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 (or 0.0491)

9709/63 Oct/Nov 2025 — Mathematics Solved Past Paper

Q1Medium-HardThe 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

(b)

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

3M Similar

DifficultyMedium

Worked solution

Approach

Since $X$ and $Y$ are independent Poisson variables, their sum is also Poisson with mean $3 + 2 = 5$ . Use the complement rule: $P(X+Y > 2) = 1 - P(X+Y \leq 2)$ .

Working

X + Y \sim \text{Po}(5)

P(X+Y > 2) = 1 - \left[P(X+Y = 0) + P(X+Y = 1) + P(X+Y = 2)\right]

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

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

= 1 - (0.0067379 + 0.0336897 + 0.0842243)

= 1 - 0.124652 = 0.875348

Answer

P(X+Y > 2) = 0.875 \quad (3 \text{ sf})

Final answer

0.875

Detailed explanation

Walkthrough

Because $X$ and $Y$ are independent Poisson random variables, the sum $X+Y$ is also Poisson. Its mean is the sum of the means: $3 + 2 = 5$ .

The event $X+Y > 2$ is the complement of $X+Y \leq 2$ . It is easier to calculate the probability of 0, 1, or 2 and subtract from 1.

For $X+Y \sim \text{Po}(5)$ ,

P(X+Y = r) = \frac{e^{-5}5^r}{r!}

So add the probabilities for $r=0$ , $r=1$ and $r=2$ , then subtract the total from 1.

Key Takeaways

The sum of independent Poisson random variables is Poisson with mean equal to the sum of the means.
Using the complement rule simplifies probabilities of the form $P(Z > k)$ .

Common Mistakes

Forgetting that $X+Y$ is Poisson with mean 5, and instead using the original means separately.
Calculating $P(X+Y > 2)$ as $1 - P(X+Y = 2)$ instead of subtracting all values up to 2.

Things to Be Careful About

The inequality is strict, so $X+Y = 2$ must be included in the subtracted probability.
Keep sufficient decimal places before giving the final answer to 3 significant figures.

Techniques used

recognise that the sum of independent Poisson variables is Poissonuse the complement rulecalculate cumulative Poisson probabilities

(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-Hard

Worked solution

Approach

The sum of 100 values of $X$ is $\text{Po}(300)$ , and the sum of 150 values of $Y$ is also $\text{Po}(300)$ . Therefore $T \sim \text{Po}(600)$ . Since $\lambda = 600$ is large, approximate $T$ by a normal distribution with mean 600 and variance 600. Use a continuity correction for $P(T < 560)$ .

Working

T \sim \text{Po}(600) \approx N(600, 600)

For a discrete variable, $P(T < 560) = P(T \leq 559)$ . With the continuity correction, use 559.5:

z = \frac{559.5 - 600}{\sqrt{600}} = \frac{-40.5}{24.4949} = -1.653

P(T < 560) = P(Z < -1.653) = 1 - \Phi(1.653)

= 1 - 0.9508 = 0.0492

Answer

P(T < 560) = 0.0492 \quad (3 \text{ sf})

(Using tables, 0.0491 is also accepted.)

Final answer

0.0492 (or 0.0491)

Detailed explanation

Walkthrough

First find the distribution of $T$ . Each of the 100 values of $X$ is $\text{Po}(3)$ , so their sum is $\text{Po}(300)$ . Each of the 150 values of $Y$ is $\text{Po}(2)$ , so their sum is also $\text{Po}(300)$ . Adding these gives $T \sim \text{Po}(600)$ .

Because $\lambda = 600$ is large, the Poisson distribution can be approximated by a normal distribution with the same mean and variance: $T \approx N(600, 600)$ .

Since $T$ is discrete, the event $T < 560$ means $T$ can be at most 559. The continuity correction uses the boundary 559.5. Standardise:

z = \frac{559.5 - 600}{\sqrt{600}}

Then use the symmetry of the normal distribution to write $P(Z < -1.653)$ as $1 - \Phi(1.653)$ and read the probability from the normal table.

Key Takeaways

The sum of independent Poisson variables is Poisson, with mean equal to the sum of the means.
A Poisson distribution with a large mean can be approximated by a normal distribution with the same mean and variance.
A continuity correction is needed when approximating a discrete distribution by a continuous one.

