A machine dispenses coffee into cups. The volume, $V \text{ cm}^3$, of coffee in a cup was measured for a random sample of 150 cups. The results were summarised as follows. $$ \sum v = 46350 \quad \sum v^2 = 14410800 $$ (i) Calculate unbiased estimates of the population mean, $\mu$, and population variance, $\sigma^2$. (ii) Calculate a 95% confidence interval for $\mu$. (b) Another random sample of $n$ cups of coffee is taken, where $100 < n < 120$. A 95% confidence interval…

(a)(i) Est(mu) = 309, Est(sigma^2) = 595 (3 s.f.) · (a)(ii) 305 to 313 cm^3 (3 s.f.) · (b) Wider, because a smaller sample size gives a larger standard error.

9709/63 May/June 2025 — Mathematics Solved Past Paper

Q1MediumThe Poisson Distribution

At a certain shop, customers arrive independently and randomly at a constant average rate of 23.4 per hour.

(a)

Find the probability that, in a randomly chosen 1-minute period, at least 2 customers arrive.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since arrivals are independent and random at a constant average rate, the number of arrivals in a fixed interval follows a Poisson distribution. Convert the hourly rate to a 1-minute rate, then find the probability of at least 2 arrivals by subtracting the probabilities of 0 and 1 arrivals from 1.

Working

\lambda = \frac{23.4}{60} = 0.39

Let $Y \sim \mathrm{Po}(0.39)$ . Then

P(Y = r) = \frac{e^{-0.39}(0.39)^r}{r!}

P(Y = 0) = e^{-0.39} = 0.67706

P(Y = 1) = 0.39e^{-0.39} = 0.26405

P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1)

P(Y \geq 2) = 1 - e^{-0.39}(1 + 0.39)

P(Y \geq 2) = 1 - (0.67706 + 0.26405) = 0.05889

Answer

0.0589

to 3 significant figures.

Final answer

0.0589

Detailed explanation

Walkthrough

The shop has a constant average arrival rate of 23.4 customers per hour. Since the arrivals are independent and random, the number of arrivals in any fixed interval is modelled by a Poisson distribution. For a 1-minute period, the average rate must be converted from hours to minutes: divide 23.4 by 60 to get $\lambda = 0.39$ .

Let $Y$ be the number of arrivals in 1 minute, so $Y \sim \mathrm{Po}(0.39)$ . The event at least 2 arrivals means $Y = 2, 3, 4, \dots$ . It is easier to use the complement: $P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1)$ .

Using the Poisson formula $P(Y = r) = e^{-\lambda}\lambda^r/r!$ , calculate $P(Y=0) = e^{-0.39}$ and $P(Y=1) = 0.39e^{-0.39}$ . Adding these gives the probability of 0 or 1 arrivals, and subtracting from 1 gives the required probability. The result is 0.0589 to 3 significant figures.

Key Takeaways

The Poisson distribution is used for counting random events over time.
The parameter $\lambda$ must match the time interval being considered.
For at least probabilities, the complement rule is often the most efficient method.

Common Mistakes

Using $\lambda = 23.4$ for a 1-minute period instead of converting to $0.39$ .
Forgetting to subtract both $P(Y=0)$ and $P(Y=1)$ when finding at least 2.
Giving an unsupported answer: the mark scheme allows only B1 for an unsupported final answer, instead of the full method marks.

Things to Be Careful About

Always check the time units before setting $\lambda$ .
Show the substitution into the Poisson formula so the method marks can be awarded.
Give the final probability to 3 significant figures.

Techniques used

convert the hourly arrival rate to the 1-minute ratecalculate Poisson probabilities for zero and one arrivalsuse the complement rule for at least two arrivals

(b)

The random variable $X$ denotes the number of customers who arrive in a randomly chosen 1-hour period.

5M

(i)

State a suitable approximating distribution for $X$ , giving the value(s) of any parameter(s).

2M Similar

DifficultyEasy

Worked solution

Approach

Since $X$ is a Poisson random variable with a large mean, use the normal approximation to the Poisson distribution. The approximating normal distribution has the same mean and variance as the Poisson distribution.

Working

X \sim \mathrm{Po}(23.4)

For a Poisson distribution, $E(X) = \mathrm{Var}(X) = \lambda = 23.4$ . Therefore, approximately,

X \sim N(23.4, 23.4)

Answer

X \sim N(23.4, 23.4)

Final answer

X ~ N(23.4, 23.4)

Detailed explanation

Walkthrough

Here $X$ is the number of customers arriving in a 1-hour period, so $X \sim \mathrm{Po}(23.4)$ . The mean is 23.4, which is large enough for the normal approximation to the Poisson distribution to be suitable.

For a Poisson distribution, both the mean and the variance are equal to $\lambda$ . Therefore the approximating normal distribution has mean $23.4$ and variance $23.4$ . The standard deviation is not needed for stating the distribution, but it is $\sqrt{23.4}$ and will be used in part (ii).

Key Takeaways

When $\lambda$ is large, a Poisson distribution can be approximated by a normal distribution.
The approximating normal distribution uses the same mean and variance as the Poisson distribution.
The notation $N(\mu, \sigma^2)$ gives the variance as the second parameter.

Common Mistakes

Writing the second parameter as the standard deviation instead of the variance.
Using a binomial or another distribution instead of the normal approximation.
Trying to recover these marks from later working; the mark scheme states that marks for part (i) cannot be recovered from part (ii).

Things to Be Careful About

State the distribution clearly as $X \sim N(23.4, 23.4)$ .
Remember that the normal approximation is only suitable because $\lambda$ is large.

Techniques used

recognise the normal approximation to a Poisson distributionstate the mean and variance of the approximating normal distribution

(ii)

Use your approximating distribution to find $P(20 < X < 30)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the normal approximation with a continuity correction because $X$ is discrete. The inequality $20 < X < 30$ becomes $20.5 < X < 29.5$ after the correction. Standardise both boundaries and find the area between them using the standard normal distribution.

