The random variables $X$ and $Y$ have independent distributions $X \sim \text{Po}(3)$ and $Y \sim \text{Po}(2)$ respectively. (a) Find $P(2 2)$. (c) The total of 100 random values of $X$ and 150 random values of $Y$ is denoted by $T$. Use a suitable approximating distribution to find $P(T < 560)$.

(a) 0.392 · (b) 0.875 · (c) 0.0492

9709/61 Oct/Nov 2025 — Mathematics Solved Past Paper

Q1MediumThe Poisson DistributionLinear Combinations of Random Variables

The random variables $X$ and $Y$ have independent distributions $X \sim \text{Po}(3)$ and $Y \sim \text{Po}(2)$ respectively.

(a)

Find $P(2 < X < 5)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $X$ is a discrete Poisson random variable, the event $2 < X < 5$ includes only the integers $X = 3$ and $X = 4$ . Write the Poisson probabilities for these two values using $\lambda = 3$ and add them.

Working

P(2 < X < 5) = P(X = 3) + P(X = 4)

= e^{-3}\frac{3^3}{3!} + e^{-3}\frac{3^4}{4!}

= e^{-3}\left(\frac{27}{6} + \frac{81}{24}\right)

= e^{-3}(4.5 + 3.375)

= 0.22404 + 0.16803 = 0.392

Answer

P(2 < X < 5) = 0.392

to 3 significant figures.

Final answer

0.392

Detailed explanation

Walkthrough

First identify which integer values of $X$ satisfy $2<X<5$ . Because $X$ is a discrete Poisson random variable, the only possibilities are $X=3$ and $X=4$ . Substitute each value into the Poisson probability formula $P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$ with $\lambda=3$ , then add the results. This is a direct application of the Poisson pmf; no approximation is needed.

Key Takeaways

This part tests the ability to translate a strict compound inequality for a discrete random variable into a sum of probabilities. It also tests accurate evaluation of the Poisson pmf and attention to factorials and powers.

Common Mistakes

A common mistake is to include $X=2$ or $X=5$ by misreading the strict inequalities. Another is to compute the probability using an incorrect range, such as $P(X\le4)-P(X\le1)$ but making an arithmetic error. Some candidates use an incorrect $\lambda$ , which the mark scheme allows for the method mark but not for the final value.

Things to Be Careful About

Remember that $2<X<5$ means only the integers 3 and 4. When using a calculator, ensure the factorial denominator is entered correctly, and give the final answer to 3 significant figures.

Techniques used

list the integer values satisfying a discrete inequalityevaluate Poisson probabilitiessum probabilities

(b)

Find $P(X + Y > 2)$ .

3M Similar

DifficultyMedium

Worked solution

Approach

The sum of independent Poisson variables is Poisson, so $X + Y \sim \mathrm{Po}(5)$ . For the event $X+Y>2$ , use the complement $X+Y\le2$ to avoid an infinite sum.

Working

X + Y \sim \mathrm{Po}(3 + 2) = \mathrm{Po}(5)

P(X+Y>2) = 1 - P(X+Y \le 2)

P(X+Y \le 2) = e^{-5}\left(1 + 5 + \frac{5^2}{2!}\right)

= e^{-5}(1 + 5 + 12.5)

= 0.0067379 + 0.0336897 + 0.0842243

P(X+Y>2) = 1 - 0.124652 = 0.875

Answer

P(X+Y>2) = 0.875

to 3 significant figures.

Final answer

0.875

Detailed explanation

Walkthrough

The sum of two independent Poisson random variables is itself Poisson with parameter equal to the sum of the parameters, so $X+Y\sim\mathrm{Po}(5)$ . The event $X+Y>2$ includes all totals 3, 4, 5, ...; it is easier to subtract from 1 the complementary event $X+Y\le2$ , which includes totals 0, 1 and 2. Write each of these three Poisson probabilities using $\lambda=5$ , sum them, and subtract the sum from 1.

Key Takeaways

The key idea is the additive property of independent Poisson distributions. Also important is the complement method, which avoids summing an infinite number of terms. Finally, expanding the Poisson probabilities up to $x=2$ is essential.

Common Mistakes

Forgetting that $P(X+Y>2)$ means $1-P(X+Y\le2)$ , not $1-P(X+Y<2)$ , changes the answer. Another common error is to use $\lambda=3$ or $\lambda=2$ rather than the combined $\lambda=5$ . A further mistake is omitting one of the probabilities for values 0, 1 or 2 when forming the complement.

Things to Be Careful About

Because the combined distribution is Poisson, not normal, use exact Poisson probabilities. The mark scheme allows an alternative combination method, but any correct method must still cover all cases producing totals 0, 1 and 2 and subtract from 1. Keep enough decimal places to round correctly to 0.875.

Techniques used

use the additive property of independent Poisson variablesset up a complement probabilitysum Poisson probabilities up to the boundaryapply total probability

(c)

The total of 100 random values of $X$ and 150 random values of $Y$ is denoted by $T$ .

Use a suitable approximating distribution to find $P(T < 560)$ .

4M Similar

DifficultyMedium

Worked solution

Approach

Find the distribution of the total $T$ . Since each value of $X$ is $\mathrm{Po}(3)$ and each value of $Y$ is $\mathrm{Po}(2)$ , and all variables are independent, the sum of all 250 values is $\mathrm{Po}(600)$ . With mean 600, use the normal approximation $T \approx N(600,600)$ . Apply a continuity correction because $T$ is discrete: $P(T<560) = P(T\le559)$ uses the boundary $559.5$ .

