9709/62 Oct/Nov 2024 — Mathematics Solved Past Paper

Q13MMedium-EasyThe Poisson Distribution

A random variable $X$ has the distribution $\text{B}\left(4500000, \frac{1}{1000000}\right)$ .

Use a Poisson distribution to calculate an estimate of $\text{P}(X \ge 4)$ .

DifficultyMedium-Easy

Worked solution

Approach

Since $n = 4500000$ is large and $p = \frac{1}{1000000}$ is small, approximate $X$ by a Poisson distribution with parameter

\lambda = np = 4500000 \times \frac{1}{1000000} = 4.5

We need $\text{P}(X \ge 4)$ ; use the complement $1 - \text{P}(X \le 3)$ and sum the Poisson probabilities for $0,1,2,3$ .

Working

Let $Y \sim \text{Po}(4.5)$ be the Poisson approximation to $X$ .

\text{P}(Y \ge 4) = 1 - \text{P}(Y \le 3)

Using the Poisson probability formula $\text{P}(Y = r) = e^{-\lambda}\frac{\lambda^r}{r!}$ ,

\text{P}(Y \le 3) = e^{-4.5}\left(1 + 4.5 + \frac{4.5^2}{2!} + \frac{4.5^3}{3!}\right)

Evaluate the factorial terms:

\frac{4.5^2}{2!} = 10.125, \qquad \frac{4.5^3}{3!} = 15.1875

So

\text{P}(Y \le 3) = e^{-4.5}(1 + 4.5 + 10.125 + 15.1875)

\text{P}(Y \le 3) \approx 0.011109 + 0.049999 + 0.112480 + 0.168720 = 0.342308

Therefore

\text{P}(X \ge 4) = 1 - 0.342308 = 0.657692

Answer

\text{P}(X \ge 4) \approx 0.658

Final answer

0.658

Detailed explanation

Walkthrough

This question tests the Poisson approximation to the binomial distribution. The random variable is $X \sim \text{B}(4500000, 1/1000000)$ . Since $n$ is very large and $p$ is very small, $X$ can be approximated by a Poisson distribution with the same mean.

First find the mean:

\lambda = np = 4500000 \times \frac{1}{1000000} = 4.5

So $X$ is approximated by $Y \sim \text{Po}(4.5)$ .

We need $\text{P}(X \ge 4)$ . Because a Poisson distribution has infinitely many possible values, we use the complement:

\text{P}(Y \ge 4) = 1 - \text{P}(Y \le 3)

Then write $\text{P}(Y \le 3)$ as the sum of probabilities for $Y = 0,1,2,3$ :

\text{P}(Y \le 3) = e^{-4.5}\left(1 + 4.5 + \frac{4.5^2}{2!} + \frac{4.5^3}{3!}\right)

The factorial terms are $4.5^2/2! = 10.125$ and $4.5^3/3! = 15.1875$ , so

\text{P}(Y \le 3) = e^{-4.5}(1 + 4.5 + 10.125 + 15.1875)

Using $e^{-4.5} \approx 0.011109$ , the four terms are approximately $0.011109, 0.049999, 0.112480, 0.168720$ . Their sum is about $0.342308$ , so

\text{P}(X \ge 4) = 1 - 0.342308 = 0.657692

which rounds to $0.658$ to 3 significant figures.

Key Takeaways

The Poisson approximation to the binomial applies when $n$ is large and $p$ is small; take $\lambda = np$ .
For a Poisson probability of the form $\text{P}(X \ge k)$ , use the complement $1 - \text{P}(X \le k-1)$ .
A Poisson probability is $\text{P}(X = r) = e^{-\lambda}\lambda^r/r!$ , so cumulative probabilities are sums of these terms.

Common Mistakes

Trying to sum from $X = 4$ to infinity directly, which is hard; the complement is much simpler.
Forgetting to include the $X = 0$ term in $\text{P}(X \le 3)$ .
Misusing the factorial, for example writing $3! = 3$ instead of $3! = 6$ .
Giving $0.658$ without showing the Poisson expression. The mark scheme says an unsupported correct answer scores only B1 B1, not full marks.

Things to Be Careful About

$\lambda$ must be $4.5$ , not $1$ or $\frac{1}{1000000}$ .
The final answer should be given to 3 significant figures: $0.658$ .
Show the full cumulative expression $e^{-4.5}(1 + 4.5 + 10.125 + 15.1875)$ ; the mark scheme requires the expression to be seen or implied by correct figures.
The mark scheme allows one end error in the sum, but full marks require the cumulative expression to be correct.

Techniques used

recognise when a binomial distribution can be approximated by a Poisson distributioncalculate the mean parameter λ = npapply the complement rule to find P(X ≥ 4)evaluate cumulative Poisson probabilities

Approach

For a 95% confidence interval for a population mean, the interval has the form

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

The half-width of the interval is therefore $1.96 \frac{\sigma}{\sqrt{50}}$ . Equate this to the half-width of the given interval, then solve for $\sigma$ .