Working

For the normal approximation, $\mu = 23.4$ and $\sigma = \sqrt{23.4}$ .

z_1 = \frac{20.5 - 23.4}{\sqrt{23.4}} = -0.5995

z_2 = \frac{29.5 - 23.4}{\sqrt{23.4}} = 1.2610

P(20 < X < 30) \approx P(-0.5995 < Z < 1.2610)

P(20 < X < 30) \approx \Phi(1.2610) - \Phi(-0.5995)

= 0.8964 - 0.2743 = 0.6221

Answer

0.622

to 3 significant figures.

Final answer

0.622

Detailed explanation

Walkthrough

Since $X$ is discrete but we are using a continuous normal approximation, a continuity correction is needed. The inequality $20 < X < 30$ means $X$ can take integer values 21 to 29 inclusive. In the continuous approximation this is represented by the interval from $20.5$ to $29.5$ .

The approximating normal distribution has $\mu = 23.4$ and $\sigma = \sqrt{23.4}$ . Standardise the two boundaries:

z_1 = \frac{20.5 - 23.4}{\sqrt{23.4}} = -0.5995

z_2 = \frac{29.5 - 23.4}{\sqrt{23.4}} = 1.2610

Using standard normal tables, $\Phi(1.2610) = 0.8964$ and $\Phi(-0.5995) = 0.2743$ . The probability is the difference between these two areas:

0.8964 - 0.2743 = 0.6221

So $P(20 < X < 30) \approx 0.622$ to 3 significant figures.

Key Takeaways

Continuity correction is essential when approximating a discrete distribution with a continuous one.
For a strict inequality such as $20 < X < 30$ , use $20.5$ and $29.5$ .
Standardising converts the normal probability into a standard normal probability, which can be read from tables.

Common Mistakes

Omitting the continuity correction and using 20 and 30 as the boundaries.
Using the wrong continuity correction, such as 19.5 and 30.5.
Subtracting the probabilities in the wrong order.
Forgetting to divide by $\sqrt{23.4}$ when standardising.
An unsupported answer can only score B2, so full working is important.

Things to Be Careful About

The variance is 23.4, so the standard deviation is $\sqrt{23.4}$ , not 23.4.
Use the standard normal table correctly: $\Phi(-0.5995)$ is read as the lower-tail probability.
Give the final answer to 3 significant figures.

Techniques used

apply a continuity correction to the discrete boundsstandardise the bounds using the normal mean and standard deviationuse standard normal tables to find the required probability

Q2Medium-EasySampling and EstimationHypothesis Tests

The lengths of pencils made at a factory are normally distributed. The standard deviation of the lengths is $\sigma$ cm, and the mean is supposed to be 10 cm. An inspector thinks that the mean is actually greater than 10 cm. He takes a random sample of 50 pencils produced at the factory and finds that the mean of these 50 lengths is 10.03 cm. He then carries out a hypothesis test.

(a)

He finds that the value of the test statistic $z$ is 1.995 correct to 3 decimal places.

6M

(i)

Calculate the value of $\sigma$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a sample from a normal population, the sample mean is normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$ . The test statistic is

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

Substitute the given values and solve for $\sigma$ .

Working

z = \frac{10.03 - 10}{\sigma/\sqrt{50}} = 1.995

So

\frac{0.03\sqrt{50}}{\sigma} = 1.995

\sigma = \frac{0.03\sqrt{50}}{1.995} = 0.106331\ldots

Answer

\sigma = 0.106\ \text{cm (3 s.f.)}

Final answer

0.106 cm (3 s.f.)

Detailed explanation

Walkthrough

The inspector uses the sample mean $\bar{x} = 10.03$ to test whether the population mean $\mu$ is greater than 10. Because the population is normal, $\bar{x}$ is normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{50}$ . The test statistic is found by standardising $\bar{x}$ :

z = \frac{\bar{x} - \mu}{\sigma/\sqrt{50}}

We are told this equals 1.995. Substitute $\bar{x} = 10.03$ and $\mu = 10$ :

1.995 = \frac{0.03}{\sigma/\sqrt{50}}

Rearrange: multiply both sides by $\sigma/\sqrt{50}$ , then multiply by $\sqrt{50}$ and divide by 1.995:

\sigma = \frac{0.03\sqrt{50}}{1.995} = 0.106331\ldots

So the population standard deviation is 0.106 cm to 3 significant figures.

Key Takeaways

The formula for the test statistic of a sample mean is $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ . The denominator is the standard error of the mean. Rearranging this formula can recover an unknown population standard deviation.

Common Mistakes

Omitting the square root of 50 in the denominator; the standard error is $\sigma/\sqrt{50}$ , not $\sigma/50$ .
Equating $z$ to a cumulative probability such as $\phi(1.995)$ ; the test statistic is the standardised value 1.995, not a probability.
Rounding too early; keep enough accuracy before giving the final 3 significant figures.

Things to Be Careful About

$\sigma$ is the population standard deviation, not the sample standard deviation.
The sample size is 50, so use $\sqrt{50}$ in the standard error.
The final answer should be stated in cm and to 3 significant figures.

Techniques used

standardise the sample meanrearrange the z formula to solve for the population standard deviation

(ii)

Carry out the hypothesis test at the 2.5% significance level.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Set up a one-tailed hypothesis test for the population mean. The null hypothesis is that the mean is still 10 cm; the alternative is that it is greater than 10 cm. Compare the test statistic $z = 1.995$ with the upper 2.5% critical value $z = 1.960$ .

Working

H_0: \mu = 10, \qquad H_1: \mu > 10

For a one-tailed test at the 2.5% significance level, the critical value is

z_{0.975} = 1.960

Since

1.995 > 1.960

or equivalently the p-value is

P(Z > 1.995) = 0.023 < 0.025

we reject $H_0$ .

Answer

There is sufficient evidence at the 2.5% significance level to suggest that the mean length of pencils is greater than 10 cm.

Final answer

Reject H0. There is sufficient evidence that the mean length is greater than 10 cm.

Detailed explanation

Walkthrough

We are testing whether the population mean length $\mu$ has increased from 10 cm. The inspector believes it is greater than 10, so this is a one-tailed test with alternative hypothesis $H_1: \mu > 10$ . The null hypothesis is always the 'no change' statement $H_0: \mu = 10$ .