Common Mistakes

Forgetting the continuity correction, or using 560 instead of 559.5.
Using the standard deviation instead of the variance when writing $N(600, 600)$ .
Forgetting to subtract the table probability from 1 when the $z$ -value is negative.

Things to Be Careful About

The variance of $T$ is 600, so the standard deviation is $\sqrt{600}$ , not 600.
The continuity correction must move the boundary in the correct direction: $T < 560$ becomes $T \leq 559$ , so use 559.5.
The final answer may be 0.0492 or 0.0491 depending on the normal table used; both are accepted.

Techniques used

sum independent Poisson variables to obtain T ~ Po(600)apply normal approximation with continuity correctionstandardise and use normal distribution tables

Q28MMediumSampling and EstimationHypothesis Tests

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

We want to test whether the true mean mass is 500 grams. Because the sample is large and the population variance is unknown, we first estimate the population variance from the sample, then use the Central Limit Theorem to standardise the sample mean as an approximately normal statistic.

Working

The sample mean is

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

The unbiased estimate of the population variance is

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

Substituting:

s^2 = \frac{1}{59}\left(14970300 - \frac{29970^2}{60}\right) = \frac{1}{59}(14970300 - 14970015) = \frac{285}{59} = 4.8305

The hypotheses are

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

Under $H_0$ , the standardised sample mean is

z = \frac{\bar{x} - 500}{s/\sqrt{n}} = \frac{499.5 - 500}{\sqrt{4.8305}/\sqrt{60}} = \frac{-0.5}{0.28374} = -1.762

For a two-tailed test at the 5% significance level, the critical values are $z = \pm 1.960$ . Since

-1.960 < -1.762 < 1.960

we do not reject $H_0$ . Equivalently, the tail probability beyond $1.762$ is approximately $0.0390$ , which is greater than $0.025$ , so the result is not significant.

Answer

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

Final answer

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

Detailed explanation

Walkthrough

This is a hypothesis test about a population mean, using a large sample. The first step is to find the sample mean from the given sums:

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

This is the point estimate of the population mean. Next we need an estimate of the variability of the sample mean. The unbiased estimate of the population variance is

s^2 = \frac{1}{n-1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{285}{59}

Using $n-1$ rather than $n$ is essential here because we are estimating a population variance from sample data.

The hypotheses are

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

The alternative is two-tailed because the inspector is testing whether the mean is correct, not whether it is specifically too high or too low.

Because $n=60$ is large, the Central Limit Theorem tells us that the sample mean is approximately normally distributed. So, assuming $H_0$ is true,

z = \frac{\bar{x} - 500}{s/\sqrt{n}} \approx -1.762

For a 5% two-tailed test, the critical region is $z < -1.960$ or $z > 1.960$ . Since $-1.762$ lies between the critical values, we do not reject the null hypothesis. Equivalently, the tail probability $0.0390$ is larger than $0.025$ .

Key Takeaways

For a large sample from a distribution with unknown variance, the sample mean is approximately normal by the Central Limit Theorem.
An unbiased estimate of the population variance uses denominator $n-1$ .
A two-tailed test at the 5% level uses tail area $0.025$ in each tail and critical value $\pm 1.96$ .
A non-significant result means there is insufficient evidence against $H_0$ , not that $H_0$ is proven true.
Summary statistics $\sum x$ and $\sum x^2$ are enough to compute both the sample mean and the estimated variance.

Common Mistakes

Dividing by $n$ rather than $n-1$ when estimating the variance. The mark scheme specifically does not allow the biased variance $4.75$ ; it should be $s^2=4.83$ .
Using a one-tailed critical value such as $1.645$ when the alternative hypothesis is $\neq$ .
Forgetting to divide by $\sqrt{n}$ when constructing the standard error.
Writing a conclusion as though the mean is 500 grams; instead, say there is insufficient evidence that it is not 500 grams.
Not giving the hypotheses with a symbol such as $\mu$ for the population mean.

Things to Be Careful About

The test statistic is negative because $499.5 < 500$ , but the comparison uses distance, so use $|z|$ or check both tails.
The 5% significance level is split between two tails, so each tail is 2.5%, giving $1.96$ .
If using the p-value approach, use $0.0390 > 0.025$ for one tail, or double it and compare $0.0780$ with $0.05$ .
The final conclusion must be in context: packets of Trueleaf tea and mass in grams.
Avoid contradictory statements such as "reject" followed by "there is no evidence"; the conclusion must be consistent with the comparison.