Working

The interval is $[245, 263]$ , so its half-width is

\frac{263-245}{2} = 9

Using $n = 50$ ,

1.96 \times \frac{\sigma}{\sqrt{50}} = 9

\sigma = \frac{9\sqrt{50}}{1.96}

Since $\sqrt{50} \approx 7.071$ ,

\sigma \approx \frac{9 \times 7.071}{1.96} = \frac{63.64}{1.96} \approx 32.47

\sigma \approx 32.5 \text{ m (3 sf)}

Answer

The estimated standard deviation of the lengths of roads in this region is $32.5$ metres.

Final answer

32.5 m

Detailed explanation

Walkthrough

The confidence interval tells us both the centre and the spread of the sampling distribution. The sample mean is the centre. For a 95% confidence interval, the margin of error is $1.96$ standard errors, where the standard error of the sample mean is $\sigma/\sqrt{n}$ and $n=50$ .

The total width of the interval is $263-245=18$ , so the half-width is $9$ . Therefore:

1.96 \times \frac{\sigma}{\sqrt{50}} = 9

Rearrange to make $\sigma$ the subject:

\sigma = \frac{9\sqrt{50}}{1.96} \approx 32.47

So the estimated population standard deviation is $32.5$ m to 3 significant figures.

Key Takeaways

A confidence interval is written as $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$ . The number $1.96$ comes from the 95% confidence level. The standard error is $\sigma/\sqrt{n}$ , so the sample size appears as a square root.

Common Mistakes

Using the full width $18$ as the margin of error instead of the half-width $9$ . Using the wrong sample size. Rounding intermediate values too early. Confusing the population standard deviation $\sigma$ with the standard error $\sigma/\sqrt{n}$ .

Things to Be Careful About

Remember to divide the interval width by 2 before equating it to the margin of error. Use $n=50$ , not $\sqrt{50}$ incorrectly. Give the final value to 3 significant figures as requested.

Techniques used

equate half-width of confidence interval to z times standard errorsolve for population standard deviationuse sample size in the standard error formula

(c)

It is now given that the lengths of roads in this region are normally distributed.

State, with a reason, whether this fact would make any difference to your calculation in part (b).

1M Similar

DifficultyEasy

Worked solution

Approach

The confidence interval in part (b) is based on the sampling distribution of the sample mean. Because $n=50$ is large, the Central Limit Theorem says the sample mean is approximately normal regardless of the population distribution.

Working

For a large sample, the sample mean $\bar{X}$ satisfies

\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \approx N(0,1)

by the Central Limit Theorem. Therefore the population itself does not need to be normal for the calculation in part (b) to be valid.

Answer

No. This fact would not make any difference, because the sample mean is approximately normally distributed for a large sample such as $n=50$ (Central Limit Theorem).

Final answer

No, because the sample mean is approximately normally distributed for large n (Central Limit Theorem).

Detailed explanation

Walkthrough

Part (b) used the fact that the sample mean is approximately normally distributed for a sufficiently large sample, whatever the underlying population distribution. Since $n=50$ is large, the Central Limit Theorem applies, and the population does not need to be normal.

Therefore, being told that the lengths are normally distributed would not change the calculation in part (b). The interval formula is justified by the approximate normality of the sample mean, not by the normality of the individual data values.

Key Takeaways

The Central Limit Theorem allows us to use normal-theory confidence intervals for the sample mean when the sample is large, even if the population is not normal. The distribution of the data and the distribution of the sample mean are different.

Common Mistakes

Answering “Yes” without considering the sample size. Saying “No” without giving a reason. Confusing the distribution of the individual roads with the distribution of the sample mean.

Things to Be Careful About

The question asks whether the normality of the population would make a difference to part (b), not part (a). Give both the conclusion and the reason. In A-Level terms, $n \geq 30$ is usually taken as large enough for the Central Limit Theorem; $n=50$ is clearly large.

Techniques used

recognise that a large sample makes the sample mean approximately normalapply the Central Limit Theorem to justify the confidence interval calculation

Q3Medium-EasyHypothesis Tests

A factory owner models the number of employees who use the factory canteen on any day by the distribution $\text{B}(25, p)$ . In the past the value of $p$ was 0.8. A new menu is introduced in the canteen and the owner wants to test whether the value of $p$ has increased.

On a randomly chosen day he notes that the number of employees who use the canteen is 23.

(a)

Use the binomial distribution to carry out the test at the 10% significance level.

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

State the null and alternative hypotheses. Since the owner wants to test whether $p$ has increased, use a one-tailed upper-tail test at the 10% significance level. Under the null hypothesis the number of canteen users is $X \sim \mathrm{B}(25, 0.8)$ . The observed value is 23, so compute the probability of 23 or more users. If this tail probability is less than 0.10, reject $H_0$ .

Working

H_0: p = 0.8, \quad H_1: p > 0.8

Assuming $H_0$ is true:

X \sim \mathrm{B}(25, 0.8)

The probability of getting 23 or more users is:

\begin{aligned} P(X \geq 23) &= {}^{25}C_{23}(0.8)^{23}(0.2)^2 + {}^{25}C_{24}(0.8)^{24}(0.2) + {}^{25}C_{25}(0.8)^{25} \\ &= 0.070835 + 0.0236118 + 0.0037779 \\ &= 0.0982 \end{aligned}

Compare with the significance level:

0.0982 < 0.10

Therefore reject $H_0$ .

