9709/63 Oct/Nov 2024 — Mathematics Solved Past Paper

Q1Medium-EasySampling and Estimation

The heights of a certain species of deer are known to have standard deviation $0.35\text{ m}$ . A zoologist takes a random sample of 150 of these deer and finds that the mean height of the deer in the sample is $1.42\text{ m}$ .

(a)

Calculate a 96% confidence interval for the population mean height.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

The population standard deviation $\sigma = 0.35\text{ m}$ is known and the sample size $n = 150$ is large, so we use the standard normal distribution to construct a confidence interval for the population mean. The formula is:

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

where $z$ is the critical value such that the central area under the standard normal curve is 0.96.

Working

For a 96% confidence level, the total tail probability is $1 - 0.96 = 0.04$ , so each tail has probability 0.02. From the standard normal tables:

z = 2.054

The standard error of the sample mean is:

\frac{\sigma}{\sqrt{n}} = \frac{0.35}{\sqrt{150}} = 0.02858

The margin of error is:

z \cdot \frac{\sigma}{\sqrt{n}} = 2.054 \times 0.02858 = 0.0587

Lower bound of the interval:

1.42 - 0.0587 = 1.3613

Upper bound of the interval:

1.42 + 0.0587 = 1.4787

Answer

(1.36, 1.48)\text{ m} \quad (3\text{ s.f.})

Final answer

(1.36, 1.48) m

Detailed explanation

Walkthrough

We are given the population standard deviation $\sigma = 0.35$ m and a sample of size $n = 150$ with sample mean $\bar{x} = 1.42$ m. Because $\sigma$ is known and the sample is large, the sample mean $\bar{x}$ is approximately normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$ . This justifies using the standard normal distribution to build the confidence interval.

First, find the critical value $z$ . For a 96% confidence level, the central area under the standard normal curve must be 0.96, leaving 0.04 in the two tails combined, i.e. 0.02 in each tail. We therefore need $z$ such that $P(Z < z) = 0.98$ . From the standard normal tables, this gives $z = 2.054$ .

Next, compute the standard error of the sample mean:

\frac{\sigma}{\sqrt{n}} = \frac{0.35}{\sqrt{150}} \approx 0.02858

Multiply by $z$ to get the margin of error:

2.054 \times 0.02858 \approx 0.0587

Finally, add and subtract this margin from the sample mean:

1.42 - 0.0587 = 1.3613, \qquad 1.42 + 0.0587 = 1.4787

Rounding to 3 significant figures gives the interval $(1.36, 1.48)$ m.

Key Takeaways

This question tests the construction of a confidence interval for a population mean when the population standard deviation is known. The key formula is $\bar{x} \pm z\,\sigma/\sqrt{n}$ . You must know how to find the correct $z$ -value for the given confidence level, and you must remember to divide $\sigma$ by $\sqrt{n}$ .

Common Mistakes

Using $z = 1.96$ (the 95% value) instead of $z = 2.054$ for 96%.
Forgetting to divide $\sigma$ by $\sqrt{n}$ .
Writing the interval without rounding to the required degree of accuracy.
Omitting units.

Things to Be Careful About

The $z$ -value must match the confidence level exactly.
The standard error is $\sigma/\sqrt{n}$ , not $\sigma$ .
Give the final answer as an interval, to 3 significant figures, with units.

Techniques used

find the critical z-value for a 96% confidence levelcompute the standard error of the sample meanconstruct the confidence interval around the sample mean

(b)

Bubay says that 96% of deer of this species are likely to have heights that are within this confidence interval.

Explain briefly whether Bubay is correct.

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Q25MMediumLinear Combinations of Random Variables

The masses, in kilograms, of small and large bags of wheat have the independent distributions $N(16.0, 0.4)$ and $N(51.0, 0.9)$ respectively.

Find the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag.

DifficultyMedium

Worked solution

Approach

Let $S_1,S_2,S_3$ be the masses of the three small bags and $L$ be the mass of the large bag. The required probability is $P(S_1+S_2+S_3>L)$ , which is the same as $P(S_1+S_2+S_3-L>0)$ . Since the variables are independent and normal, the linear combination $X=S_1+S_2+S_3-L$ is normal, so we find its mean and variance and then standardise.

Working

Define

X=S_1+S_2+S_3-L.

Then $X$ is normal. Using the given formula for expectation,

E[X]=3(16.0)-51.0=-3.

Using independence and squared coefficients for variance,

\mathrm{Var}(X)=3(0.4)+(-1)^2(0.9)=1.2+0.9=2.1.

Hence

X \sim N(-3,2.1).

Standardise to find $P(X>0)$ :

z=\frac{0-(-3)}{\sqrt{2.1}}=\frac{3}{\sqrt{2.1}}=2.070.

Therefore

P(X>0)=P(Z>2.070)=1-\Phi(2.070)=0.0192.

