar{h}^2$ .
Use the exact value $80.2^2 = 6432.04$ , then multiply by 100 to get $643204$ .
The final answer may be written as $644748.4$ , $645000$ to 3 significant figures, or $\frac{3223742}{5}$ , all of which are accepted.

Techniques used

apply the unbiased variance formularewrite the sum of squared deviations using the sample meansolve a linear equation for the raw sum of squares

Approach

For $X \sim \mathrm{Po}(\lambda)$ , the probability that $X = x$ is

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

Here $\lambda = 15$ and $x = 12$ .

Working

Substitute $\lambda = 15$ and $x = 12$ into the formula:

\mathrm{P}(X = 12) = \frac{e^{-15} \cdot 15^{12}}{12!}.

Answer

\mathrm{P}(X = 12) = \frac{15^{12} e^{-15}}{12!}

Final answer

P(X = 12) = 15^12 e^-15 / 12!

Detailed explanation

Walkthrough

For a Poisson random variable $X \sim \mathrm{Po}(\lambda)$ , the probability of exactly $x$ occurrences is

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

The question gives $X \sim \mathrm{Po}(15)$ , so $\lambda = 15$ . We need $\mathrm{P}(X = 12)$ , so we substitute $x = 12$ directly into the formula. There is no need to simplify further; the expression in terms of $e$ is exactly what is requested.

Key Takeaways

The Poisson probability formula is $\mathrm{P}(X = x) = \dfrac{e^{-\lambda} \lambda^x}{x!}$ .
The parameter $\lambda$ is both the mean and the variance of the distribution.
For a one-mark question, the final answer is the unsimplified probability expression.

Common Mistakes

Forgetting the factor $e^{-15}$ .
Writing $15^{12}$ instead of $15^{12}$ in the numerator, or using the wrong factorial in the denominator.
Trying to evaluate the expression numerically when the question only asks for an expression in terms of $e$ .

Things to Be Careful About

Keep the exact form $\dfrac{15^{12} e^{-15}}{12!}$ ; do not approximate.
The factorial is $12!$ , not $15!$ .
The order of factors does not matter, but all three of $e^{-15}$ , $15^{12}$ , and $12!$ must appear.

Techniques used

recall the Poisson probability formulasubstitute the given mean and value of X

(b)

It is given that $\text{P}(X = n) = \text{P}(X = n + 1)$ .

Write down an equation in $n$ , and hence find the value of $n$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Write the Poisson formula for both $\mathrm{P}(X = n)$ and $\mathrm{P}(X = n + 1)$ , set them equal, cancel the common factors, and solve for $n$ .

Working

For $X \sim \mathrm{Po}(15)$ ,

\mathrm{P}(X = n) = \frac{e^{-15} \cdot 15^n}{n!}

and

\mathrm{P}(X = n + 1) = \frac{e^{-15} \cdot 15^{n+1}}{(n+1)!}.

Given that $\mathrm{P}(X = n) = \mathrm{P}(X = n + 1)$ :

\frac{e^{-15} \cdot 15^n}{n!} = \frac{e^{-15} \cdot 15^{n+1}}{(n+1)!}.

Cancel $e^{-15}$ :

\frac{15^n}{n!} = \frac{15^{n+1}}{(n+1)!}.

Rewrite the right-hand side using

15^{n+1} = 15 \cdot 15^n \quad \text{and} \quad (n+1)! = (n+1)n!.

So

\frac{15^n}{n!} = \frac{15 \cdot 15^n}{(n+1)n!}.

Cancel $\dfrac{15^n}{n!}$ from both sides:

1 = \frac{15}{n+1}.

Therefore

n + 1 = 15,

so

n = 14.

Answer

n = 14

Final answer

n = 14

Detailed explanation

Walkthrough

We begin by writing the probability that $X = n$ and the probability that $X = n+1$ using the Poisson formula. Since $X \sim \mathrm{Po}(15)$ , both probabilities share the common factor $e^{-15}$ , so we can cancel it immediately.

Next, we compare the remaining expressions. The right-hand side contains $15^{n+1}$ and $(n+1)!$ . We rewrite these as

15^{n+1} = 15 \cdot 15^n

and

(n+1)! = (n+1)n!.

This allows us to cancel $15^n$ and $n!$ from both sides, leaving

1 = \frac{15}{n+1}.

Finally, multiplying through by $n+1$ gives $n+1 = 15$ , so $n = 14$ .

Key Takeaways

The Poisson probability formula is the key tool for both parts of this question.
When equating two Poisson probabilities, common factors such as $e^{-\lambda}$ can be cancelled.
The simplification uses the identities $\lambda^{n+1} = \lambda \cdot \lambda^n$ and $(n+1)! = (n+1)n!$ .
The result $n = \lambda - 1$ is a useful pattern: here $\lambda = 15$ , so $n = 14$ .

Common Mistakes

Omitting the factor $e^{-15}$ entirely.
Incorrectly simplifying $(n+1)!$ ; it is $(n+1)n!$ , not $n! + 1$ .
Trying to solve by trial and error instead of using the algebraic equation. The mark scheme allows trial and error, but it is less efficient and can lose method marks.
Forgetting to show the equation before solving; the first mark is awarded for the correct equation.