Answer

There is sufficient evidence at the 10% significance level to suggest that $p$ has increased.

Final answer

Reject H0; sufficient evidence that p has increased

Detailed explanation

Walkthrough

This is a one-tailed binomial hypothesis test. The population parameter $p$ is the probability that a randomly chosen employee uses the canteen. The owner believes the new menu has increased $p$ , so the alternative hypothesis is $H_1: p > 0.8$ . The null hypothesis states that nothing has changed: $H_0: p = 0.8$ .

Under $H_0$ , the number of users is modelled by $X \sim \mathrm{B}(25, 0.8)$ . On the chosen day 23 employees used the canteen. Since we are testing for an increase, values of 23, 24 or 25 are the upper-tail outcomes that would support the alternative. Therefore the p-value is $P(X \geq 23)$ .

To calculate this, add the probability of exactly 23 using the canteen, exactly 24, and exactly 25. Each uses ${}^{25}C_r (0.8)^r (0.2)^{25-r}$ , because success (using the canteen) has probability 0.8 and failure has probability 0.2. These three probabilities are 0.070835, 0.0236118 and 0.0037779, giving 0.0982.

Because 0.0982 is less than 0.10, the observed result is significant at the 10% level. This means that if $p$ were still 0.8, seeing 23 or more users would happen less than 10% of the time, so there is evidence to reject $H_0$ and conclude that $p$ has increased.

Key Takeaways

A one-tailed upper test uses outcomes at least as large as the observed value as evidence for the alternative.
The p-value in a binomial test is the tail probability $P(X \geq x)$ when testing for an increase.
Reject $H_0$ when the p-value is less than or equal to the significance level.
Always finish with a conclusion in context.

Common Mistakes

Writing $H_1: p < 0.8$ or $H_1: p \neq 0.8$ instead of $H_1: p > 0.8$ .
Only calculating $P(X = 23)$ instead of $P(X \geq 23)$ , forgetting that 24 and 25 are also more extreme in the direction of increase.
Using the failure probability 0.8 with success probability 0.2, i.e. swapping $p$ and $1-p$ .
Giving the probability 0.0982 without showing the binomial expression, which the mark scheme awards only limited credit.
Ending with 'do not reject $H_0$ ' when the comparison shows 0.0982 < 0.10.

Things to Be Careful About

The significance level is 10%, so the comparison value is 0.10.
The number of trials is 25, not the observed value 23.
Check that the probability term for failures is $(0.2)^{n-x}$ ; here exactly two failures occur when $x = 23$ .
Use non-definite wording in the conclusion, such as 'evidence suggests that $p$ has increased', not 'this proves the menu worked'.

Techniques used

state null and one-tailed alternative hypothesescompute upper-tail binomial probabilitycompare tail probability with significance levelstate conclusion in context

(b)

Given that there are 30 employees at the factory comment on the suitability of the owner’s model.

1M Similar

DifficultyEasy

Worked solution

Approach

The model $\mathrm{B}(25, p)$ assumes a fixed number of 25 independent trials, so it can only give outcomes from 0 to 25. Since there are 30 employees, the model should not cap the possible number of canteen users at 25. Therefore the model is not suitable unless only a sample of 25 employees was being considered.

Working

The maximum number of employees who could use the canteen is 30, but $\mathrm{B}(25, p)$ cannot represent more than 25 users. The correct binomial model would use $n = 30$ if all employees were included, or would need to specify that only 25 employees are relevant.

Answer

Not suitable; the model $\mathrm{B}(25,p)$ does not allow for more than 25 employees to use the canteen, even though there are 30 employees.

Final answer

Not suitable; it does not allow for more than 25 employees

Detailed explanation

Walkthrough

The model $\mathrm{B}(25,p)$ means that the number of canteen users is counted from exactly 25 employees. This automatically restricts the possible result to $0, 1, \dots, 25$ . However, the question tells us that there are 30 employees at the factory, so in real life up to 30 employees could use the canteen. Therefore the binomial model with $n=25$ is not suitable for modelling all 30 employees. To model the whole factory, the owner should use $n = 30$ if every employee could independently choose to use the canteen.

Another acceptable reason is that if the owner only looked at 25 of the 30 employees, the model is using a sample rather than all employees, and how the 25 were chosen would matter. Either way, the model is not suitable without justification.

Key Takeaways

In a binomial model, $n$ is the fixed number of trials, so the maximum possible count is $n$ .
A statistical model must match the real-world restriction of the situation.
A one-mark comment question expects a clear conclusion plus one valid reason.

Common Mistakes

Saying 'suitable' or 'not suitable' without giving a reason; the mark requires both.
Saying 'not suitable because 23 is greater than 25', which confuses the observed value with the number of trials.
Introducing an irrelevant reason such as 'the Poisson distribution would be better'.