Working

T \sim \mathrm{Po}(100 \times 3 + 150 \times 2) = \mathrm{Po}(600)

So

T \approx N(600, 600)

Standardise using the continuity-corrected boundary:

P(T < 560) = P\left(Z \le \frac{559.5 - 600}{\sqrt{600}}\right)

= P(Z \le -1.653)

= 1 - \Phi(1.653) = 1 - 0.9508 = 0.0492

Answer

P(T < 560) = 0.0492

to 3 significant figures.

Final answer

0.0492

Detailed explanation

Walkthrough

First find the distribution of $T$ . A single $X$ has mean and variance 3, and a single $Y$ has mean and variance 2. The total of 100 independent $X$ values has mean and variance 300; the total of 150 independent $Y$ values also has mean and variance 300. Hence $T$ has mean 600 and variance 600. Because the sum of independent Poisson variables is Poisson, $T\sim\mathrm{Po}(600)$ . Since $\lambda=600$ is large, use the normal approximation $N(600,600)$ . Because $T$ is discrete, $P(T<560)=P(T\le559)$ , so the continuity-corrected boundary is $559.5$ . Standardise by subtracting the mean and dividing by $\sqrt{600}$ , giving $z=-1.653$ . Use the standard normal table: $\Phi(-1.653)=1-\Phi(1.653)=0.0492$ .

Key Takeaways

This question combines three ideas: the sum of independent Poisson variables is Poisson, a Poisson distribution with a large mean can be approximated by a normal distribution with the same mean and variance, and a continuity correction is needed when approximating a discrete distribution by a continuous one.

Common Mistakes

Omitting the continuity correction is a common error and would give a different tail probability. Using the wrong mean or variance, for example taking the standard deviation as 600 instead of $\sqrt{600}$ , is also common. Another mistake is to approximate too early and forget that $T$ is discrete.

Things to Be Careful About

The mark scheme allows no or incorrect continuity correction for the method mark, but the final answer uses the corrected boundary 559.5. Use $\Phi(1.653)$ correctly and subtract from 1 because the z-value is negative. Give the final answer to 3 significant figures; both 0.0492 and 0.0491 are accepted.

Techniques used

combine total Poisson sumsapproximate a Poisson distribution by a normal distributionapply a continuity correctionstandardise and compute a normal tail probability

Q28MMediumHypothesis TestsSampling and Estimation

The mean mass of packets of Trueleaf tea is supposed to be 500 grams. An inspector wishes to test whether this value is correct. He weighs 60 randomly chosen packets and notes the mass, $x$ grams, of each packet. The results are summarised as follows.

n = 60 \quad \sum x = 29970 \quad \sum x^2 = 14970300

Test, at the 5% significance level, whether the population mean mass is 500 grams.

DifficultyMedium

Worked solution

Approach

Since the sample size is large ( $n = 60$ ), the sample mean is approximately normally distributed by the Central Limit Theorem. We first compute the unbiased estimates of the population mean and variance, set up a two-tailed hypothesis test at the 5% significance level, standardise the sample mean to obtain a $z$ -statistic, and compare it with the critical value $\pm 1.96$ .

Working

Unbiased estimate of the population mean:

\bar{x} = \frac{\sum x}{n} = \frac{29970}{60} = 499.5

Unbiased estimate of the population variance:

s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{59}\left(14970300 - \frac{29970^2}{60}\right)

Compute the inner term:

\frac{29970^2}{60} = \frac{898200900}{60} = 14970015

Therefore:

s^2 = \frac{14970300 - 14970015}{59} = \frac{285}{59} = 4.83

State the hypotheses (two-tailed):

H_0: \mu = 500 \qquad H_1: \mu \neq 500

Compute the test statistic:

z = \frac{\bar{x} - \mu_0}{\sqrt{s^2/n}} = \frac{499.5 - 500}{\sqrt{4.83/60}}

\sqrt{\frac{4.83}{60}} = \sqrt{0.0805} = 0.2837

z = \frac{-0.5}{0.2837} = -1.762

Compare with the critical value for a two-tailed 5% test, $z_{0.025} = 1.96$ :

|z| = 1.762 < 1.96

The test statistic lies inside the acceptance region, so we do not reject $H_0$ .

Answer

There is insufficient evidence, at the 5% significance level, that the population mean mass of packets of Trueleaf tea is not 500 grams.

Final answer

Do not reject the null hypothesis; insufficient evidence that the population mean mass is not 500 grams.

Detailed explanation

Walkthrough

The inspector wants to know whether the true mean mass of all packets of Trueleaf tea is 500g. We only have a sample of 60 packets, so we must decide whether the observed sample mean of 499.5g is just sampling variation away from 500g or genuinely different. Hypothesis testing formalises this decision.

Since $n = 60$ is large, the Central Limit Theorem tells us the sample mean is approximately normally distributed, so we can use a $z$ -test even without knowing the population variance.

Step 1: Estimate the population mean. The sample mean $\bar{x} = 29970/60 = 499.5$ is our best unbiased estimate of $\mu$ . It is slightly less than 500.

Step 2: Estimate the population variance. We must use the unbiased estimator that divides by $n - 1$ :

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

This is the correct formula because we are estimating the population variance from a sample. Substituting the given values:

s^2 = \frac{1}{59}(14970300 - 14970015) = \frac{285}{59} = 4.83

Step 3: Set up hypotheses. The phrase "whether this value is correct" means we test both directions, so the test is two-tailed:

H_0: \mu = 500, \qquad H_1: \mu \neq 500

The 5% significance level means that, in a two-tailed test, each tail has 2.5% (0.025) probability.