Things to Be Careful About

The equation must be written with the factorials in brackets: $(n+1)!$ . If the brackets are omitted, the expression is ambiguous and may lose credit.

Make sure to cancel correctly: after cancelling $e^{-15}$ , $15^n$ , and $n!$ , the remaining equation is $1 = \dfrac{15}{n+1}$ , not $1 = \dfrac{15}{n}$ .

Since $n$ is a non-negative integer for a Poisson probability, $n = 14$ is the only valid solution.

Techniques used

equate two Poisson probabilitiescancel common exponential factorssimplify powers and factorialssolve the resulting linear equation

Q44MMediumSampling and Estimation

A biased spinner has four sides. Each side is of a different colour: yellow, red, green or black. The probability, $p$ , that the spinner will land on red is unknown. The spinner was spun 200 times, and the proportion, $a$ , of times that it landed on red was noted. This proportion was used to calculate an approximate 90% confidence interval for $p$ . The width of this confidence interval was 0.1066 correct to 4 significant figures.

Find the two possible values of $a$ .

DifficultyMedium

Worked solution

Approach

For a large sample, an approximate 90% confidence interval for the population proportion $p$ is centred on the sample proportion $a$ , with margin of error $z\sqrt{\frac{a(1-a)}{n}}$ . The width of the interval is twice the margin, so equating the width to $0.1066$ gives an equation for $a$ .

Working

The critical value for a 90% confidence interval is

z = 1.645

With $n = 200$ , the width of the confidence interval is

2 \times 1.645 \times \sqrt{\frac{a(1-a)}{200}} = 0.1066

Divide by $2 \times 1.645 = 3.29$ :

\sqrt{\frac{a(1-a)}{200}} = \frac{0.1066}{3.29}

Square both sides:

\frac{a(1-a)}{200} = \left(\frac{0.1066}{3.29}\right)^2

Multiply by 200:

a(1-a) = 200\left(\frac{0.1066}{3.29}\right)^2 = 0.20997

Rearrange into a quadratic:

a^2 - a + 0.20997 = 0

Apply the quadratic formula:

a = \frac{1 \pm \sqrt{1 - 4(0.20997)}}{2}

a = \frac{1 \pm \sqrt{0.16012}}{2}

a = \frac{1 \pm 0.40015}{2}

So

a = 0.70008 \quad \text{or} \quad a = 0.29992

Rounded to 3 significant figures:

a = 0.700 \quad \text{or} \quad a = 0.300

Answer

a = 0.300 \quad \text{or} \quad a = 0.700

Final answer

a = 0.300 or a = 0.700

Detailed explanation

Walkthrough

The sample proportion $a$ is used as the estimate of the unknown probability $p$ . For a large sample, the normal approximation gives a 90% confidence interval for $p$ as

a \pm 1.645\sqrt{\frac{a(1-a)}{200}}

The width of an interval of the form $\text{estimate} \pm \text{margin}$ is twice the margin, so we set

2 \times 1.645\sqrt{\frac{a(1-a)}{200}} = 0.1066

We then solve for $a$ step by step. First divide by $3.29$ to isolate the square root. Then square both sides to remove the square root. Multiplying by 200 gives $a(1-a) = 0.20997$ , which rearranges to the quadratic

a^2 - a + 0.20997 = 0

Applying the quadratic formula gives two roots, approximately $0.700$ and $0.300$ . Both are between 0 and 1, so both are possible values of the sample proportion $a$ .

The two roots arise because $a(1-a)$ is unchanged when $a$ is replaced by $1-a$ ; a sample proportion of 0.300 and one of 0.700 give the same standard error and hence the same confidence interval width.

Key Takeaways

A confidence interval for a proportion is $\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ .
The width of the interval is twice the margin of error, not the margin itself.
For 90% confidence, $z = 1.645$ .
Solving for the sample proportion can lead to a quadratic equation, and both valid roots should be given.

Common Mistakes

Forgetting the factor of 2 when using the width of the interval.
Using the wrong critical value, such as $z = 1.96$ for 95% confidence.
Confusing the sample proportion $a$ with the number of red outcomes. If $a$ were a count, the roots would be $60$ and $140$ , but the question asks for the proportion, so the correct final answers are $0.300$ and $0.700$ . The mark scheme explicitly does not accept $a = 0.3 \times 200$ or $a = 0.7 \times 200$ .
Rounding too early, which can change the final value slightly.

Things to Be Careful About

Use $n = 200$ , not $n = 1$ , in the denominator of the standard error.
The given width is $0.1066$ , so the margin of error is half of this, $0.0533$ .
Both roots are valid because a proportion can be below or above 0.5 and still produce the same interval width.
The final answer must be given as a proportion, not as a count of spins.
Round the final answer to 3 significant figures as required by the mark scheme.