Things to Be Careful About

The key fact is that 30 employees are available but the model is based on 25.
If using the 'sample' reason, make clear that 25 is a sample from the 30 employees, not the whole factory.
Do not contradict the conclusion given in part (a); the suitability comment is about the modelling assumption only.

Techniques used

relate binomial trial count to population sizejudge model suitability with a reason

Q4Medium-EasySampling and Estimation

A population is normally distributed with mean 35 and standard deviation 8.1. A random sample of size 140 is chosen from this population and the sample mean is denoted by $\overline{X}$ .

(a)

Find $\text{P}(\overline{X} > 36)$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since the population is normal, the sample mean $\overline{X}$ is normally distributed with mean $\mu = 35$ and standard deviation $\frac{8.1}{\sqrt{140}}$ .

Working

The standard error of the sample mean is

\frac{8.1}{\sqrt{140}} = 0.6846

Standardise the value 36:

z = \frac{36 - 35}{8.1/\sqrt{140}} = \frac{1}{0.6846} = 1.461

Therefore

P(\overline{X} > 36) = P(Z > 1.461) = 1 - \Phi(1.461)

= 1 - 0.9280 = 0.0720

Answer

P(\overline{X} > 36) = 0.0720

(to 3 significant figures)

Final answer

0.0720

Detailed explanation

Walkthrough

Because the population is normal, the sample mean $\overline{X}$ is exactly normal with $\mathrm{E}(\overline{X}) = 35$ and standard deviation $8.1/\sqrt{140}$ . The standard error measures how much sample means vary from the population mean.

The probability asked for is the probability that the sample mean is greater than 36. Convert 36 to a standard normal z-score: $z = \frac{36 - 35}{8.1/\sqrt{140}} = 1.461$ . This tells us that 36 is 1.461 standard errors above the mean. Then use the standard normal table: $\Phi(1.461) \approx 0.9280$ , so the right-tail probability is $1 - 0.9280 = 0.0720$ .

Key Takeaways

For a normal population, $\overline{X}$ is normal with $\mathrm{E}(\overline{X}) = \mu$ and $\mathrm{Var}(\overline{X}) = \sigma^2/n$ .
The standard error is $\sigma/\sqrt{n}$ .
Standardise using $z = \frac{\overline{X} - \mu}{\sigma/\sqrt{n}}$ .
For a right-tail probability, use $1 - \Phi(z)$ .

Common Mistakes

Using the population standard deviation 8.1 instead of the standard error $8.1/\sqrt{140}$ .
Forgetting the square root in $\sqrt{140}$ .
Finding $\Phi(1.461)$ instead of $1 - \Phi(1.461)$ .
Including a continuity correction is not necessary here because the sample mean is continuous; the mark scheme allows ignoring it.

Things to Be Careful About

Give the final answer to 3 significant figures: 0.0720.
Show the standard error and the z-score so the method is clear.
Since 36 is above the mean, the required probability is the upper tail.

Techniques used

calculate the standard error of the sample meanstandardise the sample mean to a z-scorefind an upper-tail probability from the standard normal distribution

(b)

It is given that $\text{P}(\overline{X} < a) = 0.986$ . Find the value of $a$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $P(\overline{X} < a) = 0.986$ , find the standard normal z-score that has 0.986 to its left, then use the sampling distribution of $\overline{X}$ to solve for $a$ .

Working

The inverse normal value is

z = \Phi^{-1}(0.986) = 2.198

Using $\overline{X} \sim N\left(35, \frac{8.1^2}{140}\right)$ , the standard error is

\frac{8.1}{\sqrt{140}} = 0.6846

So

\frac{a - 35}{8.1/\sqrt{140}} = 2.198

a = 35 + 2.198 \times \frac{8.1}{\sqrt{140}}

a = 35 + 2.198 \times 0.6846 = 35 + 1.505 = 36.505

Answer

a = 36.5

(to 3 significant figures)

Final answer

36.5

Detailed explanation

Walkthrough

The statement $P(\overline{X} < a) = 0.986$ means that $a$ is the 98.6th percentile of the sampling distribution of the sample mean. Since 0.986 is greater than 0.5, $a$ must be above the mean 35.

First find the standard normal z-score that leaves 0.986 area to its left. The inverse normal gives $z = 2.198$ (or 2.197). Then standardise the sample mean: $z = \frac{a - 35}{8.1/\sqrt{140}}$ . Substitute $z = 2.198$ and solve for $a$ :

a = 35 + 2.198 \times \frac{8.1}{\sqrt{140}} = 36.505

which rounds to 36.5.

Key Takeaways

The inverse normal distribution converts a cumulative probability into a z-score.
The z-score formula for a sample mean is $z = \frac{\overline{X} - \mu}{\sigma/\sqrt{n}}$ .
Since 0.986 > 0.5, the z-score is positive and $a > 35$ .

Common Mistakes

Using $z = 2.2$ without more precision: the mark scheme awards B0 for the inverse-normal mark if only 2.2 is seen.
Using a negative z-score, which would correspond to a probability below 0.5.
Using the population standard deviation 8.1 instead of the standard error.
Not showing a z-score at all; the method mark requires a z-score.
Using a z-score such as 2.196 may still produce a final answer of 36.5, but it loses the B1 mark and can lose the final A1 mark because the working is not correct.