Step 4: Standardise. The test statistic measures how many standard errors the sample mean is from the hypothesised mean:

z = \frac{\bar{x} - \mu_0}{\sqrt{s^2/n}} = \frac{499.5 - 500}{\sqrt{4.83/60}} = -1.762

The negative value indicates the sample mean is below 500.

Step 5: Compare with the critical value. For a two-tailed test at 5%, we reject $H_0$ if $|z| > 1.96$ . Here $|z| = 1.762 < 1.96$ , so the test statistic falls in the acceptance region. Equivalently, the $p$ -value of about 0.078 is greater than 0.05.

Conclusion: There is not enough evidence to reject the claim that the mean mass is 500g. We phrase this as "insufficient evidence" rather than "proving the mean is 500g".

Key Takeaways

The unbiased estimate of the population variance uses the $n - 1$ denominator:

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

A large sample ( $n \geq 30$ ) justifies using the Central Limit Theorem to treat the sample mean as approximately normal.

The test statistic for a population mean with unknown variance is:

z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}

The wording "whether the value is correct" signals a two-tailed test.

A hypothesis test conclusion must be in context and non-definite: "insufficient evidence...".

Common Mistakes

Using the biased variance (dividing by $n$ rather than $n - 1$ ). The mark scheme explicitly notes that using the biased variance (4.75) limits the maximum score to 6/8.

Stating hypotheses without reference to the population mean (e.g. just " $H_0$ : mean = 500"). The mark scheme requires "pop mean" or a defined $\mu$ .

Omitting the $\sqrt{60}$ (i.e. $\sqrt{n}$ ) in the denominator of the test statistic. This is essential.

Concluding with a definite statement ("the mean is 500g") rather than "insufficient evidence". The mark scheme penalises definite or contradictory conclusions.

Things to Be Careful About

The sign of $z$ matters for the comparison — use $|z|$ or compare $-1.762 > -1.96$ correctly.

The critical value is $\pm 1.96$ for a two-tailed 5% test; the one-tailed value 1.645 would be wrong here.

When comparing, the mark scheme allows $0.0390 > 0.025$ (one-tail probability) or $0.0780 > 0.05$ (two-tail probability). Both are acceptable — just be consistent.

Follow-through marks: an incorrect variance estimate can still earn marks for standardisation and conclusion, but using the biased variance caps the marks at 6/8.

Techniques used

calculate unbiased estimates of the population mean and varianceformulate null and alternative hypotheses for a two-tailed teststandardise the sample mean into a z-statisticcompare the test statistic against the critical valuestate a conclusion in context

Approach

Since $X$ counts the number of incorrect characters in a large data set with a small probability of error, $X$ follows a binomial distribution. Because $n = 14500$ is large and $p = 0.0001$ is small, we approximate the binomial distribution with a Poisson distribution having parameter $\lambda = np = 1.45$ . We then compute $P(X < 4)$ by summing the Poisson probabilities for $X = 0, 1, 2, 3$ .

Working

Step 1: Find the Poisson parameter.

\lambda = np = 14500 \times 0.0001 = 1.45

So $X \sim Po(1.45)$ .

Step 2: Compute $P(X < 4)$ .

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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

P(X < 4) = e^{-1.45}\left(1 + 1.45 + \frac{1.45^2}{2!} + \frac{1.45^3}{3!}\right)

= e^{-1.45}(1 + 1.45 + 1.05125 + 0.5081041)

= e^{-1.45} \times 4.0093541

= 0.234570 \times 4.0093541

= 0.9405

Answer

P(X < 4) = 0.940 \text{ (3 sf)}

Final answer

0.940

Detailed explanation

Walkthrough

The number of incorrect characters $X$ in a data set of 14500 characters follows a binomial distribution, since each character is either correct or incorrect independently, with a constant probability $p = 0.0001$ of being incorrect. However, computing binomial probabilities directly with $n = 14500$ would be extremely tedious.

The key insight is that when $n$ is large and $p$ is small, the binomial distribution $B(n, p)$ can be approximated by a Poisson distribution with parameter $\lambda = np$ . Here $\lambda = 14500 \times 0.0001 = 1.45$ . This is a valid approximation because $n$ is large and $p$ is small, making the expected number of incorrect characters moderate.

Once we have the Poisson distribution, we compute $P(X < 4)$ by summing the probabilities for $X = 0, 1, 2, 3$ . The Poisson probability formula is $P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}$ . We factor out $e^{-1.45}$ and sum the remaining terms: $1 + 1.45 + \frac{1.45^2}{2!} + \frac{1.45^3}{3!}$ . This gives the required probability.

Key Takeaways

When $n$ is large and $p$ is small, the binomial distribution can be approximated by a Poisson distribution with $\lambda = np$ .
The Poisson probability formula $P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}$ .
$P(X < 4)$ means $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$ — the sum of probabilities for all values less than 4.

Common Mistakes

Forgetting to sum all terms from $r = 0$ to $r = 3$ (e.g., only computing $P(X = 3)$ ).
Using the binomial distribution directly without approximation — the mark scheme allows this as a special case but it scores fewer marks (B2 instead of B1M1A1).
Incorrectly computing $\lambda = np$ .
Not including the factorial in the denominator of the Poisson formula.