Techniques used

compute the sample mean from summary statisticscalculate the unbiased estimate of population variancestate null and alternative hypotheses for a two-tailed teststandardise the sample mean using the estimated standard errorcompare the test statistic with the critical value

Approach

The random variable $X$ counts incorrect characters in 14500 independent trials, so exactly $X \sim \mathrm{B}(14500, 0.0001)$ . Because $n$ is large and $p$ is small, use the Poisson approximation with

\lambda = np = 14500 \times 0.0001 = 1.45.

Then

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

Working

The Poisson distribution is $X \sim \mathrm{Po}(1.45)$ , with

P(X = r) = \frac{e^{-1.45} \, 1.45^r}{r!}.

So

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 + 0.508104\ldots)

= e^{-1.45}(4.009354\ldots)

= 0.940475\ldots

Answer

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

Final answer

0.940

Detailed explanation

Walkthrough

This is a binomial situation because each character in the data set is an independent trial with probability $p = 0.0001$ of being incorrect. With $n = 14500$ , the exact distribution is $X \sim \mathrm{B}(14500, 0.0001)$ , but the question asks for a suitable approximating distribution. Since $n$ is large and $p$ is very small, the Poisson approximation applies, with mean $\lambda = np = 1.45$ .

In a Poisson distribution, $P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!}$ . The event $X < 4$ means $X = 0, 1, 2$ or $3$ . Substituting $\lambda = 1.45$ and adding the four probabilities gives $0.940$ to 3 significant figures.

Key Takeaways

A binomial distribution with large $n$ and small $p$ can be approximated by a Poisson distribution with $\lambda = np$ .
For cumulative probabilities, sum the individual Poisson probabilities over the required values.
The smaller the value of $p$ and the larger the value of $n$ , the better the Poisson approximation becomes.

Common Mistakes

Not stating that a Poisson approximation is being used; the mark scheme requires an indication of Poisson.
Using $P(X < 4)$ as $P(X = 0)+\cdots+P(X = 4)$ instead of stopping at $3$ .
Forgetting the factorial in the denominator when writing $P(X = r)$ .
Trying to use a normal approximation with a continuity correction here; this is a Poisson situation.

Things to Be Careful About

$X < 4$ is strict, so exclude $X = 4$ .
The value $\lambda = 1.45$ must come from $np = 14500 \times 0.0001$ .
Rounding: the final answer may be given as $0.940$ or $0.941$ .

Techniques used

recognise a binomial contextapply the Poisson approximation to the binomial distributionsum Poisson probabilities for a strict inequality

(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

For a new proportion $p$ , the Poisson approximation has mean $14500p$ , so the probability of no incorrect characters is $e^{-14500p}$ . The original mean is $1.45$ , so the original probability of no incorrect characters is $e^{-1.45}$ . Setting the new probability equal to double this gives an equation in $p$ ; solve it by taking natural logarithms.

Working

Original:

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

New:

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

The requirement is:

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

Taking natural logarithms of both sides:

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

Therefore

p = \frac{1.45 - \ln 2}{14500}.

Using $\ln 2 = 0.693147\ldots$ ,

p = \frac{0.756852\ldots}{14500} = 0.000052196\ldots

Answer

p = 0.0000522 \quad (3 \text{ sf})

Final answer

0.0000522

Detailed explanation

Walkthrough

From part (a), the original Poisson parameter is $\lambda = 1.45$ , so the original probability of no incorrect characters is $P(X = 0) = e^{-1.45}$ .

If the new proportion of incorrect characters is $p$ , then the same Poisson approximation gives a new mean of $14500p$ , so the new probability of no incorrect characters is

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

The aim is that this new probability should be double the original probability, giving

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

Taking natural logarithms uses $\ln(ab) = \ln a + \ln b$ and $\ln(e^{-1.45}) = -1.45$ , so

-14500p = \ln 2 - 1.45.

Rearranging gives

p = \frac{1.45 - \ln 2}{14500}.

Since $\ln 2 \approx 0.693147$ , this gives $p \approx 0.0000522$ .

Key Takeaways

The Poisson parameter for $X$ is always $np$ , so when $p$ changes, the mean changes proportionally.
The probability of zero events in a Poisson distribution is simply $e^{-\lambda}$ .
Exponential equations of the form $e^{k} = c$ can be solved by taking natural logarithms of both sides.