Things to Be Careful About

Use $z = 2.197$ or $z = 2.198$ for the inverse normal value.
Give the final answer to 3 significant figures: 36.5.
The final answer is CWO (correct working only), so a wrong z-score cannot be rescued by a coincidentally correct rounded answer.
Keep $\sqrt{140}$ in the standard error expression.

Techniques used

use the inverse normal distribution to find a z-scoreset up the z-score equation for the sample meansolve for the unknown value a

Q5Medium-HardThe Poisson DistributionLinear Combinations of Random Variables

A machine puts sweets into bags at random. The numbers of lemon and orange sweets in a bag have the independent distributions $\text{Po}(3.7)$ and $\text{Po}(2.6)$ respectively.

A bag of sweets is chosen at random.

(a)

Find the probability that the number of lemon sweets in the bag is more than 2 but not more than 5.

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The number of lemon sweets in a bag is $X \sim \text{Po}(3.7)$ . The condition "more than 2 but not more than 5" means $X = 3, 4$ or $5$ . Use the Poisson probability formula

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

for each of these values and add the results.

Working

For $\lambda = 3.7$ :

P(X = 3) = e^{-3.7} \frac{3.7^3}{3!} = e^{-3.7}(8.44217) = 0.20872

P(X = 4) = e^{-3.7} \frac{3.7^4}{4!} = e^{-3.7}(7.80900) = 0.19307

P(X = 5) = e^{-3.7} \frac{3.7^5}{5!} = e^{-3.7}(5.77866) = 0.14287

Therefore

P(3 \leq X \leq 5) = e^{-3.7}\left(\frac{3.7^3}{3!} + \frac{3.7^4}{4!} + \frac{3.7^5}{5!}\right)

= 0.20872 + 0.19307 + 0.14287 = 0.54466

Answer

0.545 \ (3 \text{ s.f.})

Final answer

0.545

Detailed explanation

Walkthrough

We are told the number of lemon sweets in a single bag follows a Poisson distribution with mean $3.7$ . The phrase "more than 2 but not more than 5" means we want exactly 3, 4 or 5 lemon sweets. Since the Poisson distribution is discrete, we calculate each of these three probabilities separately with $P(X=x)=e^{-\lambda}\lambda^x/x!$ and add them. The mark scheme requires the expression to be seen or implied by correct figures, so it is important to write the sum of the three terms before evaluating.

Key Takeaways

Poisson probabilities are calculated term by term using the formula. A range probability such as "more than 2 but not more than 5" is the sum of the individual probabilities for $x=3,4,5$ . Showing the full expression is essential for method marks.

Common Mistakes

Misreading the range and including $x=2$ or $x=6$ .
Forgetting the factor $e^{-3.7}$ .
Giving only the final answer $0.545$ without showing the expression; this loses the method mark and scores only B1.

Things to Be Careful About

The boundaries are strict: "more than 2" excludes 2, and "not more than 5" includes 5. Use 3 significant figures for the final answer. The expression must be visible in the working.

Techniques used

identify the required Poisson valuesapply the Poisson probability formulasum the individual probabilities

(b)

Find the probability that the total number of lemon and orange sweets in the bag is less than 4.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $L \sim \text{Po}(3.7)$ and $O \sim \text{Po}(2.6)$ be the numbers of lemon and orange sweets in one bag. Because the two distributions are independent, the total $T = L + O$ is also Poisson, with mean

\lambda = 3.7 + 2.6 = 6.3.

"Less than 4" means $T = 0, 1, 2$ or $3$ . Sum the Poisson probabilities for these values.

Working

Using

P(T = t) = e^{-6.3} \frac{6.3^t}{t!},

we have

P(T < 4) = e^{-6.3}\left(1 + 6.3 + \frac{6.3^2}{2!} + \frac{6.3^3}{3!}\right)

= e^{-6.3}(1 + 6.3 + 19.845 + 41.6745)

= 0.0018363 + 0.011569 + 0.0364415 + 0.076527

= 0.1263738.

Answer

0.126 \ (3 \text{ s.f.})

Final answer

0.126

Detailed explanation

Walkthrough

The key step is that the sum of two independent Poisson variables is Poisson, with mean equal to the sum of the means. Here $L \sim \text{Po}(3.7)$ and $O \sim \text{Po}(2.6)$ , so $T = L+O \sim \text{Po}(6.3)$ . The event "less than 4" includes $T=0,1,2,3$ . We substitute each value into the Poisson formula and add. The mark scheme gives B1 for identifying $\lambda = 6.3$ , M1 for the summed expression, and A1 for the final value, so the expression must be shown.

Key Takeaways

Independent Poisson variables add to give another Poisson variable whose mean is the sum of the means. Range probabilities are found by summing individual Poisson probabilities.

Common Mistakes

Using $\lambda = 3.7$ or $2.6$ instead of $6.3$ .
Including $T=4$ when the question says "less than 4".
Omitting the $e^{-6.3}$ factor.
Giving an unsupported $0.126$ .