Things to Be Careful About

The mark scheme notes that an unjustified answer of 0.94 or 0.941 only scores B1B1, so showing the full working is essential to earn all 3 marks.
The approximation is valid because $n$ is large and $p$ is small; the Poisson distribution is a good model here.

Techniques used

approximate a binomial distribution with a Poisson distributioncalculate the Poisson parameter as λ = npsum Poisson probabilities for X = 0 to 3

(b)

The firm’s management wishes to decrease the value of $p$ by giving their employees some training. Their aim is that, for a data set containing 14500 characters, the value of $P(X = 0)$ for the new value of $p$ should be double the value of $P(X = 0)$ when $p = 0.0001$ .

Use a suitable approximating distribution to find the new value of $p$ .

3M Similar

DifficultyMedium

Worked solution

Approach

When $p = 0.0001$ , the Poisson parameter is $\lambda = 1.45$ , so $P(X = 0) = e^{-1.45}$ . For a new value of $p$ , the Poisson parameter becomes $\lambda = 14500p$ , so $P(X = 0) = e^{-14500p}$ . We set this equal to twice the original value and solve for $p$ using logarithms.

Working

Step 1: Write the original $P(X = 0)$ .

With $p = 0.0001$ , $\lambda = 1.45$ , so

P(X = 0) = e^{-1.45}

Step 2: Write $P(X = 0)$ for the new value of $p$ .

For the new $p$ , $\lambda = 14500p$ , so

P(X = 0) = e^{-14500p}

Step 3: Form the equation.

We require the new $P(X = 0)$ to be double the original:

e^{-14500p} = 2e^{-1.45}

Step 4: Solve for $p$ .

Taking natural logarithms of both sides:

-14500p = \ln(2e^{-1.45})

-14500p = \ln 2 - 1.45

-14500p = 0.693147 - 1.45 = -0.756853

p = \frac{0.756853}{14500} = 0.0000522

Answer

p = 0.0000522 \text{ (3 sf)}

Final answer

0.0000522

Detailed explanation

Walkthrough

We start by noting that with the original $p = 0.0001$ , the Poisson parameter is $\lambda = 14500 \times 0.0001 = 1.45$ , so $P(X = 0) = e^{-1.45}$ .

For the new value of $p$ , the Poisson parameter becomes $\lambda = 14500p$ , so $P(X = 0) = e^{-14500p}$ .

The requirement is that the new $P(X = 0)$ should be double the original. This gives the equation $e^{-14500p} = 2e^{-1.45}$ .

To solve this, we take natural logarithms of both sides. The left side becomes $-14500p$ . The right side becomes $\ln(2e^{-1.45}) = \ln 2 + \ln(e^{-1.45}) = \ln 2 - 1.45$ .

Then we solve for $p$ : $-14500p = \ln 2 - 1.45$ , so $p = \frac{1.45 - \ln 2}{14500} = 0.0000522$ .

Key Takeaways

The Poisson parameter changes proportionally with $p$ : $\lambda = np$ .
Using logarithms to solve exponential equations.
Understanding how to set up an equation from a word problem.

Common Mistakes

Forgetting to take the logarithm of both sides.
Making sign errors when simplifying $\ln(2e^{-1.45}) = \ln 2 - 1.45$ .
Placing the factor of 2 on the wrong side of the equation — the mark scheme allows this for M1 but it leads to a wrong final answer.

Things to Be Careful About

The mark scheme applies follow-through (FT): if part (a) used a different $\lambda$ , the value of $2e^{-\lambda}$ would change accordingly.
The final answer must be given to 3 significant figures.
The answer $p = 0.0000522$ is smaller than the original $p = 0.0001$ , which makes sense — training should reduce the proportion of incorrect characters.

Techniques used

approximate a binomial distribution with a Poisson distributionset up an equation for the new probability conditionsolve an exponential equation using logarithms

Q4MediumSampling and Estimation

The masses of a certain species of animal are known to be normally distributed with standard deviation $\sigma$ kg. A researcher obtains the masses of a random sample of $n$ animals of this species and uses these masses to find two confidence intervals ( $\alpha$ % and 90%) for the population mean. The width of the $\alpha$ % confidence interval is $1.414 \times$ the width of the 90% confidence interval.

(a)

Find $\alpha$ .

3M Similar

DifficultyMedium

Worked solution

Approach

For a normal population with known $\sigma$ , a confidence interval for the mean is

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

so its width is $2z\frac{\sigma}{\sqrt{n}}$ . Since $n$ and $\sigma$ are the same for both intervals, the ratio of widths is the ratio of the critical $z$ -values.

Working

The 90% confidence interval uses $z = 1.645$ . Let $z$ be the critical value for the $\alpha\%$ interval.

2z\frac{\sigma}{\sqrt{n}} = 1.414 \times 2(1.645)\frac{\sigma}{\sqrt{n}}

Cancel the common factor $2\frac{\sigma}{\sqrt{n}}$ :

z = 1.414 \times 1.645 = 2.326

Convert this $z$ -value to a two-tailed confidence percentage:

\alpha = \left(2\Phi(2.326) - 1\right) \times 100 = (2(0.99)-1) \times 100 = 98

Answer

\alpha = 98

So the interval is a 98% confidence interval.

Final answer

α = 98 (98%)

Detailed explanation

Walkthrough

A confidence interval for a population mean when $\sigma$ is known has the form

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

so its width is $2z\frac{\sigma}{\sqrt{n}}$ . The 90% confidence interval uses the critical value $z = 1.645$ , because 90% of the standard normal distribution lies between $-1.645$ and $1.645$ .