Techniques used

write the confidence interval for a population proportionset the interval width equal to the given valuesolve the resulting quadratic equationinterpret both roots as valid sample proportions

Approach

This is a hypothesis test for the population mean. The manager wants to see whether the new mean $\mu$ is greater than 10.5, so

H_0: \mu = 10.5, \quad H_1: \mu > 10.5

Since the sample comes from a normal distribution, $\bar{x}$ is normally distributed with standard deviation $\sigma/\sqrt{n}$ . The necessary assumption is that the population standard deviation $\sigma$ remains 3.8 after the move. The smallest value that gives sufficient evidence is the value of $\bar{x}$ that just reaches the upper 5% critical value of the standard normal distribution.

Working

Assumption: $\sigma = 3.8$ remains unchanged.

For a one-tailed test at the 5% significance level, the critical value is

z = 1.645

The test statistic is

z = \frac{\bar{x} - 10.5}{3.8/\sqrt{10}}

Sufficient evidence is found when $z \geq 1.645$ . The smallest possible value of $\bar{x}$ therefore occurs when

\frac{\bar{x} - 10.5}{3.8/\sqrt{10}} = 1.645

Solving for $\bar{x}$ :

\bar{x} = 10.5 + 1.645 \times \frac{3.8}{\sqrt{10}}

\frac{3.8}{\sqrt{10}} = 1.20166\ldots

1.645 \times 1.20166\ldots = 1.9767\ldots

\bar{x} = 10.5 + 1.9767\ldots = 12.4767\ldots

Answer

The smallest possible value of the sample mean is

\bar{x} = 12.5 \quad (3 \text{ s.f.})

so $\bar{x} = 12.5$ minutes, to 3 significant figures.

Final answer

Assumption: sigma remains 3.8. Smallest xbar = 12.5 (3 s.f.)

Detailed explanation

Walkthrough

The manager wants to know whether the new mean is greater than 10.5, so this is a one-tailed hypothesis test. We need to find the boundary value of the sample mean that would make the test significant at the 5% level.

First, state the necessary assumption. The test statistic for a sample mean from a normal population uses the population standard deviation. Since the only value given is the past value $\sigma = 3.8$ , we must assume that the standard deviation has not changed after the move.

Next, identify the critical value. For a one-tailed test at the 5% significance level, the critical value of the standard normal distribution is $z = 1.645$ . This is the value the test statistic must reach or exceed to reject the null hypothesis.

Then form the test statistic. Since $X \sim \text{N}(\mu, \sigma^2)$ and the sample size is $n = 10$ , the sample mean has standard error $\sigma/\sqrt{10} = 3.8/\sqrt{10}$ . The test statistic is

z = \frac{\bar{x} - 10.5}{3.8/\sqrt{10}}

The smallest possible value of $\bar{x}$ that gives sufficient evidence is the value for which this statistic equals exactly 1.645. Solving gives $\bar{x} = 12.4767\ldots$ , which rounds to 12.5 to 3 significant figures.

Key Takeaways

A one-tailed test for a higher mean uses the normal distribution of the sample mean when the population is normal and the population standard deviation is known. The critical value depends on the direction of the test and the significance level. The boundary of the critical region is found by setting the test statistic equal to the critical value and solving for the sample statistic.

Common Mistakes

Forgetting to state the assumption that $\sigma$ remains 3.8; this is explicitly required and earns a mark.
Using the two-tailed critical value $z = 1.96$ instead of the one-tailed value $z = 1.645$ .
Forgetting the $\sqrt{10}$ in the denominator of the test statistic.
Giving an unsupported answer without showing the critical value and the equation.

Things to Be Careful About

The test is one-tailed, so the critical region is $z > 1.645$ , not $z > 1.96$ . The value $\sigma = 3.8$ is the standard deviation, not the variance, so the standard error is $3.8/\sqrt{10}$ . The exact boundary is $12.4767\ldots$ , so the final answer is 12.5 to 3 significant figures; answers such as $\bar{x} > 12.476$ or $\bar{x} > 12.5$ are also acceptable. The units are minutes.

Techniques used

state necessary assumption that population standard deviation is unchangedidentify the one-tailed critical valueform the sample-mean test statisticset the test statistic equal to the critical value and solve

Approach

Since $n = 700$ is large and $p = 0.005$ is small, approximate $W \sim \text{B}(700, 0.005)$ by a Poisson distribution with $\lambda = 700 \times 0.005 = 3.5$ . Then find $\text{P}(W \geq 4)$ using the complement $1 - \text{P}(W \leq 3)$ .

Working

\lambda = 700 \times 0.005 = 3.5

Using $W \sim \text{Po}(3.5)$ ,

\begin{aligned} \text{P}(W \leq 3) &= e^{-3.5}\left(1 + 3.5 + \frac{3.5^2}{2!} + \frac{3.5^3}{3!}\right) \\ &= e^{-3.5}(1 + 3.5 + 6.125 + 7.1458) \end{aligned}

Therefore

\begin{aligned} \text{P}(W \geq 4) &= 1 - \text{P}(W \leq 3) \\ &= 1 - e^{-3.5}(1 + 3.5 + 6.125 + 7.1458) \\ &= 0.463 \quad (3\text{ sf}) \end{aligned}

Answer

\text{P}(W \geq 4) = 0.463

Final answer

0.463

Detailed explanation

Walkthrough

The distribution $\text{B}(700, 0.005)$ has a large number of trials and a small probability of success, so the Poisson approximation is appropriate. The mean of the approximating Poisson distribution is $\lambda = np = 700 \times 0.005 = 3.5$ . We need $\text{P}(W \geq 4)$ , which is easier to find by first calculating $\text{P}(W \leq 3)$ and subtracting from 1. Using the Poisson formula $\text{P}(W = r) = e^{-\lambda}\frac{\lambda^r}{r!}$ , we sum the probabilities for $r = 0, 1, 2, 3$ . Subtracting this sum from 1 gives the required probability.