Things to Be Careful About

"Less than 4" means $0,1,2,3$ , not $1,2,3,4$ . The mean of the sum is $6.3$ , and the variance is also $6.3$ because the sum is Poisson. Final answer to 3 significant figures.

Techniques used

add means of independent Poisson variablesapply the Poisson probability formulasum probabilities below a threshold

(c)

10 bags of sweets are chosen at random.

Use approximating distributions to find the probability that the total number of lemon sweets in the 10 bags is less than the total number of orange sweets in the 10 bags.

6M Similar

DifficultyMedium-Hard

Worked solution

Approach

For 10 bags, the total number of lemon sweets is the sum of 10 independent $\text{Po}(3.7)$ variables, so

L_{10} \sim \text{Po}(37).

Similarly,

O_{10} \sim \text{Po}(26).

Because the means are large, approximate each by a normal distribution with the same mean and variance:

L_{10} \approx N(37, 37), \qquad O_{10} \approx N(26, 26).

We need $P(L_{10} < O_{10})$ , which is $P(O_{10} - L_{10} > 0)$ . Since the totals are independent, their difference is normal with mean $26 - 37 = -11$ and variance $26 + 37 = 63$ . Apply a continuity correction because the difference is discrete, then standardise and use the normal tail.

Working

Let $D = O_{10} - L_{10}$ . Then

D \sim N(-11, 63).

The event $L_{10} < O_{10}$ means $D \geq 1$ , since $D$ is an integer. Using the continuity correction,

P(D > 0) \approx P\left(Z > \frac{0.5 - (-11)}{\sqrt{63}}\right)

= P\left(Z > \frac{11.5}{7.937}\right) = P(Z > 1.449)

= 1 - \Phi(1.449) = 0.0737.

So the required probability is $0.0737$ (3 s.f.).

If the continuity correction is not applied,

P(D > 0) \approx P\left(Z > \frac{11}{\sqrt{63}}\right) = P(Z > 1.386) = 0.0828,

which is also accepted by the mark scheme.

Answer

0.0737 \ (3 \text{ s.f.}) \text{ with continuity correction; } 0.0828 \text{ without continuity correction}

Final answer

0.0737 (with continuity correction; 0.0828 without)

Detailed explanation

Walkthrough

For 10 bags, the total number of lemon sweets is the sum of ten independent Poisson variables, each with mean $3.7$ , so it is Poisson with mean $10 \times 3.7 = 37$ . Similarly, the total number of orange sweets is Poisson with mean $10 \times 2.6 = 26$ . Since these means are large, the normal approximation is appropriate: a Poisson distribution with mean $\lambda$ is approximated by $N(\lambda, \lambda)$ . We need $P(L_{10} < O_{10})$ . It is convenient to consider $D = O_{10} - L_{10}$ . Because the two totals are independent, $D$ is normal with mean $26 - 37 = -11$ and variance $26 + 37 = 63$ (variances add for a difference of independent variables). The event $L_{10} < O_{10}$ is $D \geq 1$ , since $D$ is an integer. The continuity correction replaces the discrete boundary $1$ by $0.5$ , giving $z = (0.5 - (-11))/\sqrt{63} = 1.449$ . The required probability is the upper tail $1 - \Phi(1.449) = 0.0737$ . Without the continuity correction, the boundary would be $0$ , giving $z = 1.386$ and probability $0.0828$ ; the mark scheme accepts both.

Key Takeaways

The sum of independent Poisson variables is Poisson; for 10 bags the means are multiplied by 10.
A Poisson distribution with large mean can be approximated by a normal distribution with the same mean and variance.
The difference of two independent normal variables is normal, and variances add.
A continuity correction improves the normal approximation for a discrete variable.

Common Mistakes

Using $N(37, \sqrt{37})$ instead of $N(37, 37)$ ; the second parameter of the normal distribution is the variance.
Subtracting variances instead of adding them: $\text{Var}(O-L) = 26 + 37$ , not $26 - 37$ .
Using the wrong sign for the mean of $D$ .
Forgetting the continuity correction or applying it with the wrong boundary.
Using the mean of one bag ( $3.7$ and $2.6$ ) instead of the totals for 10 bags.

Things to Be Careful About

For a difference of independent variables, the mean is the difference of means but the variance is the sum of variances. The event "less than" for integer counts becomes a $\geq 1$ condition, so the continuity-corrected boundary is $0.5$ . The mark scheme accepts both $0.0737$ (with continuity correction) and $0.0828$ (without), but full working must show the standardisation and the tail probability.

Techniques used

approximate Poisson totals by normal distributionscombine independent normal variablesapply continuity correctionstandardise and find tail probability

Q6MediumContinuous Random Variables

The time, $X$ hours, taken by a large number of people to complete a challenge is modelled by the probability density function given by

f(x) = \begin{cases} \frac{1}{x^2} & a \le x \le b, \\ 0 & \text{otherwise,} \end{cases}

where $a$ and $b$ are constants.

(a)

State what the constants $a$ and $b$ represent in this context.