The $\alpha\%$ interval is 1.414 times wider than the 90% interval. Because the sample size $n$ and the population standard deviation $\sigma$ are the same for both intervals, the factor $2\frac{\sigma}{\sqrt{n}}$ cancels. Therefore the critical value for the $\alpha\%$ interval is

z = 1.414 \times 1.645 = 2.326

To convert this $z$ -value into a confidence percentage, use the two-tailed probability:

\alpha = (2\Phi(2.326) - 1) \times 100

Since $\Phi(2.326) = 0.99$ , this gives $\alpha = (2(0.99)-1) \times 100 = 98$ . Thus the interval is a 98% confidence interval.

Key Takeaways

The width of a confidence interval for a known- $\sigma$ normal mean is proportional to the critical $z$ -value.
When the sample size and population standard deviation are fixed, comparing widths is equivalent to comparing critical values.
A two-tailed confidence level is $2\Phi(z)-1$ expressed as a percentage.

Common Mistakes

Using $z = 1.282$ for the 90% interval. The value 1.282 is a one-tailed 90% point; the two-tailed 90% critical value is 1.645. (The mark scheme condones 1.282 for method, but it will not give the correct final answer.)
Putting the factor 1.414 on the wrong side of the equation. This can still earn method credit, but the final value of $z$ will be wrong.
Forgetting to multiply the probability by 100 when stating $\alpha$ .

Things to Be Careful About

The factor 2 in the width cancels, so do not double-count it.
The value $z = 2.326$ corresponds to tail probability 0.01 in each tail, so the total confidence is 0.98, not 0.2326 or 2.326%.
Use the same $\sigma$ and $n$ for both intervals; they are from the same sample and population.

Techniques used

equate confidence interval widthssolve for the critical z-valueconvert the z-value to a confidence percentage

(b)

Find the probability that the 90% confidence interval contains the population mean given that the $\alpha$ % confidence interval contains the population mean.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $A$ be the event that the $\alpha\%$ (98%) confidence interval contains $\mu$ , and let $B$ be the event that the 90% confidence interval contains $\mu$ . We want $P(B \mid A)$ . Since the 98% interval is wider than the 90% interval, $B$ is a subset of $A$ , so $P(B \cap A) = P(B)$ .

Working

P(B \mid A) = \frac{P(B \cap A)}{P(A)} = \frac{P(B)}{P(A)} = \frac{0.90}{0.98} = \frac{45}{49}

\frac{45}{49} = 0.918 \text{ (3 sf)}

Answer

\frac{45}{49} \approx 0.918

Final answer

45/49 ≈ 0.918

Detailed explanation

Walkthrough

Let $A$ be the event that the 98% confidence interval contains the population mean $\mu$ , and let $B$ be the event that the 90% confidence interval contains $\mu$ . We are asked for $P(B \mid A)$ .

Since both intervals are centred at the same sample mean and use the same $\sigma$ and $n$ , the 98% interval is wider than the 90% interval. Therefore, if the 90% interval contains $\mu$ , the 98% interval must also contain $\mu$ . This means $B$ is a subset of $A$ , so $P(B \cap A) = P(B)$ .

Using the conditional probability formula:

P(B \mid A) = \frac{P(B \cap A)}{P(A)} = \frac{P(B)}{P(A)} = \frac{0.90}{0.98} = \frac{45}{49}

\frac{45}{49} = 0.918 \text{ (3 sf)}

So, given that the wider interval contains the mean, the probability that the narrower interval also contains it is about 0.918.

Key Takeaways

Confidence intervals with higher confidence levels are wider.
For nested intervals based on the same sample, the narrower interval being successful implies the wider interval is successful.
Conditional probability for a subset event simplifies to $P(B)/P(A)$ .

Common Mistakes

Quoting 0.90 as the answer. The condition changes the sample space, so we must divide by $P(A)=0.98$ .
Treating the two intervals as independent. They are based on the same sample and are nested, not independent.
If $\alpha < 90$ , the 90% interval would not be a subset of the $\alpha\%$ interval, so this simplification would not apply.

Things to Be Careful About

Use the follow-through value of $\alpha$ from part (a); here it is 98.
Give the answer as an exact fraction $\frac{45}{49}$ or as 0.918 to 3 significant figures.
The mark scheme says no follow-through if $\alpha < 90$ in part (a), because then the subset relationship would be reversed.

Techniques used

recognise nested confidence intervalsapply conditional probability formula

Q5MediumHypothesis Tests

It is known that 20% of households in a certain country contain more than 4 people. Laxmi believes that, in her town, the percentage is lower than 20%. She chooses a random sample of 40 households in her town and notes the number which contain more than 4 people. She then carries out a test at the 2.5% significance level using a binomial distribution.

(a)

Find the probability of a Type I error.

4M Similar

DifficultyMedium

Worked solution

Approach

Set up the hypotheses for the one-tailed test. Under $H_0$ , $X \sim B(40, 0.2)$ . The probability of a Type I error is the probability that $X$ falls in the rejection region when $H_0$ is true. To find the rejection region, compare cumulative probabilities $P(X \leq c)$ with the 2.5% significance level.

Working

The hypotheses are:

H_0: p = 0.2, \quad H_1: p < 0.2

Under $H_0$ , $X \sim B(40, 0.2)$ . Since $H_1$ is $p < 0.2$ , the rejection region is of the form $X \leq c$ .