Key Takeaways

This question tests the Poisson approximation to the binomial distribution. The key signal is a large $n$ and a small $p$ , so the binomial distribution can be replaced by $\text{Po}(np)$ . It also tests the ability to calculate Poisson tail probabilities efficiently using the complement rule.

Common Mistakes

Using the binomial distribution directly instead of the Poisson approximation; the mark scheme gives only partial credit for this.
Writing an unsupported final answer; the method and intermediate terms must be shown.
Forgetting to subtract $\text{P}(W \leq 3)$ from 1.

Things to Be Careful About

Use $\lambda = 3.5$ , not $700 \times 0.005$ incorrectly. The event $W \geq 4$ excludes $W = 0, 1, 2, 3$ , so the complement is $\text{P}(W \leq 3)$ . Give the final answer to 3 significant figures.

Techniques used

approximate a binomial distribution with a Poisson distributionuse the complement rulecalculate Poisson probabilities

(ii)

Two values of $W$ are chosen at random.

Find the probability that the sum of these two values is less than 3.

3M Similar

DifficultyMedium

Worked solution

Approach

Each value of $W$ is approximately $\text{Po}(3.5)$ . Since the two values are chosen independently, their sum $S$ is also Poisson with parameter $3.5 + 3.5 = 7$ . The event "less than 3" means $S = 0$ , $S = 1$ or $S = 2$ , so add the three Poisson probabilities.

Working

Let $S = W_1 + W_2$ . Since $W_1, W_2 \sim \text{Po}(3.5)$ independently,

S \sim \text{Po}(7)

Now

\begin{aligned} \text{P}(S < 3) &= \text{P}(S = 0) + \text{P}(S = 1) + \text{P}(S = 2) \\ &= e^{-7}\left(1 + 7 + \frac{7^2}{2!}\right) \\ &= e^{-7}(1 + 7 + 24.5) \\ &= 0.0009119 + 0.006383 + 0.0223411 \\ &= 0.0296 \quad (3\text{ sf}) \end{aligned}

Answer

\text{P}(S < 3) = 0.0296

Final answer

0.0296

Detailed explanation

Walkthrough

First, each value of $W$ is approximated by $\text{Po}(3.5)$ . When two independent Poisson random variables are added, the sum is also Poisson and its parameter is the sum of the two parameters, so $S = W_1 + W_2 \sim \text{Po}(7)$ . The event $S < 3$ means $S$ can only be 0, 1 or 2. Use the Poisson probability formula for each of these values and add them together. The three terms are $e^{-7}$ , $7e^{-7}$ and $\frac{49}{2}e^{-7}$ , which sum to 0.0296.

Key Takeaways

This question tests the Poisson approximation to the binomial and an important property of Poisson distributions: the sum of independent Poisson variables is Poisson. It also tests the ability to convert a phrase such as "less than 3" into the correct set of discrete values.

Common Mistakes

Using the sum as $\text{Po}(3.5)$ instead of $\text{Po}(7)$ .
Forgetting that "less than 3" includes 0, 1 and 2, not 3.
Writing an unsupported answer; the mark scheme requires the Poisson expression or terms to be seen.
Using the binomial distribution for the sum instead of the Poisson approximation; this only earns partial credit.

Things to Be Careful About

Remember that the sum of two independent $\text{Po}(3.5)$ variables has parameter $7$ , not $3.5$ . Keep all three probability terms visible in the working. Give the final answer to 3 significant figures.

Techniques used

approximate a binomial with a Poisson distributionadd independent Poisson ratescalculate Poisson probabilities

(b)

The random variable $X$ has the distribution $\text{Po}(200)$ .

Use a suitable approximating distribution to find $\text{P}(X > 205)$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Since $\lambda = 200$ is large, approximate $X \sim \text{Po}(200)$ by the normal distribution $N(200, 200)$ . Apply a continuity correction because $X$ is discrete: $\text{P}(X > 205)$ is the same as $\text{P}(X \geq 206)$ , so use the boundary $205.5$ . Then standardise and find the upper-tail probability.

Working

Approximate

X \sim \text{Po}(200) \approx N(200, 200)

Using the continuity correction,

\text{P}(X > 205) = \text{P}(X \geq 206) \approx \text{P}(Y > 205.5)

where $Y \sim N(200, 200)$ . Standardise:

z = \frac{205.5 - 200}{\sqrt{200}} = \frac{5.5}{\sqrt{200}} = 0.38891

Therefore

\text{P}(Y > 205.5) = 1 - \Phi(0.38891) = 0.349 \quad (3\text{ sf})

Answer

\text{P}(X > 205) = 0.349

Final answer

0.349

Detailed explanation

Walkthrough

Since $\lambda = 200$ is large, the Poisson distribution $\text{Po}(200)$ can be approximated by a normal distribution with the same mean and variance, namely $N(200, 200)$ . Because $X$ is discrete, a continuity correction is needed when using the normal approximation. The inequality $X > 205$ is equivalent to $X \geq 206$ for integer values, so the normal boundary is $205.5$ . Standardise by subtracting the mean and dividing by the standard deviation $\sqrt{200}$ . Then find the upper-tail probability $1 - \Phi(z)$ , which gives 0.349.