1M Similar

(b)

Show that $a = \frac{b}{b+1}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a probability density function, the total area under the curve over its support must equal $1$ . Integrate $f(x)$ from $a$ to $b$ , set the result equal to $1$ , and solve for $a$ .

Working

\int_a^b \frac{1}{x^2}\,dx = 1

Evaluate the integral:

\left[-\frac{1}{x}\right]_a^b = -\frac{1}{b} + \frac{1}{a} = 1

So

\frac{1}{a} - \frac{1}{b} = 1

Combine into a single fraction:

\frac{b-a}{ab} = 1

Hence

b-a = ab

Add $a$ to both sides:

b = a + ab = a(1+b)

Therefore

a = \frac{b}{b+1}

Answer

a = \frac{b}{b+1}

Final answer

a = b/(b+1)

Detailed explanation

Walkthrough

For any valid PDF, the total probability, which is the integral of the PDF over all possible values, must equal $1$ . Here the possible values are from $a$ to $b$ , so we integrate $\frac{1}{x^2}$ over $[a,b]$ . The antiderivative is $-\frac{1}{x}$ . Substituting the limits gives $\frac{1}{a} - \frac{1}{b}$ . Setting this equal to $1$ and combining the fractions gives $\frac{b-a}{ab}=1$ , so $b-a=ab$ . Rearranging gives $b=a(b+1)$ , and hence $a=\frac{b}{b+1}$ .

Key Takeaways

The total-integral property $\int f(x)\,dx = 1$ is essential for PDFs.
Clear algebraic rearrangement is needed to isolate the requested variable.

Common Mistakes

Forgetting to set the integral equal to $1$ .
Using the wrong antiderivative, e.g. $\ln x$ instead of $-\frac{1}{x}$ .
Making a sign error when substituting the limits.

Things to Be Careful About

The mark scheme requires the result to be convincingly obtained with no errors. Show the line $\frac{1}{a}-\frac{1}{b}=1$ before rearranging to $a=\frac{b}{b+1}$ .

Techniques used

integrate the PDF over its supportset total probability to 1rearrange to express a in terms of b

(c)

It is given that $\text{E}(X) = \ln 3$ .

Show that $b = 2$ and find the value of $a$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Use $\mathrm{E}(X) = \int_a^b x f(x)\,dx$ . Since $f(x)=\frac{1}{x^2}$ , the integrand becomes $\frac{1}{x}$ . Then use $a=\frac{b}{b+1}$ from part (b) to obtain an equation in $b$ only.

Working

\mathrm{E}(X) = \int_a^b x\cdot\frac{1}{x^2}\,dx = \int_a^b \frac{1}{x}\,dx = \left[\ln x\right]_a^b

So

\mathrm{E}(X) = \ln b - \ln a

Using $a = \frac{b}{b+1}$ :

\mathrm{E}(X) = \ln b - \ln\left(\frac{b}{b+1}\right)

= \ln b - \left(\ln b - \ln(b+1)\right) = \ln(b+1)

Given $\mathrm{E}(X) = \ln 3$ :

\ln(b+1) = \ln 3 \Rightarrow b+1 = 3 \Rightarrow b = 2

Then

a = \frac{b}{b+1} = \frac{2}{3}

Answer

b = 2, \quad a = \frac{2}{3}

Final answer

b = 2, a = 2/3

Detailed explanation

Walkthrough

To find the expectation of a continuous random variable, integrate $x$ times the PDF over the support. Here $x \cdot \frac{1}{x^2} = \frac{1}{x}$ , whose antiderivative is $\ln x$ . Substituting the limits gives $\ln b - \ln a$ . Using $a=\frac{b}{b+1}$ eliminates $a$ , so the expression becomes $\ln(b+1)$ . Setting this equal to $\ln 3$ gives $b+1=3$ , so $b=2$ . Finally, $a=\frac{2}{2+1}=\frac{2}{3}$ .

Key Takeaways

$\mathrm{E}(X)=\int x f(x)\,dx$ for a continuous random variable.
Two unknown constants in a PDF can be found using the total-probability condition and the expectation condition.
Logarithm laws simplify expressions such as $\ln b - \ln\left(\frac{b}{b+1}\right)$ .

Common Mistakes

Verifying the answer instead of deriving it: checking that $a=2/3$ and $b=2$ give $\mathrm{E}(X)=\ln 3$ does not earn full marks.
Forgetting to use part (b) to eliminate $a$ , leaving an equation in two variables.
Incorrectly simplifying $\ln b - \ln\left(\frac{b}{b+1}\right)$ .

Things to Be Careful About

The mark scheme requires a correct equation in $b$ only before solving. Ensure $b>a>0$ so the logarithms are defined.

Techniques used

use the integral of xf(x) to find the expectationsubstitute the relation between a and bsolve a logarithmic equationdetermine a from b

(d)

Find the median of $X$ .

3M Similar

Q7Medium-HardHypothesis TestsSampling and Estimation

The heights of one-year-old trees of a certain variety are known to have mean 2.3 m. A scientist believes that, on average, trees of this age and variety in her region are slightly taller than in other places. She plans to carry out a hypothesis test, at the 2% significance level, in order to test her belief.