Compute $P(X \leq 2)$ :

P(X \leq 2) = 0.8^{40} + 40 \times 0.8^{39} \times 0.2 + {}^{40}C_2 \times 0.8^{38} \times 0.2^2

= 0.0001329 + 0.0013292 + 0.0064799 = 0.00794

Since $0.00794 < 0.025$ , $X = 2$ is in the rejection region.

Now check $X = 3$ :

P(X \leq 3) = 0.0079421 + {}^{40}C_3 \times 0.8^{37} \times 0.2^3

= 0.0079421 + 0.0205199 = 0.0285

Since $0.0285 > 0.025$ , $X = 3$ is not in the rejection region.

Therefore the rejection region is $X \leq 2$ , and the probability of a Type I error is:

P(\text{Type I error}) = P(X \leq 2) = 0.00794 \quad (3 \text{ sf})

Answer

P(\text{Type I error}) = 0.00794

Final answer

0.00794

Detailed explanation

Walkthrough

This is a hypothesis test about a proportion. Laxmi believes the percentage of households with more than 4 people in her town is lower than the national 20%. So we set up:

$H_0: p = 0.2$ (the town's percentage is the same as the country's)
$H_1: p < 0.2$ (the town's percentage is lower)

Since $H_1$ uses "less than", this is a one-tailed test in the lower tail. Under $H_0$ , the number of households $X$ with more than 4 people follows a binomial distribution $X \sim B(40, 0.2)$ .

A Type I error means rejecting $H_0$ when it is actually true. This happens exactly when $X$ falls in the rejection region while the true probability is $p = 0.2$ .

To find the rejection region, we need the largest value $c$ such that $P(X \leq c) \leq 0.025$ . We start by computing $P(X \leq 2)$ using the binomial probability formula. Each term is of the form ${}^{40}C_k \times 0.8^{40-k} \times 0.2^k$ because the probability of "success" (a household with more than 4 people) is 0.2 and "failure" is 0.8.

Adding the three terms for $k = 0, 1, 2$ gives $0.00794$ . Since this is less than 0.025, the value $X = 2$ is in the rejection region.

We then check whether $X = 3$ should also be included. We add $P(X = 3) = {}^{40}C_3 \times 0.8^{37} \times 0.2^3 = 0.0205199$ to get $P(X \leq 3) = 0.0285$ . Since this exceeds 0.025, $X = 3$ cannot be in the rejection region.

So the rejection region is exactly $X \leq 2$ , and the probability of a Type I error is the probability of this region under $H_0$ , which is $0.00794$ .

Key Takeaways

A Type I error is rejecting $H_0$ when it is true; its probability is the probability of the rejection region under $H_0$ .
For a one-tailed binomial test, the critical region is found by comparing cumulative probabilities with the significance level.
You must show the binomial expressions when computing probabilities — the mark scheme requires the expressions to be seen.

Common Mistakes

Giving an unjustified answer of 0.00794 without showing the binomial expressions — the mark scheme scores this M0B1.
Forgetting to check both $P(X \leq 2)$ and $P(X \leq 3)$ to justify the critical region — you need to show both comparisons.
Using the wrong tail (e.g. $X \geq c$ instead of $X \leq c$ ) — this is a lower-tail test.
Rounding intermediate values too early, which can change the final comparison.

Things to Be Careful About

The test is one-tailed in the lower tail because Laxmi believes the percentage is lower.
Both comparisons must be shown: $P(X \leq 2) = 0.00794 < 0.025$ and $P(X \leq 3) = 0.0285 > 0.025$ .
The final answer should be given to 3 significant figures: 0.00794.

Techniques used

set up hypotheses for one-tailed binomial testcompute binomial cumulative probabilitiescompare cumulative probabilities with significance levelidentify probability of Type I error

(b)

State the rejection region for the test.

1M Similar

(c)

Laxmi finds that exactly 2 households in her sample contain more than 4 people.

Explain why it is impossible for Laxmi to make a Type II error.

1M Similar

Q6MediumLinear Combinations of Random Variables

The masses, in kilograms, of large and small bags of potatoes have the independent distributions $N(2.5, 0.05)$ and $N(0.8, 0.02)$ respectively.

(a)

Find the probability that the total mass of a randomly chosen large bag of potatoes and a randomly chosen small bag of potatoes is more than 3.55 kg.

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $L$ and $S$ be the masses of a large and a small bag. Since $L$ and $S$ are independent normal variables, their sum is normal with mean equal to the sum of the means and variance equal to the sum of the variances. Standardise the total and use the normal tail probability.

Working

Let

L \sim N(2.5, 0.05), \quad S \sim N(0.8, 0.02)

with $L$ and $S$ independent. For the total mass $T = L + S$ ,

T \sim N(2.5 + 0.8, 0.05 + 0.02) = N(3.3, 0.07)

We need $P(T > 3.55)$ . Standardising:

z = \frac{3.55 - 3.3}{\sqrt{0.07}} = \frac{0.25}{0.264575\ldots} = 0.9449\ldots \approx 0.945

Therefore

P(T > 3.55) = 1 - \Phi(0.945) = 1 - 0.8276 = 0.1724\ldots \approx 0.172

Answer

0.172 \quad (3 \text{ s.f.})

Final answer

0.172

Detailed explanation

Walkthrough

We have two independent normal distributions: large bags have mean 2.5 kg and variance 0.05, small bags have mean 0.8 kg and variance 0.02. The total mass is the sum of these two independent normal variables. A key result is that the sum of independent normal variables is also normal, with mean equal to the sum of the means and variance equal to the sum of the variances. This gives $T \sim N(3.3, 0.07)$ .