Key Takeaways

This question tests the normal approximation to the Poisson distribution. When $\lambda$ is large, $\text{Po}(\lambda)$ can be approximated by $N(\lambda, \lambda)$ . It also tests the continuity correction, which is essential when approximating a discrete distribution by a continuous one.

Common Mistakes

Forgetting the continuity correction and using $205$ instead of $205.5$ .
Using the wrong tail: $\text{P}(X > 205)$ requires the upper tail, so $1 - \Phi(z)$ is needed.
Using the standard deviation incorrectly; the variance is $200$ , so the standard deviation is $\sqrt{200}$ .
Writing an unsupported answer; the normal approximation and standardisation must be shown.

Things to Be Careful About

The mean and variance of a Poisson distribution are both equal to $\lambda$ , so $N(200, 200)$ is the correct approximation. Remember that for a discrete variable, $X > 205$ means $X \geq 206$ , making the continuity-corrected boundary $205.5$ . Give the final answer to 3 significant figures.

Techniques used

approximate a Poisson distribution with a normal distributionapply a continuity correctionstandardise a normal variablecalculate a normal tail probability

Q7MediumContinuous Random Variables

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

f(x) = \begin{cases} \frac{kx^2}{a^2} & 0 \leq x \leq a, \\ 0 & \text{otherwise,} \end{cases}

where $k$ and $a$ are positive constants.

(a)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the defining property of a probability density function: the total area under the curve over its full domain must equal 1. Integrate $f(x)$ from 0 to $a$ , set the result equal to 1, and solve for $k$ .

Working

The PDF must satisfy $\int_0^a f(x)\,dx = 1$ :

\int_0^a \frac{kx^2}{a^2}\,dx = 1

Factor out the constant $\frac{k}{a^2}$ :

\frac{k}{a^2}\int_0^a x^2\,dx = 1

Integrate:

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

Evaluate at the limits:

\frac{k}{a^2}\cdot\frac{a^3}{3} = 1

Simplify:

\frac{ka}{3} = 1

Hence:

k = \frac{3}{a}

Answer

k = \frac{3}{a}

Final answer

k = 3/a

Detailed explanation

Walkthrough

The total area under any probability density function must be exactly 1. Since X only takes values between 0 and $a$ , the area under $f(x)$ between these limits equals 1. We integrate $\frac{kx^2}{a^2}$ from 0 to $a$ . Because $\frac{k}{a^2}$ is constant, pull it out of the integral. Integrating $x^2$ gives $\frac{x^3}{3}$ ; applying the limits 0 and $a$ gives $\frac{a^3}{3} - 0 = \frac{a^3}{3}$ . Multiplying by $\frac{k}{a^2}$ leaves $\frac{ka}{3}$ . Setting this equal to 1 and dividing by $a$ (which is positive, so nonzero) gives $k = \frac{3}{a}$ . The intermediate line $\frac{ka}{3} = 1$ must be shown — it proves the derivation is correct.

Key Takeaways

A PDF must integrate to 1 over its support. Constant factors can be factored out before integrating. To find a normalisation constant, set the integral of the PDF equal to 1 and solve.

Common Mistakes

Forgetting to set the integral equal to 1.
Omitting the limits and writing only an indefinite integral.
Skipping the intermediate step $\frac{ka}{3} = 1$ — the mark scheme requires it ("must see an intermediate step"), so jumping straight to $k = \frac{3}{a}$ loses the final mark.
Inverting the algebra: $\frac{ka}{3} = 1$ gives $k = \frac{3}{a}$ , not $k = \frac{a}{3}$ .

Things to Be Careful About

Since $k$ and $a$ are positive constants, $a \neq 0$ , so division by $a$ is legitimate. The statement $k = \frac{3}{a}$ is given in the question (AG), so a convincing chain of working is mandatory; simply writing the answer earns no marks.

Techniques used

integrate the PDF over its full domainset the total probability equal to 1solve for the normalisation constant

(b)

It is given that $\text{E}(X) = 1$ .

Find the value of $a$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the formula $\text{E}(X) = \int_0^a x f(x)\,dx$ . Substitute the expression for $f(x)$ with $k$ written in terms of $a$ , integrate, set the result equal to 1, and solve for $a$ .