The test statistic has already been calculated as $z = 1.995$ . At the 2.5% significance level, a one-tailed upper-tail test rejects $H_0$ if the test statistic exceeds the critical value. For an upper tail of 0.025, the critical value is $z = 1.960$ (because $P(Z \leq 1.960) = 0.975$ ).

Since $1.995 > 1.960$ , the result lies in the critical region. Equivalently, the probability of getting a $z$ -value at least 1.995 when $H_0$ is true is about 0.023, which is less than 0.025. Therefore we reject $H_0$ .

The conclusion must be in context and must not be over-stated: there is sufficient evidence to suggest that the mean length is greater than 10 cm, but this is not a proof.

Key Takeaways

A one-tailed test at significance level $\alpha$ uses the critical value that leaves area $\alpha$ in the relevant tail.
The decision is made by comparing the test statistic with the critical value, or by comparing the p-value with $\alpha$ .
The conclusion must be written in the context of the problem.

Common Mistakes

Using a two-tailed test: for $H_1: \mu > 10$ the test is one-tailed; a two-tailed test would score no marks for hypotheses or conclusion.
Using the wrong critical value: at 2.5% one-tailed the critical value is 1.960, not 1.645 (which is for 5% one-tailed).
Writing hypotheses in terms of the sample mean instead of the population mean $\mu$ .
Stating that the mean 'is' greater than 10 cm; the correct language is 'there is sufficient evidence to suggest'.

Things to Be Careful About

The mark scheme accepts comparing $0.977 > 0.975$ or $0.023 < 0.025$ as an alternative to comparing $1.995 > 1.960$ .
The conclusion must be in context and must not be definite.
Hypotheses must use the population mean, not just 'mean'.

Techniques used

state null and alternative hypothesescompare test statistic with one-tailed critical valuewrite the conclusion in context

(b)

Explain whether it was necessary to use the Central Limit Theorem in carrying out the test.

1M Similar

DifficultyEasy

Worked solution

Approach

Decide whether the normal distribution of the sample mean needed the Central Limit Theorem. Since the original population of pencil lengths is stated to be normally distributed, the sample mean is exactly normally distributed for any sample size.

Working

The population distribution is normal, so

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

exactly, without needing a large sample. The Central Limit Theorem would only be needed to justify approximate normality if the population were not normal and the sample were large.

Answer

No. It was not necessary to use the Central Limit Theorem because the population distribution of lengths is normal.

Final answer

No; the population distribution is normal, so the sample mean is exactly normal.

Detailed explanation

Walkthrough

The test statistic $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$ is valid only if $\bar{x}$ is normally distributed. There are two ways this can happen:

The population itself is normal, in which case the sample mean is exactly normal for any sample size.
The population is not normal, but the sample is large enough for the Central Limit Theorem to make the sample mean approximately normal.

Here the question states that the lengths of pencils are normally distributed. Therefore the sample mean is exactly normal, and the Central Limit Theorem is not needed.

Key Takeaways

The Central Limit Theorem is useful for large samples from non-normal populations, but it is unnecessary when the population is already normal.

Common Mistakes

Saying the Central Limit Theorem is needed because the sample size is 50. This is incorrect when the population is normal.
Confusing the exact normality of the sample mean from a normal population with the approximate normality provided by the Central Limit Theorem.

Things to Be Careful About

The answer is a one-mark explanation. It must clearly state that the population distribution is normal, not just mention the large sample size.

Techniques used

recognise exact normality of the sample mean from a normal populationdistinguish exact normality from Central Limit Theorem approximation

Q3Medium-EasySampling and Estimation

A machine dispenses coffee into cups. The volume, $V \text{ cm}^3$ , of coffee in a cup was measured for a random sample of 150 cups. The results were summarised as follows.

\sum v = 46350 \quad \sum v^2 = 14410800

(a)

6M

(i)

Calculate unbiased estimates of the population mean, $\mu$ , and population variance, $\sigma^2$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the sample mean to estimate the population mean. To estimate the population variance, use the unbiased formula with $n-1$ in the denominator.

Working

\bar{v} = \frac{\sum v}{n} = \frac{46350}{150} = 309

So $Est(\mu) = 309$ .

Unbiased estimate of the population variance:

Est(\sigma^2) = \frac{1}{n-1}\left[\sum v^2 - \frac{(\sum v)^2}{n}\right]

Substitute $n=150$ , $\sum v = 46350$ , $\sum v^2 = 14410800$ :

Est(\sigma^2) = \frac{1}{149}\left[14410800 - \frac{46350^2}{150}\right]

Compute the bracket:

\frac{46350^2}{150} = \frac{2148322500}{150} = 14322150

Therefore

Est(\sigma^2) = \frac{1}{149}\left[14410800 - 14322150\right] = \frac{88650}{149} = 595 \quad (3\text{ sf})

Answer

$Est(\mu) = 309$ and $Est(\sigma^2) = 595$ (3 sf).

Final answer

Est(mu) = 309, Est(sigma^2) = 595 (3 s.f.)

Detailed explanation

Walkthrough

We have a sample of 150 cups, with $\sum v = 46350$ and $\sum v^2 = 14410800$ . The mean is the total divided by the sample size:

\bar{v} = \frac{46350}{150} = 309

This is the unbiased estimate of the population mean.

For the variance, we use the unbiased estimator

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

The term $\frac{(\sum v)^2}{n}$ corrects for using the sample mean instead of the true population mean. Substituting the values:

s^2 = \frac{1}{149}(14410800 - 14322150) = \frac{88650}{149} = 595

Key Takeaways

The sample mean is an unbiased estimate of the population mean.
The unbiased variance uses $n-1$ in the denominator.
The formula $\frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right)$ is the standard way to find it from summary data.

Common Mistakes

Using $n$ instead of $n-1$ in the denominator, giving the biased variance.
Forgetting to divide $(\sum v)^2$ by $n$ before subtracting.
Arithmetic errors in squaring 46350.

Things to Be Careful About

The unbiased estimate is $595$ (3 s.f.), not the biased value $591$ .
Keep the exact fraction $88650/149$ if you need it in later parts.
Variance has units $(\text{cm}^3)^2$ .