The question asks for the probability that this total is more than 3.55 kg. To use the standard normal table, we standardise: subtract the mean and divide by the standard deviation. The standard deviation is $\sqrt{0.07}$ , not 0.07, because the variance is 0.07. This gives $z \approx 0.945$ . Since the question asks for 'more than', we need the upper tail, so the probability is $1 - \Phi(0.945)$ . Reading the normal table gives $\Phi(0.945) \approx 0.8276$ , so the required probability is about 0.172.

Key Takeaways

The main idea is the linear combination of independent normal variables. If $X$ and $Y$ are independent normal variables, then $aX + bY$ is normal with mean $aE(X) + bE(Y)$ and variance $a^2Var(X) + b^2Var(Y)$ . For a sum $X + Y$ , this simplifies to mean $\mu_X + \mu_Y$ and variance $\sigma_X^2 + \sigma_Y^2$ . Always remember to take the square root of the variance when standardising.

Common Mistakes

Adding variances but then forgetting to take the square root when standardising.
Using $\Phi(0.945)$ directly instead of $1 - \Phi(0.945)$ for a 'more than' probability.
Mixing up variance and standard deviation when writing the normal distribution.

Things to Be Careful About

The variance of the sum is $0.05 + 0.02 = 0.07$ , so the standard deviation is $\sqrt{0.07}$ . The probability must be the upper tail because the question asks for 'more than 3.55 kg'. Give the final answer to 3 significant figures as required.

Techniques used

form the sum of independent normal variablesstandardise a normal variablefind an upper-tail normal probability

(b)

Find the probability that the mass of a randomly chosen large bag of potatoes is less than 3 times the mass of a randomly chosen small bag of potatoes.

5M Similar

DifficultyMedium

Worked solution

Approach

Let $D = L - 3S$ . The event that the large bag is less than 3 times the small bag is $L < 3S$ , i.e. $D < 0$ . Since $L$ and $S$ are independent, the mean and variance of $D$ are found using $E(aX + bY)$ and $Var(aX + bY)$ . Then standardise and find the lower-tail probability.

Working

Let

D = L - 3S

The expectation is

E(D) = 2.5 - 3(0.8) = 0.1

The variance is

Var(D) = Var(L) + (-3)^2 Var(S) = 0.05 + 9(0.02) = 0.05 + 0.18 = 0.23

So

D \sim N(0.1, 0.23)

We require

P(L < 3S) = P(L - 3S < 0) = P(D < 0)

Standardising:

z = \frac{0 - 0.1}{\sqrt{0.23}} = \frac{-0.1}{0.479583\ldots} = -0.2085\ldots

Therefore

P(D < 0) = \Phi(-0.2085) = 1 - \Phi(0.2085) = 1 - 0.5826 = 0.4174\ldots \approx 0.417

Answer

0.417 \quad (3 \text{ s.f.})

Final answer

0.417

Detailed explanation

Walkthrough

We need the probability that a large bag's mass is less than 3 times a small bag's mass. This is not a simple sum; it is a comparison. Define $D = L - 3S$ . Then the event $L < 3S$ is exactly $D < 0$ .

Because $L$ and $S$ are independent normal variables, $D$ is also normal. Its mean is $E(L) - 3E(S) = 2.5 - 3(0.8) = 0.1$ . Its variance is $Var(L) + (-3)^2 Var(S) = 0.05 + 9(0.02) = 0.23$ . Note that the coefficient $-3$ is squared when it is taken out of the variance.

Now we need $P(D < 0)$ . Standardise:

z = \frac{0 - 0.1}{\sqrt{0.23}} = -0.2085\ldots

Because this is a less-than probability, we use the lower tail directly: $P(D < 0) = \Phi(-0.2085)$ . Using symmetry, $\Phi(-0.2085) = 1 - \Phi(0.2085)$ . From the table, $\Phi(0.2085) \approx 0.5826$ , so the probability is $1 - 0.5826 = 0.4174 \approx 0.417$ .

Key Takeaways

When comparing two quantities such as $L < 3S$ , form a single normal variable $D = L - 3S$ and translate the inequality into a condition on $D$ . For independent variables, $Var(aX + bY) = a^2Var(X) + b^2Var(Y)$ , so the coefficient of $S$ must be squared. Then standardise and use the correct tail.

Common Mistakes

Forgetting to square the coefficient 3 when calculating the variance, giving $0.05 + 3(0.02) = 0.11$ instead of $0.23$ .
Using $E(D) = 2.5 - 0.8$ instead of $2.5 - 3(0.8)$ .
Confusing the direction of the inequality and using the wrong tail.
Not realising that $P(D < 0)$ is a lower-tail probability, so $\Phi(-0.2085)$ is needed.

Things to Be Careful About

The variance must include $(-3)^2Var(S)$ , not $-3Var(S)$ . The standard deviation is $\sqrt{0.23}$ , not $0.23$ . The final probability can be written as $0.417$ or $0.418$ depending on the rounding of the z-value, and either is accepted.

Techniques used

form a difference of independent normal variablescompute expectation and variance of a linear combinationstandardise a normal variablefind a lower-tail normal probability

Q7Medium-EasyContinuous Random Variables

The time, in minutes, taken by students to complete a test is modelled by the random variable $X$ with probability density function

f(x) = \begin{cases} -\frac{3}{4}(x-3)(x-5) & 3 \le x \le 5, \\ 0 & \text{otherwise.} \end{cases}

(a)

Find the probability that a randomly chosen student takes longer than 4.5 minutes to complete the test.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

We need the probability that $X>4.5$ . Since $X$ is continuous with PDF $f$ , this probability is the area under the curve between $x=4.5$ and $x=5$ :

P(X>4.5)=\int_{4.5}^{5} f(x)\,dx.