Working

First substitute $k = \frac{3}{a}$ into $f(x)$ :

f(x) = \frac{3x^2}{a^3}

The expectation is:

\text{E}(X) = \int_0^a x\cdot\frac{3x^2}{a^3}\,dx = \frac{3}{a^3}\int_0^a x^3\,dx

Integrate:

\frac{3}{a^3}\left[\frac{x^4}{4}\right]_0^a = \frac{3}{a^3}\cdot\frac{a^4}{4}

Simplify:

\text{E}(X) = \frac{3a}{4}

Given $\text{E}(X) = 1$ :

\frac{3a}{4} = 1

a = \frac{4}{3}

Answer

a = \frac{4}{3}

Final answer

a = 4/3 ≈ 1.33

Detailed explanation

Walkthrough

The expectation of a continuous random variable is $\text{E}(X) = \int_0^a x\,f(x)\,dx$ . Replace $k$ using part (a): since $k = \frac{3}{a}$ , $f(x) = \frac{3x^2}{a^3}$ . Then $\text{E}(X) = \int_0^a x\cdot\frac{3x^2}{a^3}\,dx$ ; combining the powers of $x$ gives $x^3$ , so we integrate $\frac{3}{a^3}x^3$ from 0 to $a$ . Pulling out $\frac{3}{a^3}$ and integrating $x^3$ gives $\frac{x^4}{4}$ ; evaluating at the limits yields $\frac{a^4}{4}$ , and multiplying by $\frac{3}{a^3}$ gives $\frac{3a}{4}$ . Setting this equal to the given value 1 and solving gives $a = \frac{4}{3}$ .

Key Takeaways

For a continuous random variable, $\text{E}(X) = \int x\,f(x)\,dx$ . Substituting a previously found constant (here $k = \frac{3}{a}$ ) before integrating keeps the working clean. Definite integrals from 0 to $a$ follow the same evaluation pattern as in part (a).

Common Mistakes

Writing $\text{E}(X) = \int f(x)\,dx$ , forgetting the factor of $x$ .
Using incorrect limits (anything other than 0 and $a$ ).
Failing to reach $\frac{3a}{4} = 1$ — the mark scheme requires the correct expression after integration.
Arithmetic slips in $\frac{3}{a^3}\cdot\frac{a^4}{4} = \frac{3a}{4}$ .

Things to Be Careful About

The mark scheme condones keeping $k$ as a symbol as long as $k = \frac{3}{a}$ is substituted at some point, but substituting first is cleaner and less error-prone. Solve $\frac{3a}{4} = 1$ correctly: multiply by 4 and divide by 3 to get $a = \frac{4}{3}$ , or 1.33 (3 sf).

Techniques used

multiply the PDF by x to form the expectation integrandintegrate x f(x) over the supportsubstitute the normalisation constant and solve the equation

(c)

Find the median of $X$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The median $m$ is the value such that $\text{P}(X \leq m) = 0.5$ . Integrate $f(x)$ from 0 to $m$ with the found values $k = \frac{3}{a}$ and $a = \frac{4}{3}$ substituted, set the result equal to 0.5, and solve for $m$ .

Working

From parts (a) and (b), $k = \frac{3}{a}$ and $a = \frac{4}{3}$ . Hence:

k = \frac{3}{4/3} = \frac{9}{4}

and the coefficient:

\frac{k}{a^2} = \frac{9/4}{(4/3)^2} = \frac{9}{4}\cdot\frac{9}{16} = \frac{81}{64}

So $f(x) = \frac{81}{64}x^2$ for $0 \leq x \leq \frac{4}{3}$ .

The median $m$ satisfies:

\int_0^m f(x)\,dx = 0.5

\int_0^m \frac{81}{64}x^2\,dx = 0.5

\frac{81}{64}\left[\frac{x^3}{3}\right]_0^m = 0.5

\frac{81}{64}\cdot\frac{m^3}{3} = 0.5

\frac{27}{64}m^3 = 0.5

m^3 = 0.5\times\frac{64}{27} = \frac{32}{27}

m = \sqrt[3]{\frac{32}{27}} \approx 1.06

Since $m \approx 1.06 < a = \frac{4}{3}$ , the median lies within the domain, so it is valid.

Answer

m = \sqrt[3]{\frac{32}{27}} \approx 1.06

Final answer

m = (32/27)^(1/3) ≈ 1.06

Detailed explanation

Walkthrough

The median $m$ is the value at which half the probability lies below it, i.e. $\text{P}(X \leq m) = 0.5$ . Since X only takes values from 0 to $a$ , integrate $f(x)$ from 0 to $m$ . First collect the needed constants: from parts (a) and (b), $k = \frac{3}{a}$ and $a = \frac{4}{3}$ . Substituting gives $k = \frac{3}{4/3} = \frac{9}{4}$ and $\frac{k}{a^2} = \frac{9/4}{(4/3)^2} = \frac{9}{4} \cdot \frac{9}{16} = \frac{81}{64}$ , so $f(x) = \frac{81}{64}x^2$ . Integrating from 0 to $m$ gives $\frac{81}{64}\cdot\frac{m^3}{3} = \frac{27}{64}m^3$ . Set this equal to 0.5: $m^3 = 0.5 \times \frac{64}{27} = \frac{32}{27}$ , so $m = \sqrt[3]{\frac{32}{27}} \approx 1.06$ . Check $m < a = \frac{4}{3} \approx 1.33$ : the median lies inside the domain, confirming the answer.

Key Takeaways

The median of a continuous random variable is found by integrating the PDF to $m$ and setting the result equal to 0.5. Full substitution of both parameters ( $a$ and $k$ ) is required before integrating. The result must be checked against the domain $0 \leq m \leq a$ .