Techniques used

calculate the sample meanapply the unbiased variance formulasimplify the arithmetic expression

(ii)

Calculate a 95% confidence interval for $\mu$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Use the 95% confidence interval formula

\bar{x} \pm z \times \sqrt{\frac{s^2}{n}}

with $z = 1.96$ for a 95% two-sided interval, and $s^2 = 595$ from part (a)(i).

Working

Standard error:

\sqrt{\frac{s^2}{n}} = \sqrt{\frac{595}{150}} = \sqrt{3.9667} = 1.9916

Margin of error:

1.96 \times 1.9916 = 3.9036

Confidence interval:

309 \pm 3.9036

Lower bound:

309 - 3.9036 = 305.0964

Upper bound:

309 + 3.9036 = 312.9036

So to 3 significant figures:

305 \text{ to } 313

Answer

The 95% confidence interval is $305$ to $313$ cm $^3$ (3 sf).

Final answer

305 to 313 cm^3 (3 s.f.)

Detailed explanation

Walkthrough

The 95% confidence interval for the mean is

\bar{x} \pm z \times \sqrt{\frac{s^2}{n}}

where $z = 1.96$ is the critical value for a 95% two-sided interval. We use the unbiased estimate $s^2 = 595$ from part (a)(i) and $n = 150$ .

First find the standard error:

\sqrt{\frac{s^2}{n}} = \sqrt{\frac{595}{150}} = 1.9916

The margin of error is:

1.96 \times 1.9916 = 3.9036

So the interval is

309 \pm 3.9036

Lower limit:

309 - 3.9036 = 305.0964

Upper limit:

309 + 3.9036 = 312.9036

Rounded to 3 significant figures, the interval is 305 to 313.

Key Takeaways

A 95% confidence interval uses $z = 1.96$ .
The standard error is $\sqrt{s^2/n}$ .
The interval is centred at the sample mean.

Common Mistakes

Using the biased variance $591$ instead of the unbiased $595$ .
Forgetting to take the square root of $s^2/n$ .
Using the wrong critical value, e.g. $1.64$ for 90% or $2.575$ for 99%.
Giving an unsupported answer without showing the formula.

Things to Be Careful About

The final interval must be given to 3 significant figures: 305 to 313.
The width of the interval is approximately $2 \times 3.9036 \approx 7.81$ .
Units are $\text{cm}^3$ .

Techniques used

identify the critical value for 95% confidencecompute the standard errorconstruct the confidence interval around the sample mean

(b)

Another random sample of $n$ cups of coffee is taken, where $100 < n < 120$ . A 95% confidence interval for $\mu$ is calculated using this sample. You may assume that, for large samples, unbiased estimates of $\sigma^2$ are very similar.

Without calculation, state whether this confidence interval would be wider or narrower than the confidence interval found in part (a)(ii). Give a reason for your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The width of a confidence interval depends on the standard error $\frac{s}{\sqrt{n}}$ . Since the new sample has a smaller size than 150, its standard error is larger, so the interval is wider.

Working

For a 95% confidence interval, the half-width is

1.96 \times \frac{s}{\sqrt{n}}

The new sample has $100 < n < 120$ , so $n < 150$ . Therefore

\frac{1}{\sqrt{n}} > \frac{1}{\sqrt{150}}

so the standard error is larger for the new sample, and the confidence interval is wider.

Answer

The interval would be wider, because the smaller sample size gives a larger standard error and therefore less precision.

Final answer

Wider, because a smaller sample size gives a larger standard error.

Detailed explanation

Walkthrough

The width of a confidence interval is controlled by the standard error $s/\sqrt{n}$ . A smaller sample size gives a larger standard error, because dividing by a smaller number of observations gives a larger value. Here the new sample has $100 < n < 120$ , which is smaller than $150$ . Therefore

\frac{1}{\sqrt{n}} > \frac{1}{\sqrt{150}}

so the standard error is larger and the interval is wider.

Key Takeaways

Confidence interval width is inversely proportional to the square root of the sample size.
Smaller samples produce wider intervals because they give less information about the population.
The sample size affects the standard error even when the variance estimate is similar.

Common Mistakes

Saying the interval is narrower because a smaller sample is "more precise".
Giving only the conclusion without mentioning the standard error or $1/\sqrt{n}$ .

Things to Be Careful About

The question asks for no calculation, just a comparison and a reason.
You must mention both the conclusion (wider) and the reason (smaller $n$ gives a larger standard error).

Techniques used

relate confidence interval width to sample sizecompare standard errors for different sample sizesdraw a conclusion about the interval width

Q4Medium-EasySampling and Estimation

Emma needs to choose one person at random from three people, $P$ , $Q$ and $R$ . She plans to throw two fair coins and note the number, $n$ , of heads. If $n$ is 0, she will choose $P$ . If $n$ is 1, she will choose $Q$ . If $n$ is 2, she will choose $R$ .

(a)

By considering probabilities, show that the choice made by this method is not random.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

A random choice means each of $P$ , $Q$ and $R$ has the same probability. Count the number of heads from two fair coins.

Working

The two coins have four equally likely outcomes:

\begin{array}{ccl} TT &\Rightarrow & n = 0 \\ HT, TH &\Rightarrow & n = 1 \\ HH &\Rightarrow & n = 2 \end{array}

So

P(n = 0) = \frac{1}{4}, \quad P(n = 1) = \frac{2}{4} = \frac{1}{2}, \quad P(n = 2) = \frac{1}{4}

These probabilities are not equal; in particular $P(n=1) = 2P(n=0)$ . Therefore $P$ , $Q$ and $R$ are not chosen with equal probability.

Answer

The probabilities are not equal, so the choice is not random.

Final answer

P(n=0)=1/4, P(n=1)=1/2, P(n=2)=1/4; probabilities are not equal, so the choice is not random.

Detailed explanation

Walkthrough

A fair coin has two equally likely outcomes, so two fair coins have four equally likely outcomes: TT, HT, TH and HH. Count the number of heads in each outcome. Only TT gives 0 heads, both HT and TH give 1 head, and HH gives 2 heads. Therefore the probabilities are $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{1}{4}$ respectively. Since these are not equal, the method is not random: Q is more likely to be chosen than P or R.