(a)

State the probability that she will make a Type I error.

1M Similar

(b)

She takes a random sample of 100 such trees in her region and measures their heights, $h$ m. Her results are summarised below.

n = 100 \quad \sum h = 238 \quad \sum h^2 = 580

Carry out the test at the 2% significance level.

7M Similar

DifficultyMedium-Hard

Worked solution

Approach

Set up a one-tailed hypothesis test for the population mean. Since the sample size is large ( $n = 100$ ), the Central Limit Theorem ensures that the sample mean is approximately normally distributed. First estimate the unbiased population variance, then compute the test statistic and compare it with the critical value at the 2% significance level.

Working

Step 1: Hypotheses

H_0: \mu = 2.3 \qquad H_1: \mu > 2.3

Step 2: Sample mean

\bar{x} = \frac{\sum h}{n} = \frac{238}{100} = 2.38

Step 3: Unbiased estimate of the population variance

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

s^2 = \frac{1}{99}\left(580 - \frac{238^2}{100}\right) = \frac{1}{99}(580 - 566.44)

s^2 = \frac{13.56}{99} = 0.137 \text{ (3 sf)}

s = \sqrt{0.137} = 0.370 \text{ (3 sf)}

Step 4: Test statistic

z = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{2.38 - 2.3}{\sqrt{0.137/100}}

\sqrt{\frac{0.137}{100}} = \sqrt{0.00137} = 0.0370

z = \frac{0.08}{0.0370} = 2.16 \text{ (3 sf)}

Step 5: Critical value

At the 2% significance level for a one-tailed test, the critical value is $z = 2.054$ (or 2.055).

Step 6: Comparison

Since $2.16 > 2.054$ , we reject $H_0$ .

The corresponding probability is $P(Z > 2.16) = 0.0153$ (or 0.0154), which is less than 0.02, confirming the rejection.

Answer

There is sufficient evidence to suggest that the mean height of one-year-old trees in the scientist's region is greater than 2.3 m.

Final answer

Reject H0; sufficient evidence that the mean height in the scientist's region is greater than 2.3 m

Detailed explanation

Walkthrough

Step 1 — State the hypotheses. The null hypothesis is $H_0: \mu = 2.3$ (the mean height is 2.3 m). The alternative hypothesis is $H_1: \mu > 2.3$ (the mean height is greater than 2.3 m), because the scientist believes the trees are taller. This is a one-tailed test.

Step 2 — Compute the sample mean. $\bar{x} = \sum h / n = 238/100 = 2.38$ .

Step 3 — Estimate the population variance. Since only a sample is available, we use the unbiased estimate $s^2 = \frac{1}{n-1}(\sum h^2 - (\sum h)^2/n)$ . Substituting the given values gives $s^2 = 0.137$ (3 sf), so $s = 0.370$ .

Step 4 — Compute the test statistic. Using $z = (\bar{x} - \mu)/(s/\sqrt{n})$ , we get $z = (2.38 - 2.3)/(0.370/\sqrt{100}) = 2.16$ (3 sf).

Step 5 — Compare with the critical value. At the 2% significance level for a one-tailed test, the critical value is $z = 2.054$ (or 2.055). Since $2.16 > 2.054$ , we reject $H_0$ . Equivalently, the p-value $P(Z > 2.16) = 0.0153$ is less than 0.02, confirming the rejection.

Step 6 — Conclude in context. There is sufficient evidence to suggest that the mean height of trees in the scientist's region is greater than 2.3 m.

Key Takeaways

For large samples ( $n \geq 30$ ), the Central Limit Theorem allows the sample mean to be treated as approximately normal regardless of the population distribution.
The unbiased estimate of the population variance uses $n-1$ in the denominator.
The test statistic for a one-sample z-test is $z = (\bar{x} - \mu)/(s/\sqrt{n})$ .
The critical value at the 2% level for a one-tailed test is approximately 2.054.

Common Mistakes

Using the biased variance estimator with $n$ in the denominator instead of $n-1$ — this loses the M1 and A1 marks.
Using a two-tailed test instead of a one-tailed test — the scientist believes the trees are taller, so the test is one-tailed. A two-tailed test scores at most 5 of the 7 marks.
Using the wrong critical value (e.g., 1.96 for a 5% two-tailed test).
Forgetting to compare the test statistic with the critical value.

Things to Be Careful About

The unbiased variance must use $n-1$ in the denominator.
The conclusion must be in context and non-definite (e.g., "sufficient evidence to suggest..."), not a definitive proof.
The mark scheme allows the comparison either via the test statistic ( $2.16 > 2.054$ ) or via the probability ( $0.0153 < 0.02$ ).

Techniques used

set up null and alternative hypothesescompute unbiased variance estimatecompute the sample meancalculate the z-test statisticcompare with the critical value

(c)

The scientist carries out the test correctly, but another scientist claims that she has made a Type II error.

Comment on this claim.

1M Similar

Mathematics 9709/62 — October/November 2024

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