Expand the quadratic and integrate term by term.

Working

The PDF is

f(x)=-\frac{3}{4}(x-3)(x-5)=-\frac{3}{4}(x^2-8x+15).

Therefore

P(X>4.5)=-\frac{3}{4}\int_{4.5}^{5}(x^2-8x+15)\,dx.

Integrating gives

-\frac{3}{4}\left[\frac{x^3}{3}-4x^2+15x\right]_{4.5}^{5}.

At $x=5$ :

\frac{125}{3}-100+75=\frac{50}{3}.

At $x=\frac{9}{2}$ :

\frac{1}{3}\left(\frac{9}{2}\right)^3-4\left(\frac{9}{2}\right)^2+15\left(\frac{9}{2}\right) =\frac{243}{8}-81+\frac{135}{2}=\frac{135}{8}.

So

P(X>4.5)=-\frac{3}{4}\left(\frac{50}{3}-\frac{135}{8}\right) =-\frac{3}{4}\left(-\frac{5}{24}\right)=\frac{5}{32}.

Answer

P(X>4.5)=\frac{5}{32}.

Final answer

5/32

Detailed explanation

Walkthrough

We are asked for the probability that a student takes longer than $4.5$ minutes. Because $X$ is continuous with a probability density function, this is the area under the PDF curve from $4.5$ to $5$ , the upper endpoint of the sample space.

First expand the product in the PDF:

(x-3)(x-5)=x^2-8x+15.

Then write the probability as a definite integral:

P(X>4.5)=-\frac{3}{4}\int_{4.5}^{5}(x^2-8x+15)\,dx.

The integral of $x^2-8x+15$ is $\frac{x^3}{3}-4x^2+15x$ . Substitute $x=5$ and $x=\frac{9}{2}$ , take the difference, and multiply by $-\frac{3}{4}$ . This gives $\frac{5}{32}$ .

Key Takeaways

A probability from a continuous PDF is the area under the curve over the required interval.
To integrate a polynomial PDF, expand into standard powers first.
Care must be taken when substituting fractional limits.

Common Mistakes

Forgetting the factor $-\frac{3}{4}$ when setting up the integral. The mark scheme condones this at the integration step but requires it when substituting.
Using the wrong limits, such as $3$ to $4.5$ , instead of $4.5$ to $5$ .
Arithmetic errors in evaluating fractions such as $\frac{50}{3}-\frac{135}{8}$ .

Things to Be Careful About

The PDF is zero outside $3\le x\le 5$ , so for "longer than $4.5$ " integrate from $4.5$ to $5$ only.
The final substitute must include the $\frac{3}{4}$ factor; otherwise the mark for correct substitution is not awarded.

Techniques used

expand the quadratic form of the probability density functionset up a definite integral for a tail probabilityintegrate the polynomialevaluate the definite integral at its limits

(b)

Write down the median of $X$ .

1M Similar

(c)

Without performing an integration, use your answer to part (a) to find $P(3.5 < X < 4.5)$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The PDF is symmetric about $x=4$ , so the right tail $X>4.5$ has the same probability as the left tail $X<3.5$ . Since the total probability is $1$ , subtract both tails from $1$ .

Working

From part (a),

P(X>4.5)=\frac{5}{32}.

By symmetry,

P(X<3.5)=\frac{5}{32}.

Hence

P(3.5<X<4.5)=1-2\left(\frac{5}{32}\right) =1-\frac{10}{32}=\frac{22}{32}=\frac{11}{16}.

Equivalently,

P(3.5<X<4.5)=2\left(\frac12-\frac{5}{32}\right)=\frac{11}{16}.

Answer

P(3.5<X<4.5)=\frac{11}{16}.

Final answer

11/16

Detailed explanation

Walkthrough

We are asked to find $P(3.5<X<4.5)$ without performing an integration. This is the central region of the distribution.

The distribution is symmetric about $x=4$ . The interval $X>4.5$ is the right-hand tail; from part (a) its probability is $\frac{5}{32}$ . By symmetry the left-hand tail $X<3.5$ has the same probability, $\frac{5}{32}$ .

The whole probability is $1$ , so the required middle region is $1$ minus the two equal tails:

P(3.5<X<4.5)=1-2\cdot\frac{5}{32}.

This simplifies to $\frac{11}{16}$ .

Key Takeaways

Symmetry of the PDF lets you transfer a known probability from one tail to the other.
The total probability of $1$ can be split into complementary regions.
Always look for ways to reuse previous results rather than re-integrating a PDF.

Common Mistakes

Not showing working; the mark scheme requires the working for the method mark.
Forgetting that both tails must be subtracted, so calculating only $1-\frac{5}{32}$ .
Attempting to integrate from $3.5$ to $4.5$ despite the instruction not to perform integration; if done correctly it earns only a special-case mark, not full marks.

Things to Be Careful About

The interval $3.5<X<4.5$ is centered at the median $4$ , so symmetry applies directly.
Use exact fractions such as $\frac{5}{32}$ to avoid rounding errors before the final answer.

Techniques used

use the result from the previous partapply symmetry to pair equal tail probabilitiessubtract tail probabilities from the total probability

Mathematics 9709/61 — October/November 2025

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