Key Takeaways

Random means each choice has the same probability. A fair coin produces equally likely outcomes, and to test whether a method is random we compare the probabilities of all possible outcomes.

Common Mistakes

Forgetting that HT and TH are two different outcomes. Also, saying that the probabilities are not equal without giving any numerical justification may lose marks.

Things to Be Careful About

$P(n=1) = \frac{1}{2}$ , not $\frac{1}{3}$ . The mark scheme awards one mark for stating that the probabilities are not equal and one mark for a numerical justification.

Techniques used

list all equally likely outcomescalculate probabilities of each number of headscompare probabilities to assess randomness

(b)

Later, Emma has to choose two people at random from three people.

Describe how Emma could use a single throw of a fair six-sided dice to make this random choice.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

There are three possible pairs: $PQ$ , $QR$ and $RP$ . A single fair six-sided die has six equally likely outcomes, so use the six outcomes in three groups of two.

Working

Using one die score to choose a pair:

\begin{array}{c|c} \text{Score} & \text{Pair chosen} \\ \hline 1 \text{ or } 2 & \text{PQ} \\ 3 \text{ or } 4 & \text{QR} \\ 5 \text{ or } 6 & \text{RP} \end{array}

Each pair is chosen for two of the six equally likely scores, so each has probability $\frac{2}{6} = \frac{1}{3}$ .

Equivalent method: use the score to reject one person.

1 \text{ or } 2 \Rightarrow \text{reject } R, \text{ choose } PQ

3 \text{ or } 4 \Rightarrow \text{reject } P, \text{ choose } QR

5 \text{ or } 6 \Rightarrow \text{reject } Q, \text{ choose } RP

Again each pair is chosen with probability $\frac{1}{3}$ .

Answer

Throw one fair six-sided die once. If the score is 1 or 2, choose $PQ$ ; if 3 or 4, choose $QR$ ; if 5 or 6, choose $RP$ . (Or equivalently reject $R$ , $P$ or $Q$ respectively.)

Final answer

Use one fair die: scores 1-2 choose PQ, 3-4 choose QR, 5-6 choose RP (or equivalently reject R, P, Q respectively).

Detailed explanation

Walkthrough

A single throw of a fair six-sided die has six equally likely outcomes. Since there are three possible pairs, group the six outcomes into three blocks of two. Each block has probability $\frac{2}{6} = \frac{1}{3}$ , so each pair is equally likely. Alternatively, use the score to reject one person: scores 1 or 2 reject R and choose PQ, scores 3 or 4 reject P and choose QR, scores 5 or 6 reject Q and choose RP. This also gives each pair probability $\frac{1}{3}$ .

Key Takeaways

A random choice can be made by grouping equally likely outcomes. A fair die has equally likely outcomes, and mapping groups of outcomes onto choices makes each choice equally likely.

Common Mistakes

Using more than one throw of the dice scores B0B0. Also, choosing a single person instead of a pair is incorrect because two people must be chosen.

Things to Be Careful About

The method must use a single die, not multiple throws. Each pair must be assigned exactly two scores, and all three pairs must be covered.

Techniques used

identify the three possible pairsmap dice outcomes to pairs in equal groupscheck equal probabilities

Q54MMediumLinear Combinations of Random VariablesSampling and Estimation

In Urberia, the masses, in kilograms, of men have the distribution $N(70.3, 5.9^2)$ . A certain footbridge in Urberia can take a maximum safe load of 1500 kg. When $n$ men stand on the bridge, the probability that the bridge is unsafe is less than 0.01.

Stating a necessary assumption, find the maximum value of $n$ .

DifficultyMedium

Worked solution

Approach

Let $X$ be the mass of one man, so $X \sim N(70.3, 5.9^2)$ . For $n$ independent men, the total mass $T = X_1 + \cdots + X_n$ is normally distributed with mean $70.3n$ and variance $n(5.9^2)$ , so its standard deviation is $5.9\sqrt{n}$ . The bridge is unsafe if $T > 1500$ . We need $P(T > 1500) < 0.01$ . Since $P(Z > 2.326) = 0.01$ , we solve

\frac{1500 - 70.3n}{5.9\sqrt{n}} = 2.326

to find the boundary value of $n$ , then choose the largest integer for which the probability is below $0.01$ .

Working

Assume that the men on the bridge are a random sample, so their masses are independent.

The total mass is

T \sim N(70.3n,\; n(5.9)^2).

Standardising the unsafe value 1500 kg:

\frac{1500 - 70.3n}{5.9\sqrt{n}} = 2.326.

Multiply through:

1500 - 70.3n = 2.326 \times 5.9\sqrt{n} = 13.7234\sqrt{n}.

Rearrange:

70.3n + 13.7234\sqrt{n} - 1500 = 0.

Let $y = \sqrt{n}$ :

70.3y^2 + 13.7234y - 1500 = 0.

Using the quadratic formula:

y = \frac{-13.7234 + \sqrt{13.7234^2 + 4(70.3)(1500)}}{2(70.3)}.

y = 4.5226 \quad (3\text{ sf}).

So $\sqrt{n} = 4.5226$ , giving $n = 20.45$ . Since $n$ must be an integer and the probability must be strictly less than $0.01$ , the maximum value is $n = 20$ .

Answer

n = 20

Final answer

n = 20

Detailed explanation

Walkthrough

The question gives the distribution of the mass of one man. When $n$ men stand on the bridge, the total mass is the sum of $n$ independent normal variables. We must assume that the men are a random sample, so their masses are independent. Then the total mass $T$ is also normal, with mean $70.3n$ and variance $n(5.9^2)$ ; hence its standard deviation is $5.9\sqrt{n}$ .

The bridge is unsafe exactly when $T > 1500$ . We need this probability to be less than $0.01$ . For a standard normal variable $Z$ , $P(Z > 2.326) = 0.01$ , so we find the value of $n$ for which the standardized value of 1500 equals 2.326. This gives the boundary; any smaller integer $n$ makes the probability smaller.

Standardizing gives

\frac{1500 - 70.3n}{5.9\sqrt{n}} = 2.326.