Common Mistakes

Using the old parameter $\lambda = 1.45$ for the new situation instead of $14500p$ .
Writing $\ln(2e^{-1.45})$ incorrectly as $\ln 2 \times (-1.45)$ ; the correct value is $\ln 2 - 1.45$ .
Setting the factor of 2 on the wrong side of the equation, which gives an incorrect value of $p$ .
Forgetting to divide by 14500 after taking logarithms.

Things to Be Careful About

When rearranging, note that $e^{-14500p} = 2e^{-1.45}$ gives $-14500p = \ln 2 - 1.45$ . Be careful with signs.
The final answer must be given to 3 significant figures: $0.0000522$ .
The numerical size of $p$ should be a small proportion, so a result near $0.0000522$ is sensible.

Techniques used

model the new probability of zero with parameter 14500pset up a doubling equationsolve an exponential equation using natural 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-distribution confidence interval, the width is twice the critical value multiplied by the standard error. Equate the width of the $\alpha\%$ interval to $1.414$ times the width of the 90% interval to find the critical value, then convert this critical value to a confidence percentage.

Working

The width of a confidence interval for a population mean, using a known standard deviation $\sigma$ and sample size $n$ , is

\text{width} = 2z \cdot \frac{\sigma}{\sqrt{n}}

where $z$ is the two-tailed standard normal critical value for the confidence level.

For a 90% confidence interval, $z = 1.645$ . Let $z_{\alpha}$ be the critical value for the $\alpha\%$ confidence interval. Since the $\alpha\%$ interval has width $1.414$ times the 90% interval:

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

Cancelling the common factor $2\sigma/\sqrt{n}$ :

z_{\alpha} = 1.414 \times 1.645 = 2.326

Now convert this critical value to a two-tailed confidence level:

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

Since $\Phi(2.326) = 0.9900$ :

\alpha\% = (2(0.9900) - 1) \times 100\% = 0.9800 \times 100\% = 98\%

Answer

\alpha = 98

Final answer

98

Detailed explanation

Walkthrough

The confidence interval here has the form

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

so its total width is $2z\sigma/\sqrt{n}$ . We are told the $\alpha\%$ interval is $1.414$ times as wide as the 90% interval. Because the sample size and standard deviation are the same for both intervals, the factor $2\sigma/\sqrt{n}$ appears in both widths and cancels. This leaves a simple relation between the two critical values: the $\alpha\%$ critical value is $1.414$ times the 90% critical value.

Using the 90% critical value $1.645$ , the $\alpha\%$ critical value is $1.414 \times 1.645 = 2.326$ . A critical value of $2.326$ means that the central area of the standard normal distribution between $-2.326$ and $2.326$ is $0.98$ , so the confidence level is 98%.

Key Takeaways

The width of a confidence interval scales with the critical value, so common factors such as $2$ , $\sigma$ , and $\sqrt{n}$ cancel when comparing intervals from the same sample.
A two-tailed confidence level is the total central probability, found from $2\Phi(z) - 1$ .
The standard normal critical value can be converted into a percentage confidence level.

Common Mistakes

Forgetting the factor of 2 in the width. It cancels here, but it must be included in the formula.
Using a one-tailed probability for the critical value instead of the two-tailed confidence level.
Not recognising that both intervals use the same standard error, and unnecessarily trying to solve for $\sigma$ or $n$ .
Approximating $1.414 \times 1.645$ incorrectly instead of using $2.326$ .

Things to Be Careful About

The confidence level is expressed as a percentage, so the decimal probability $0.98$ corresponds to $\alpha = 98\%$ . The mark scheme accepts either 98 or 98%. Also, the critical value for 90% confidence is $1.645$ , not $1.282$ (which would be for an 80% confidence interval); although the method mark could be awarded with 1.282, the final relation is based on 90%.

Techniques used

relate confidence interval width to the standard normal critical valuecancel the common standard-error factorconvert a critical 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

Since both intervals have the same centre, the 90% interval is contained inside the wider 98% interval. The event that the 90% interval contains the mean is therefore a subset of the event that the 98% interval contains the mean, so use the conditional probability formula for nested events.