Common Mistakes

Integrating from $m$ to $a$ instead of 0 to $m$ without adjusting the equation accordingly (the mark scheme allows either as long as it is set to 0.5, but the 0-to- $m$ form is standard).
Forgetting to substitute $a = \frac{4}{3}$ and $k = \frac{3}{a}$ , or mistakenly assuming a value such as $a = 1$ .
Inverting the fraction when solving: $m^3 = 0.5 \times \frac{64}{27} = \frac{32}{27}$ , not $\frac{27}{32}$ .
Not simplifying the cube root or omitting the 3 sf value.

Things to Be Careful About

Show the substitution explicitly: $(4/3)^2 = 16/9$ and $(9/4)/(16/9) = (9/4)(9/16) = 81/64$ . The mark scheme requires $a$ and $k$ to be correctly substituted at some point. The final mark requires correct working only (CWO), so present the median as $m = \sqrt[3]{\frac{32}{27}}$ or 1.06 (3 sf).

Techniques used

substitute the found parameter values into the PDFintegrate the PDF from 0 to the medianset the result to 0.5 and solve for the median

Q8MediumHypothesis Tests

Birgitte has a six-sided dice. She suspects that the dice is biased so that the probability, $p$ , that it will show a six on one throw is less than $\frac{1}{6}$ . She throws the dice 30 times and finds that it shows a six on exactly 2 throws.

(a)

Use a binomial distribution with a 5% significance level to test Birgitte’s suspicion.

5M Similar

DifficultyMedium

Worked solution

Approach

Set up a one-tailed lower test. Since Birgitte suspects the probability is less than $\frac{1}{6}$ , the alternative hypothesis is one-tailed. Under the null hypothesis, $X \sim \mathrm{B}\left(30, \frac{1}{6}\right)$ , where $X$ is the number of sixes in 30 throws. The observed value is $X = 2$ , so calculate $P(X \le 2)$ and compare it with the 5% significance level.

Working

H_0: p = \frac{1}{6}, \quad H_1: p < \frac{1}{6}

Under $H_0$ , $X \sim \mathrm{B}\left(30, \frac{1}{6}\right)$ .

\begin{aligned} P(X \le 2) &= P(X=0) + P(X=1) + P(X=2) \\ &= \left(\frac{5}{6}\right)^{30} + 30\left(\frac{5}{6}\right)^{29}\left(\frac{1}{6}\right) + {}^{30}C_2\left(\frac{5}{6}\right)^{28}\left(\frac{1}{6}\right)^2 \\ &= 0.00421272 + 0.0252763 + 0.07330 \\ &= 0.1028... \\ &= 0.103 \quad (3\text{ s.f.}) \end{aligned}

Compare with the 5% significance level:

0.103 > 0.05

The result is not significant.

Answer

Accept $H_0$ ; there is insufficient evidence to suggest that the probability of showing a six is less than $\frac{1}{6}$ .

Final answer

Accept H0; insufficient evidence that the probability of a six is less than 1/6.

Detailed explanation

Walkthrough

Birgitte suspects the probability is biased so that $p < \frac{1}{6}$ , so this is a one-tailed lower-tail test. We set up the null hypothesis as the fair-dice assumption $p = \frac{1}{6}$ and the alternative as $p < \frac{1}{6}$ . Under the null hypothesis, the number of sixes follows $X \sim B(30, \frac{1}{6})$ . Since the observed value is exactly 2, we calculate the probability of getting 2 or fewer sixes. This is the $p$ -value. We add the binomial probabilities for 0, 1 and 2 sixes. We then compare this probability with the significance level 0.05. Because $0.103 > 0.05$ , the result could reasonably have occurred by chance, so we do not reject $H_0$ .

Key Takeaways

This question asks for a one-tailed hypothesis test using the binomial distribution. It is important to identify the correct tail from the direction of the suspicion, calculate the full tail probability, and state the conclusion in context without claiming certainty.

Common Mistakes

Using a two-tailed test instead of a one-tailed test when the suspicion is $p < \frac{1}{6}$ .
Calculating only $P(X = 2)$ instead of $P(X \le 2)$ .
Not showing the binomial expression; the mark scheme requires the expression or terms to be seen.
Giving the unsupported answer $0.103$ without working; this only earns limited credit.
Comparing the wrong probability to 0.05, for example using $P(X \le 1)$ .

Things to Be Careful About

Use $p = \frac{1}{6}$ in the null distribution, not the observed proportion.
Include all three terms when calculating $P(X \le 2)$ .
Compare the tail probability to 0.05, not to 0.95.
The conclusion must be non-definite and in context: there is insufficient evidence that the probability is less than $\frac{1}{6}$ .

Techniques used

set up null and alternative hypothesesidentify a one-tailed lower binomial testcompute the binomial tail probabilitycompare the p-value with the significance level

(b)

Later, Birgitte carries out a similar test at the 5% significance level, using another 30 throws of the dice.