Multiplying by $5.9\sqrt{n}$ and rearranging produces a quadratic in $\sqrt{n}$ :

70.3n + 13.7234\sqrt{n} - 1500 = 0.

Writing $y = \sqrt{n}$ and using the quadratic formula gives $y = 4.5226$ , so $n = 20.45$ . Since $n$ must be a whole number and the probability must be strictly below $0.01$ , we take the largest integer below 20.45, namely $n = 20$ .

Key Takeaways

This question combines the distribution of a sum of independent normal variables with the normal distribution's upper-tail percentage points. It also tests the ability to solve a quadratic in $\sqrt{n}$ and to interpret an inequality for an integer variable. The key facts are:

If $X_i \sim N(\mu, \sigma^2)$ independently, then $\sum X_i \sim N(n\mu, n\sigma^2)$ .
The upper 1% point of the standard normal distribution is $z = 2.326$ .
A condition like $P(T > 1500) < 0.01$ translates into a $z$ -value greater than 2.326.

Common Mistakes

Omitting the assumption of a random sample or independence loses the first mark.
Using $5.9n$ instead of $5.9\sqrt{n}$ for the standard deviation of the total.
Using the wrong critical value, such as 1.96 (for a two-tailed 5% test) or 2.576 (for 1% two-tailed), instead of 2.326.
Rounding $n = 20.45$ up to 21 without checking the inequality. Since the probability must be less than 0.01, $n = 21$ would make it greater than 0.01.
When squaring the equation to obtain a quadratic in $n$ , introducing an extraneous root; always check that $\sqrt{n}$ is positive.

Things to Be Careful About

The bridge is unsafe if the total load exceeds 1500 kg, so the inequality is $T > 1500$ .
The condition is strict: $< 0.01$ . At the boundary the probability equals 0.01, so the integer value must be strictly below the boundary.
Use the correct $z$ -value for the upper tail: $P(Z > 2.326) = 0.01$ .
The final answer must be an integer; $n = 20$ , not 20.45.
The mark scheme requires the square root of $n$ to appear in the standardisation step.

Techniques used

assume a random sample of independent menform the distribution of the sum of independent normal variablesstandardise using the upper 1% normal critical valuesolve a quadratic equation in sqrt(n)round down to the largest integer satisfying the probability condition

Approach

Since $f(x)$ is a probability density function, the total area under the curve over its support must equal 1. The curve $y = ax$ is a straight line through the origin, so the area under it from $x = 0$ to $x = b$ is a triangle with base $b$ and height $ab$ . Equate this area to 1 and solve for $a$ .

Working

The area under the PDF is:

\int_0^b ax \,dx = 1

Evaluating the integral:

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

\frac{ab^2}{2} = 1

Solving for $a$ :

a = \frac{2}{b^2}

Alternatively, using the triangle area: $\frac{1}{2} \times b \times ab = \frac{ab^2}{2} = 1$ , giving the same result.

Answer

a = \frac{2}{b^2}

Final answer

a = 2/b^2

Detailed explanation

Walkthrough

The key property of any probability density function is that the total area under the curve over its entire range must equal 1, because the total probability of all possible outcomes is 1. Here the PDF is $f(x) = ax$ for $0 \leq x \leq b$ and 0 elsewhere, so we only need to consider the interval from 0 to b.

The graph of $y = ax$ is a straight line passing through the origin, rising to the point $(b, ab)$ . The area under this line from $x = 0$ to $x = b$ forms a right triangle with base $b$ and height $ab$ . The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$ , so the area is $\frac{1}{2} \times b \times ab = \frac{ab^2}{2}$ .

Setting this area equal to 1 gives $\frac{ab^2}{2} = 1$ , and multiplying both sides by 2 and dividing by $b^2$ gives $a = \frac{2}{b^2}$ .

Alternatively, we can integrate: $\int_0^b ax\,dx = a\left[\frac{x^2}{2}\right]_0^b = \frac{ab^2}{2}$ , and setting this equal to 1 gives the same result.

Key Takeaways

A probability density function must integrate to 1 over its entire support.
For a linear PDF of the form $f(x) = ax$ , the area under the curve is a triangle, which can be computed using the triangle area formula as a shortcut.
The constant in a PDF is determined by the normalisation condition.

Common Mistakes

Forgetting to set the integral (or area) equal to 1.
Using the wrong limits — the integral must run over the full support $0$ to $b$ .
Dropping the factor of $\frac{1}{2}$ when using the triangle area formula.
Confusing the height of the triangle: at $x = b$ , the height is $f(b) = ab$ , not $a$ .

Things to Be Careful About

The mark scheme requires a convincing derivation with no errors — show the integration step $\int_0^b ax\,dx = 1$ or the triangle area statement explicitly.
Make sure the final expression is $a = \frac{2}{b^2}$ with no algebraic slips.

Techniques used

integrate the PDF over its support and set equal to 1use the triangle area formula for a linear PDF

(b)

Show that $P(X < E(X)) = \frac{4}{9}$ .

6M Similar

DifficultyMedium

Worked solution

Approach

First compute the expectation $E(X) = \int_0^b xf(x)\,dx$ using $a = \frac{2}{b^2}$ from part (a). Then compute the probability $P(X < E(X))$ by integrating $f(x)$ from 0 up to the value of $E(X)$ .

Working

Compute the expectation:

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

E(X) = \frac{2}{b^2}\left[\frac{x^3}{3}\right]_0^b = \frac{2}{b^2} \cdot \frac{b^3}{3} = \frac{2b}{3}

Now compute the probability that $X$ is less than this value:

P(X < E(X)) = P\left(X < \frac{2b}{3}\right) = \int_0^{\frac{2b}{3}} \frac{2}{b^2}x \,dx

P(X < E(X)) = \frac{2}{b^2}\left[\frac{x^2}{2}\right]_0^{\frac{2b}{3}}

P(X < E(X)) = \frac{2}{b^2} \cdot \frac{\left(\frac{2b}{3}\right)^2}{2} = \frac{2}{b^2} \cdot \frac{\frac{4b^2}{9}}{2}

P(X < E(X)) = \frac{2}{b^2} \cdot \frac{2b^2}{9} = \frac{4}{9}

Answer

P(X < E(X)) = \frac{4}{9}

Final answer

P(X < E(X)) = 4/9

Detailed explanation

Walkthrough

To find $P(X < E(X))$ , we need two things: the mean $E(X)$ and then the probability that $X$ falls below it.