Working

Let $Z = \dfrac{\bar{X} - \mu}{\sigma/\sqrt{n}}$ . The 90% interval contains $\mu$ when

|Z| \leq 1.645

The 98% interval contains $\mu$ when

|Z| \leq 2.326

Since $1.645 < 2.326$ , the event $|Z| \leq 1.645$ is a subset of $|Z| \leq 2.326$ . Therefore

P(90\% \text{ interval contains } \mu \mid 98\% \text{ interval contains } \mu) = \frac{P(90\% \text{ interval contains } \mu)}{P(98\% \text{ interval contains } \mu)}

Substituting the confidence levels as probabilities:

= \frac{0.90}{0.98} = \frac{45}{49} = 0.918\text{ (3 s.f.)}

Answer

\frac{45}{49} \text{ or } 0.918

Final answer

45/49 or 0.918

Detailed explanation

Walkthrough

Both confidence intervals are centred on the same sample mean $\bar{x}$ ; only their widths differ. A 90% interval uses a smaller critical value ( $1.645$ ) than the 98% interval ( $2.326$ ), so whenever the 90% interval contains the population mean, the 98% interval must also contain it. This makes the 90% event a subset of the 98% event.

For conditional probability, when one event is a subset of another, $P(A \mid B) = P(A)/P(B)$ . Here $A$ is the event that the 90% interval contains $\mu$ , whose probability is 0.90, and $B$ is the event that the 98% interval contains $\mu$ , whose probability is 0.98. So the required conditional probability is $0.90/0.98$ .

Key Takeaways

A lower-confidence interval is narrower and is contained within a higher-confidence interval centred at the same sample mean.
If event $A$ is contained in event $B$ , then $P(A \mid B) = P(A)/P(B)$ .
Confidence level probabilities can be used directly in probability calculations.

Common Mistakes

Assuming the probability is 1 because the wider interval already contains the mean.
Dividing in the wrong order, i.e. $0.98/0.90$ instead of $0.90/0.98$ .
Forgetting to use the confidence probabilities (0.90 and 0.98) as decimals.

Things to Be Careful About

The mark scheme allows follow-through from the value of $\alpha$ found in part (a), but if $\alpha$ were less than 90, the subset relationship would not hold and the conditional-probability ratio would not apply. Here $\alpha = 98$ , so the calculation is valid. Express the answer as a fraction $45/49$ or as $0.918$ to 3 significant figures.

Techniques used

recognise that a narrower confidence interval event is nested inside a wider oneapply conditional probability to nested confidence interval events

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

Let $X$ be the number of households in the sample with more than 4 people. Under $H_0$ , $X \sim \mathrm{B}(40,0.2)$ . Since Laxmi believes the percentage is lower, this is a one-tailed lower-tail test:

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

The rejection region is the largest set $X \le c$ for which $P(X \le c) \le 0.025$ . The probability of a Type I error is $P(\text{reject } H_0 \mid H_0 \text{ true}) = P(X \le c)$ .

Working

Calculate the binomial probabilities:

P(X=0) = 0.8^{40} = 0.0001329

P(X=1) = \binom{40}{1}(0.8)^{39}(0.2) = 0.0013292

P(X=2) = \binom{40}{2}(0.8)^{38}(0.2)^2 = 0.0064799

So

P(X \le 2) = 0.0001329 + 0.0013292 + 0.0064799 = 0.00794

This is less than $0.025$ . Check the next value:

P(X=3) = \binom{40}{3}(0.8)^{37}(0.2)^3 = 0.0205199

P(X \le 3) = 0.00794 + 0.0205199 = 0.0285 > 0.025

Thus the critical region is $X \le 2$ , and

P(\text{Type I error}) = P(X \le 2) = 0.00794

Answer

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

Final answer

0.00794 (3 sf)

Detailed explanation

Walkthrough

We start by defining the random variable $X$ as the number of households in the sample of 40 that contain more than 4 people. The null hypothesis is $H_0: p = 0.2$ , because the national percentage is 20%. Laxmi believes the percentage in her town is lower, so the alternative hypothesis is $H_1: p < 0.2$ . This is a one-tailed lower-tail test.

A Type I error is made when the test rejects $H_0$ even though $H_0$ is true. Its probability is therefore the probability that the test statistic falls in the rejection region when $X \sim \mathrm{B}(40,0.2)$ . We need to find the largest number $c$ such that $P(X \le c) \le 0.025$ , because the test is at the 2.5% significance level.