Calculate the probability of a Type I error.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

A Type I error occurs when $H_0$ is true but the test rejects it. The critical region is the set of values of $X$ for which the rejection-tail probability is below 5%. From part (a), $P(X \le 2) = 0.103 > 0.05$ , while $P(X \le 1) = 0.0295 < 0.05$ , so the critical region is $X \le 1$ . Therefore the probability of a Type I error is $P(X \le 1)$ under $H_0$ .

Working

Under $H_0$ , $X \sim \mathrm{B}\left(30, \frac{1}{6}\right)$ .

\begin{aligned} P(\text{Type I error}) &= P(X \le 1) \\ &= \left(\frac{5}{6}\right)^{30} + 30\left(\frac{5}{6}\right)^{29}\left(\frac{1}{6}\right) \\ &= 0.00421272 + 0.0252763 \\ &= 0.029489... \\ &= 0.0295 \quad (3\text{ s.f.}) \end{aligned}

Answer

The probability of a Type I error is $0.0295$ .

Final answer

0.0295

Detailed explanation

Walkthrough

A Type I error occurs when the null hypothesis is rejected even though it is true. In this test, we reject $H_0$ only when the observed number of sixes lies in the critical region. The critical region is the lower tail whose probability is below 5%. Since $P(X \le 2) = 0.103$ and $P(X \le 1) = 0.0295$ , the critical region is $X \le 1$ . Thus, under $H_0$ , the probability of a Type I error is exactly $P(X \le 1)$ . We compute this by adding the probabilities for $X = 0$ and $X = 1$ under $B(30, \frac{1}{6})$ .

Key Takeaways

A Type I error is the probability of falling into the rejection region when $H_0$ is true. It is not the same as the observed $p$ -value from part (a). The critical region must be identified correctly before the Type I error probability can be calculated.

Common Mistakes

Using $P(X \le 2)$ from part (a) as the Type I error probability.
Using the significance level 0.05 directly as the Type I error probability instead of the actual tail probability.
Forgetting that the critical region is $X \le 1$ because $P(X \le 2)$ exceeds 0.05.
Not showing the binomial expression; an unsupported answer only earns limited credit.

Things to Be Careful About

The Type I error probability is calculated under $H_0$ , so use $p = \frac{1}{6}$ .
Use the lower tail, not the upper tail.
Give the final answer to 3 significant figures: 0.0295.

Techniques used

identify the critical regioncalculate the probability of a Type I errorevaluate the binomial tail probability under the null hypothesis

(c)

Given that the value of $p$ is actually 0.02, calculate the probability of a Type II error.

3M Similar

DifficultyMedium

Worked solution

Approach

A Type II error occurs when $H_0$ is false but the test fails to reject it. Here the true value is $p = 0.02$ , so the true distribution is $X \sim B(30, 0.02)$ . From part (b), the critical region is $X \le 1$ , so the test rejects $H_0$ when $X \le 1$ and fails to reject when $X \ge 2$ . Therefore the probability of a Type II error is $P(X \ge 2 \mid p = 0.02) = 1 - P(X \le 1 \mid p = 0.02)$ .

Working

With $p = 0.02$ , $X \sim \mathrm{B}(30, 0.02)$ .

\begin{aligned} P(X \le 1) &= 0.98^{30} + 30 \times 0.98^{29} \times 0.02 \\ &= 0.54548 + 0.33397 \\ &= 0.87945 \end{aligned}

Therefore

P(\text{Type II error}) = 1 - 0.87945 = 0.12055 = 0.121 \quad (3\text{ s.f.})

Answer

The probability of a Type II error is $0.121$ .

Final answer

0.121

Detailed explanation

Walkthrough

A Type II error occurs when $H_0$ is false but the test fails to reject it. We are told the true probability is $p = 0.02$ , so the true distribution is $B(30, 0.02)$ . The rejection region from part (b) is $X \le 1$ ; if $X \ge 2$ , the test does not reject $H_0$ . Thus a Type II error happens exactly when $X \ge 2$ under $p = 0.02$ . We calculate $P(X \le 1)$ under $p = 0.02$ , then subtract it from 1. The probability of 0 or 1 sixes is $0.98^{30} + 30(0.98)^{29}(0.02)$ , which is about 0.879. Subtracting from 1 gives 0.121.

Key Takeaways

Type II error depends on the true alternative value of $p$ . It is the probability of falling outside the rejection region when the alternative is true. It is often calculated as $1 - P(\text{reject} \mid p_{\text{true}})$ .

Common Mistakes

Confusing Type II error with Type I error. Type II error uses the alternative distribution, not the null distribution.
Using $P(X \le 1)$ as the Type II error probability without subtracting from 1; $P(X \le 1)$ is the rejection probability, not the failure-to-reject probability.
Using $p = \frac{1}{6}$ instead of $p = 0.02$ in the calculation.
Not showing the expression or terms; the mark scheme requires them to be seen.

Things to Be Careful About

The critical region is $X \le 1$ , so the non-rejection region is $X \ge 2$ .
Use $p = 0.02$ for the binomial probabilities in this part.
Give the final answer to 3 significant figures: 0.121.

Techniques used

recognise the Type II error conditioncompute the binomial probability under the alternative valuesubtract from 1 to obtain the Type II error probability

Mathematics 9709/62 — May/June 2025

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