First, the mean of a continuous random variable is given by $E(X) = \int_{-\infty}^{\infty} xf(x)\,dx$ . Since $f(x) = 0$ outside $0 \leq x \leq b$ , this becomes $E(X) = \int_0^b xf(x)\,dx$ . Substituting $f(x) = \frac{2}{b^2}x$ from part (a), we get $E(X) = \frac{2}{b^2}\int_0^b x^2\,dx$ .

Evaluating: $\int_0^b x^2\,dx = \left[\frac{x^3}{3}\right]_0^b = \frac{b^3}{3}$ . So $E(X) = \frac{2}{b^2} \cdot \frac{b^3}{3} = \frac{2b}{3}$ .

Now, the probability that $X < \frac{2b}{3}$ is the area under the PDF from 0 to $\frac{2b}{3}$ : $P\left(X < \frac{2b}{3}\right) = \int_0^{\frac{2b}{3}} f(x)\,dx = \frac{2}{b^2}\int_0^{\frac{2b}{3}} x\,dx$ .

Evaluating: $\int_0^{\frac{2b}{3}} x\,dx = \left[\frac{x^2}{2}\right]_0^{\frac{2b}{3}} = \frac{\left(\frac{2b}{3}\right)^2}{2} = \frac{\frac{4b^2}{9}}{2} = \frac{2b^2}{9}$ .

So $P\left(X < \frac{2b}{3}\right) = \frac{2}{b^2} \cdot \frac{2b^2}{9} = \frac{4}{9}$ .

Alternatively, using triangle areas: the PDF is a triangle with base $b$ and height $\frac{2}{b^2} \cdot b = \frac{2}{b}$ . The region $X < \frac{2b}{3}$ is a smaller similar triangle with base $\frac{2b}{3}$ and height $\frac{2}{b^2} \cdot \frac{2b}{3} = \frac{4}{3b}$ . Its area is $\frac{1}{2} \times \frac{2b}{3} \times \frac{4}{3b} = \frac{4}{9}$ .

Key Takeaways

The mean of a continuous random variable is found by integrating $xf(x)$ over the support.
Probabilities for continuous random variables are areas under the PDF, found by integrating $f(x)$ between the relevant limits.
When the PDF is linear, triangle area ratios can provide a quick check.

Common Mistakes

Forgetting to multiply by $x$ when computing $E(X)$ — a common error is to integrate $f(x)$ instead of $xf(x)$ .
Using the wrong upper limit in the probability integral — it must be $E(X) = \frac{2b}{3}$ , not $b$ .
Forgetting the factor of $\frac{1}{2}$ in the triangle area approach.
Losing the $\frac{2}{b^2}$ factor when substituting the PDF.

Things to Be Careful About

The mark scheme allows using the triangle area approach for the probability, but requires the correct expression after substituting the limits.
When using the "area" method, the height of the triangle at $x = \frac{2b}{3}$ is $f\left(\frac{2b}{3}\right) = \frac{2}{b^2} \cdot \frac{2b}{3} = \frac{4}{3b}$ .
The result must be shown convincingly with no errors — write out the substitution of limits explicitly.

Techniques used

compute the expectation by integrating xf(x)compute a probability by integrating the PDF up to a limitsubstitute the normalisation constant

Q7Medium-HardHypothesis Tests

In the past, one quarter of job applicants at a certain firm had first-class degrees. A change is made in the job description and a director of the firm believes that, on average, the proportion of job applicants with first class degrees has decreased.

In the month following the change, there were 35 job applicants, and $r$ of these had first-class degrees. The firm carried out a hypothesis test at the 4% significance level to test the director's belief.

(a)

Use a binomial distribution to find the largest value of $r$ that would provide sufficient evidence that the director's belief is correct.

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

Set up a one-tailed test. The null hypothesis is $H_0: p = 0.25$ and the alternative is $H_1: p < 0.25$ , where $p$ is the proportion of applicants with first-class degrees. The test statistic is $X \sim B(35, 0.25)$ , the number of applicants with first-class degrees. The critical region is the set of values $r$ with $P(X \leq r) \leq 0.04$ . Compute $P(X \leq 3)$ and $P(X \leq 4)$ and compare each with $0.04$ .

Working

With $X \sim B(35, 0.25)$ :

\begin{aligned} P(X=0) &= 0.75^{35} = 0.000042378 \\ P(X=1) &= 35 \times 0.75^{34} \times 0.25 = 0.0004944 \\ P(X=2) &= {}^{35}C_2 \times 0.75^{33} \times 0.25^2 = 0.00280168 \\ P(X=3) &= {}^{35}C_3 \times 0.75^{32} \times 0.25^3 = 0.01027283 \end{aligned}

So

P(X \leq 3) = 0.000042378 + 0.0004944 + 0.00280168 + 0.01027283 = 0.0136 \quad (< 0.04)

Now add the next term:

P(X=4) = {}^{35}C_4 \times 0.75^{31} \times 0.25^4 = 0.0273942

P(X \leq 4) = 0.0136 + 0.0273942 = 0.0410 \quad (> 0.04)

Since $P(X \leq 3) < 0.04$ but $P(X \leq 4) > 0.04$ , the critical region is $X \leq 3$ .

Answer

The largest value of $r$ that provides sufficient evidence is $r = 3$ .

Final answer

r = 3

Detailed explanation

Walkthrough

We want to find the largest number of first-class degree holders $r$ such that, if we observed $r$ or fewer, we would reject the null hypothesis at the 4% level. Under the null hypothesis the proportion is 0.25, so with 35 applicants the number with first-class degrees follows $B(35, 0.25)$ . Because the director believes the proportion has decreased, this is a one-tailed test in the lower tail. The critical region consists of small values of $r$ for which $P(X \leq r) \leq 0.04$ . We first compute $P(X \leq 3)$ by adding the binomial probabilities for 0, 1, 2 and 3 successes. This sum is 0.0136, which is less than 0.04, so observing 3 or fewer is significant. We then check whether including 4 would still be significant: $P(X \leq 4) = 0.0410$ , which exceeds 0.04. So 4 is not in the critical region. The largest value of $r$ that is significant is therefore 3.