We calculate $P(X=0)$ , $P(X=1)$ and $P(X=2)$ using the binomial formula. Their sum is $P(X \le 2)=0.00794$ , which is less than 0.025. We must also check the next value: adding $P(X=3)$ gives $P(X \le 3)=0.0285$ , which is greater than 0.025. This shows that the rejection region cannot include 3, so the critical region is $X \le 2$ . The probability of a Type I error is the actual probability of being in this critical region, namely 0.00794.

Key Takeaways

This question tests how to find the critical region for a one-tailed binomial test and how to convert that region into the probability of a Type I error. The probability of a Type I error is not automatically the nominal significance level; it is the actual probability of the rejection region under $H_0$ . You must compare cumulative binomial probabilities with the significance level to decide where the critical region ends.

Common Mistakes

Giving only $P(X \le 2)=0.00794$ without showing the binomial expressions. The mark scheme requires the expressions and the comparison with $P(X \le 3)$ for full marks.
Forgetting to check $P(X \le 3)>0.025$ . Without this comparison, you have not justified that the critical region is $X \le 2$ .
Using the wrong tail. Since Laxmi believes the percentage is lower, the test must be lower-tailed.
Using $p=0.8$ instead of $p=0.2$ . The probability of a household containing more than 4 people is 0.2 under $H_0$ .

Things to Be Careful About

The significance level is 2.5%, so the cumulative probability of the rejection region must be at most 0.025. Because $P(X \le 2)=0.00794$ is well below 0.025 while $P(X \le 3)=0.0285$ is above, the boundary is at $X=2$ . Round only at the end: use the full values when adding the terms. The final probability should be given as 0.00794 (3 sf).

Techniques used

set up a one-tailed binomial hypothesis testcalculate individual binomial probabilitiescompare cumulative probabilities with the significance levelidentify the critical region and the Type I error probability

(b)

State the rejection region for the test.

1M Similar

DifficultyMedium-Easy

Worked solution

Approach

The rejection region is the set of values of $X$ for which $H_0$ is rejected. For a lower-tail binomial test, it has the form $X \le c$ , where $c$ is the largest value such that $P(X \le c) \le 0.025$ when $X \sim \mathrm{B}(40,0.2)$ .

Working

From part (a), $P(X \le 2) = 0.00794 \le 0.025$ and $P(X \le 3) = 0.0285 > 0.025$ . Therefore the largest value in the rejection region is 2.

Answer

Rejection region: $X \le 2$ .

Final answer

X ≤ 2

Detailed explanation

Walkthrough

The rejection region is the set of values of $X$ for which Laxmi will reject $H_0$ . In a lower-tail binomial test, this region has the form $X \le c$ . From part (a), $P(X \le 2)=0.00794 \le 0.025$ , but $P(X \le 3)=0.0285>0.025$ . Therefore the largest value that can be included is 2, and the rejection region is $X \le 2$ .

Key Takeaways

The rejection region is determined by the cumulative probability being no greater than the significance level. It must be stated in terms of the test statistic $X$ , not just as a probability.

Common Mistakes

Writing $X<2$ instead of $X \le 2$ .
Writing the probability 0.00794 as the rejection region. The rejection region is the set of observed values, not the probability.
Including 3 in the rejection region because $P(X=3)$ is small; the correct condition is on the cumulative probability $P(X \le 3)$ , which exceeds 0.025.

Things to Be Careful About

The rejection region for a lower-tail test is always of the form $X \le c$ . Make sure the boundary value is included when its cumulative probability is below the significance level.

Techniques used

use cumulative binomial probabilities to determine the critical regioncompare P(X ≤ 2) and P(X ≤ 3) with the significance level

(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

DifficultyMedium-Easy

Worked solution

Approach

A Type II error occurs when $H_0$ is false but the test does not reject $H_0$ . If the observed value lies in the rejection region, $H_0$ is rejected, so a Type II error cannot occur.

Working

Laxmi observed $X = 2$ . The rejection region from part (b) is $X \le 2$ , so the observed value lies in the rejection region. Therefore she rejects $H_0$ . A Type II error would require failing to reject $H_0$ , which is not the case.

Answer

Because $X = 2$ is in the rejection region, $H_0$ is rejected; a Type II error is only possible when $H_0$ is not rejected.

Final answer

Because X = 2 lies in the rejection region, H0 is rejected, so a Type II error cannot occur.

Detailed explanation

Walkthrough

A Type II error occurs when $H_0$ is false but the test fails to reject $H_0$ . Laxmi observed $X=2$ . From part (b), the rejection region is $X \le 2$ , so the observed value lies in the rejection region. This means she rejects $H_0$ . Since she rejects $H_0$ , she cannot fail to reject a false $H_0$ , so a Type II error is impossible in this outcome.

Key Takeaways

Type II error is conditional on $H_0$ being false and on the test not rejecting it. If the observed value falls in the rejection region, the test rejects $H_0$ , so a Type II error cannot occur.

Common Mistakes

Saying a Type II error is impossible because the sample proportion is low. The reason is specifically that the observed value lies in the rejection region.
Confusing Type I and Type II errors. A Type I error is rejecting a true $H_0$ ; a Type II error is failing to reject a false $H_0$ .

Things to Be Careful About

The observed value $X=2$ is exactly the boundary of the rejection region, so it is included. Because it is included, $H_0$ is rejected and no Type II error can be made.

Techniques used

check the observed value against the rejection regionapply the definition of a Type II error

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 3.55kg.

5M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $L$ be the mass of a large bag and $S$ the mass of a small bag. The total mass is $T = L + S$ . Since $L$ and $S$ are independent normal variables, $T$ is also normal with mean equal to the sum of the means and variance equal to the sum of the variances. Standardise $T$ and use the standard normal table to find the required probability.

Working

Given:

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

The total mass is $T = L + S$ . Since $L$ and $S$ are independent:

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

Standardise:

P(T > 3.55) = P\left(Z > \frac{3.55 - 3.3}{\sqrt{0.07}}\right) = P(Z > 0.945)

Using the normal distribution table:

P(Z > 0.945) = 1 - \Phi(0.945) = 1 - 0.8276 = 0.1724

Answer

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

Final answer

0.172 (3 s.f.)

Detailed explanation

Walkthrough

We are given the masses of two independent normal distributions. The key idea is that when we add two independent normal random variables, the result is also normal. The mean of the sum is the sum of the means, and the variance of the sum is the sum of the variances (this is because the variables are independent, so the covariance is zero).

Once we know the distribution of the total mass $T \sim N(3.3, 0.07)$ , we need to find $P(T > 3.55)$ . To use the standard normal table, we standardise by subtracting the mean and dividing by the standard deviation. The standard deviation is $\sqrt{0.07}$ , not $0.07$ — this is a common error.

The standardised value is $z = 0.945$ . Since we want the probability that $T$ is greater than 3.55, we want the area to the right of $z = 0.945$ under the standard normal curve. The table gives the area to the left, so we subtract from 1.

Key Takeaways

The sum of independent normal variables is normal. For independent variables, $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$ . To find probabilities for a normal distribution, always standardise first. The normal table gives $\Phi(z) = P(Z \le z)$ , the area to the left.

Common Mistakes

Using the variance (0.07) instead of the standard deviation ( $\sqrt{0.07}$ ) when standardising. Forgetting to subtract from 1 when the question asks for the probability of being "more than" a value. Adding variances incorrectly or forgetting that independence is required.

Things to Be Careful About

The variance of the sum is the sum of the variances only because $L$ and $S$ are independent. The final answer must be given to 3 significant figures as requested. Make sure to standardise using the standard deviation, not the variance.

Techniques used

form the sum of independent normal variablescalculate the mean and variance of the sumstandardise to the standard normal distributionuse the normal distribution table

(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

We need $P(L < 3S)$ . Rearrange to $P(L - 3S < 0)$ and define $D = L - 3S$ . Since $L$ and $S$ are independent normal variables, $D$ is normal. Find its mean and variance, then standardise and use the normal table.

Working

Define $D = L - 3S$ . Since $L$ and $S$ are independent:

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

\text{Var}(D) = \text{Var}(L) + 3^2\text{Var}(S) = 0.05 + 9(0.02) = 0.23

So $D \sim N(0.1, 0.23)$ .

We need:

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

Standardise:

P(D < 0) = P\left(Z < \frac{0 - 0.1}{\sqrt{0.23}}\right) = P(Z < -0.2085)

By symmetry of the normal distribution:

P(Z < -0.2085) = 1 - \Phi(0.2085) = 1 - 0.5826 = 0.4174

Answer

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

Final answer

0.417 (3 s.f.)