Key Takeaways

This question tests how to perform a one-tailed binomial hypothesis test and how to determine the critical region by comparing cumulative probabilities with the significance level. It also reinforces that the critical region is the set of values whose cumulative probability under the null hypothesis is at most the significance level.

Common Mistakes

Using a two-tailed test instead of a one-tailed test — the director's belief is one-directional (decreased).
Forgetting to compare both $P(X \leq 3)$ and $P(X \leq 4)$ with 0.04; you must check the boundary to find the largest value.
Using the wrong parameter (e.g. 0.75 instead of 0.25) — the null proportion is 0.25, the probability of a first-class degree.
Rounding intermediate probabilities too early, which can change the comparison.
The mark scheme notes that an unsupported answer of 0.0136 or 0.041 gets only B1 — full working must be shown.

Things to Be Careful About

The significance level is 4%, written as 0.04.
The comparison must be strict: the critical region includes values with $P \leq 0.04$ ; $P(X \leq 4) = 0.0410 > 0.04$ so 4 is excluded.
Keep enough decimal places in intermediate terms so the final comparisons are reliable.

Techniques used

set up a one-tailed binomial hypothesis testcompute cumulative binomial probabilitiescompare cumulative probabilities with the significance levelidentify the critical region

(b)

In another month, the director carries out a similar test at the 4% significance level using the 35 job applicants from that month.

Explain the meaning of a Type I error in this context, and state the probability of a Type I error.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

A Type I error is rejecting the null hypothesis when it is actually true. In this context, that means concluding that the proportion of applicants with first-class degrees has decreased when in fact it has not. The probability of a Type I error is the probability of falling in the critical region under the null hypothesis, which is $P(X \leq 3)$ with $X \sim B(35, 0.25)$ .

Working

From part (a), the critical region is $X \leq 3$ . Under $H_0$ , $X \sim B(35, 0.25)$ , so

P(\text{Type I error}) = P(X \leq 3) = 0.0136

Answer

A Type I error means concluding that the proportion of applicants with first-class degrees has decreased when in fact it has not. The probability of a Type I error is $0.0136$ .

Final answer

Type I error: concluding the proportion has decreased when it has not; probability = 0.0136

Detailed explanation

Walkthrough

A Type I error occurs when we reject the null hypothesis even though it is true. Here the null hypothesis is that the proportion is still 0.25, so a Type I error means we conclude the proportion has decreased when actually it has not changed. The probability of a Type I error is exactly the probability of the critical region under the null hypothesis, which we found in part (a) to be $P(X \leq 3) = 0.0136$ .

Key Takeaways

The probability of a Type I error equals the significance level only when the critical region is chosen so that its cumulative probability under the null is exactly the significance level; here it is the actual probability of the critical region, 0.0136, which is below 0.04.

Common Mistakes

Confusing Type I and Type II errors. Type I is rejecting a true null; Type II is failing to reject a false null.
Stating the significance level (0.04) as the probability of a Type I error instead of the actual probability of the critical region (0.0136).

Things to Be Careful About

The probability of a Type I error is the probability of the critical region under the null hypothesis, not the significance level itself.
Use the value from part (a): 0.0136.

Techniques used

define a Type I error in contextstate the probability of a Type I error from the critical region

(c)

Given that the proportion of job applicants with first class degrees this year is actually 0.05, find the probability of a Type II error.

2M Similar

DifficultyMedium

Worked solution

Approach

A Type II error is failing to reject the null hypothesis when it is false. The critical region from part (a) is $X \leq 3$ , so a Type II error occurs when $X > 3$ . With the true proportion $p = 0.05$ , $X \sim B(35, 0.05)$ , and we need $P(X > 3) = 1 - P(X \leq 3)$ .

Working

With $X \sim B(35, 0.05)$ :

\begin{aligned} P(X=0) &= 0.95^{35} = 0.16608 \\ P(X=1) &= 35 \times 0.95^{34} \times 0.05 = 0.305943 \\ P(X=2) &= {}^{35}C_2 \times 0.95^{33} \times 0.05^2 = 0.2737385 \\ P(X=3) &= {}^{35}C_3 \times 0.95^{32} \times 0.05^3 = 0.158480 \end{aligned}

So

P(X \leq 3) = 0.16608 + 0.305943 + 0.2737385 + 0.158480 = 0.9042

Therefore

P(\text{Type II error}) = 1 - P(X \leq 3) = 1 - 0.9042 = 0.0958

Answer

The probability of a Type II error is $0.0958$ (3 sf).

Final answer

0.0958

Detailed explanation

Walkthrough

A Type II error happens when the test fails to reject the null hypothesis even though it is false. The critical region from part (a) is $X \leq 3$ , so the test rejects $H_0$ only when $X \leq 3$ . A Type II error therefore occurs when $X > 3$ . Now the true proportion is given as 0.05, so under the alternative the number of first-class degree holders follows $B(35, 0.05)$ . We compute $P(X \leq 3)$ by adding the binomial probabilities for 0, 1, 2 and 3 successes, which gives 0.9042. The probability of a Type II error is the complement: $1 - 0.9042 = 0.0958$ .

Key Takeaways

A Type II error is the failure to reject a false null hypothesis. Its probability is computed under the true (alternative) distribution, using the critical region determined under the null hypothesis.

Common Mistakes

Computing $P(X \leq 3)$ instead of $1 - P(X \leq 3)$ — the Type II error is the complement.
Using the null proportion 0.25 instead of the true proportion 0.05 when computing the probability.
Using the wrong critical region (e.g. $X \leq 4$ instead of $X \leq 3$ ).

Things to Be Careful About

The critical region is $X \leq 3$ , so the Type II error region is $X > 3$ .
Use $p = 0.05$ for the binomial distribution in this part.
Give the final answer to 3 significant figures: 0.0958.

Techniques used

identify the Type II error conditioncompute binomial probabilities under the alternative proportionuse the complement rule

Mathematics 9709/63 — May/June 2025

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