Detailed explanation

Walkthrough

The inequality $L < 3S$ can be rearranged to $L - 3S < 0$ . This is a linear combination of the two independent normal variables. The key difference from part (a) is that $S$ is multiplied by 3, so when we find the variance, the coefficient 3 must be squared: $\text{Var}(3S) = 3^2\text{Var}(S) = 9 \times 0.02 = 0.18$ .

The mean of $D = L - 3S$ is $E(D) = E(L) - 3E(S) = 2.5 - 2.4 = 0.1$ . The variance is $\text{Var}(D) = \text{Var}(L) + 3^2\text{Var}(S) = 0.05 + 0.18 = 0.23$ . Note that the variance of $-3S$ is the same as the variance of $3S$ because variance is unaffected by sign.

We then standardise: $z = (0 - 0.1)/\sqrt{0.23} = -0.2085$ . Since we want $P(D < 0)$ , we want the area to the left of $z = -0.2085$ . By symmetry, this equals $1 - \Phi(0.2085)$ .

Key Takeaways

An inequality like $L < 3S$ can be converted to a linear combination $L - 3S < 0$ . When a variable is multiplied by a constant, its variance is multiplied by the square of that constant. The variance of $aX + bY$ for independent $X, Y$ is $a^2\text{Var}(X) + b^2\text{Var}(Y)$ . The standard normal distribution is symmetric about 0, so $\Phi(-z) = 1 - \Phi(z)$ .

Common Mistakes

Forgetting to square the coefficient 3 when finding the variance of $3S$ . Using $\text{Var}(L) - \text{Var}(S)$ or $\text{Var}(L) - 3\text{Var}(S)$ instead of adding the squared-coefficient terms. Confusing the direction of the inequality when standardising.

Things to Be Careful About

The variance of a linear combination is always a sum of squared-coefficient times variance terms (for independent variables) — the minus sign on $3S$ does not make the variance negative. The standardised value is negative here; use the symmetry property of the normal distribution correctly. The answer may be given as 0.417 or 0.418 depending on rounding; the mark scheme allows both.

Techniques used

rearrange the inequality into a linear combinationcalculate the mean and variance of the linear combinationstandardise to the standard normal distributionapply the symmetry property of the normal distribution

Q7MediumContinuous 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

Worked solution

Approach

We want $P(X > 4.5)$ . Since the pdf is zero outside $3 \le x \le 5$ , this is the integral of $f(x)$ from $4.5$ to $5$ .

Working

Expand the integrand:

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

So

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

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

Evaluate at the limits:

\frac{5^3}{3} - 4(5)^2 + 15(5) = \frac{125}{3} - 100 + 75 = \frac{50}{3}

\frac{4.5^3}{3} - 4(4.5)^2 + 15(4.5) = \frac{135}{8}

Thus

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} \approx 0.156

Final answer

5/32 or 0.156

Detailed explanation

Walkthrough

We need the probability that $X$ exceeds $4.5$ . For a continuous random variable, probabilities are found by integrating the pdf over the required interval. Because the pdf is defined only for $3 \le x \le 5$ , the event $X > 4.5$ corresponds to the interval $4.5 \le x \le 5$ . First expand $(x-3)(x-5)$ to make the integral easier. Then integrate term by term: $\int x^2\,dx = x^3/3$ , $\int -8x\,dx = -4x^2$ , and $\int 15\,dx = 15x$ . Substitute $x=5$ and $x=4.5$ , subtract, and multiply by $-3/4$ . The negative sign in the pdf is important; the bracket difference is negative, so the final probability comes out positive.

Key Takeaways

A continuous probability is an area under the pdf. When the pdf is a quadratic, expand it before integrating. Always use the correct limits for the event described.

Common Mistakes

Forgetting the factor $-3/4$ when integrating. Substituting only one limit. Confusing $P(X > 4.5)$ with $P(X < 4.5)$ . Not expanding the quadratic correctly.

Things to Be Careful About

The pdf is only non-zero on $3 \le x \le 5$ , so the upper limit must be $5$ , not infinity. Keep the negative factor throughout; it is essential for the final positive probability.

Techniques used

expand the quadratic integrandintegrate the pdf over the tail intervalevaluate the definite integral

(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

Mathematics 9709/63 — October/November 2025

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