Answer

0.0192

Final answer

0.0192

Detailed explanation

Walkthrough

We need $P(S_1+S_2+S_3>L)$ . Because the three small bags and the large bag are independent and normal, the combination $X=S_1+S_2+S_3-L$ is also normal. The mean of $X$ is the same linear combination of the means: $E[X]=16+16+16-51=-3$ . The variance is found by adding the variances, with each coefficient squared: $\mathrm{Var}(X)=0.4+0.4+0.4+(-1)^2(0.9)=2.1$ . So $X\sim N(-3,2.1)$ . To find $P(X>0)$ , standardise: $z=(0-(-3))/\sqrt{2.1}=2.070$ . The required probability is the upper tail of the standard normal distribution, $1-\Phi(2.070)=0.0192$ . If instead you use $Y=L-(S_1+S_2+S_3)$ , then $E[Y]=51-48=3$ , $\mathrm{Var}(Y)=2.1$ , and the event $S_1+S_2+S_3>L$ becomes $Y<0$ , giving $P(Y<0)=\Phi(-2.070)=0.0192$ .

Key Takeaways

A linear combination of independent normal variables is itself normal.
The expectation of a linear combination is the same linear combination of the expectations.
The variance of a linear combination of independent variables uses squared coefficients; covariance terms are zero because of independence.
To find a probability for a normal distribution, standardise to $Z\sim N(0,1)$ and use the standard normal CDF.

Common Mistakes

Forgetting that the coefficient of $L$ in $S_1+S_2+S_3-L$ is $-1$ , so its variance contribution is $(-1)^2(0.9)=0.9$ , not $-0.9$ .
Adding only the variances of the three small bags and forgetting the large bag's variance.
Using variance instead of standard deviation in the standardisation formula.
Choosing the wrong tail: for $X=S_1+S_2+S_3-L$ the event is $X>0$ , so the probability is $1-\Phi(z)$ . For $Y=L-(S_1+S_2+S_3)$ the event is $Y<0$ , so the probability is $\Phi(-z)$ .
Giving the answer without rounding to 3 significant figures.

Things to Be Careful About

The distributions are independent, so no covariance terms are needed.
Because the distributions are continuous, $P(X>0)=P(X\geq 0)$ ; the strictness of the inequality does not affect the probability.
Standardise using the mean and variance of the whole linear combination, not of an individual bag.
Check the standard normal value for $z=2.070$ ; the tail probability is $0.0192$ .

Techniques used

form a linear combination of independent normal variablescompute the expectation and variance of the linear combinationstandardise to the standard normal distributionfind an upper-tail probability

Approach

Use the sample sums to estimate the population mean. To obtain an unbiased estimate of the population variance, use the divisor $n-1$ .

Working

The sample mean is

\bar{x} = \frac{\sum t}{n} = \frac{230}{75} = 3.0667

The unbiased estimate of the population variance is

s^2 = \frac{1}{n-1}\left(\sum t^2 - \frac{(\sum t)^2}{n}\right) = \frac{1}{74}\left(930 - \frac{230^2}{75}\right)

Now

\frac{230^2}{75} = \frac{52900}{75} = 705.3333

so

s^2 = \frac{930 - 705.3333}{74} = \frac{224.6667}{74} = 3.0360

Equivalently, $s^2 = \frac{337}{111}$ .

Answer

\bar{x} = \frac{230}{75} = 3.07 \ (3 \text{ sf}), \qquad s^2 = 3.04 \ (3 \text{ sf})

Final answer

Mean estimate = 230/75 = 3.07 (3 sf); variance estimate = 337/111 = 3.04 (3 sf)

Detailed explanation

Walkthrough

The sample data are summarised by $\sum t = 230$ and $\sum t^2 = 930$ . The natural estimate of the population mean is the sample mean:

\bar{x} = \frac{\sum t}{n} = \frac{230}{75}.

For the variance, using a divisor of $n$ would give a biased estimate. The unbiased estimate uses a divisor of $n-1$ :

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

Here $\frac{(\sum t)^2}{n}$ is the correction that adjusts for the sample mean. Substituting the values gives $s^2 = \frac{337}{111} = 3.0360$ .

Key Takeaways

The sample mean is an unbiased estimator of the population mean.
The unbiased estimator of the population variance uses $n-1$ rather than $n$ .
Summary statistics $\sum t$ and $\sum t^2$ are sufficient to compute both estimates.

Common Mistakes

Using $n$ instead of $n-1$ as the divisor, which gives a biased variance estimate.
Forgetting to subtract $\frac{(\sum t)^2}{n}$ when using the computational formula.
Using $\sum t^2/n$ alone as the variance.

Things to Be Careful About

The correction term uses $n=75$ , while the variance divisor is $n-1=74$ .
Keep enough precision, or use the exact fraction $\frac{337}{111}$ , because part (b) uses these estimates.
The variance estimate to 3 significant figures is $3.04$ .

Techniques used

calculate the sample mean from summary statisticsapply the unbiased variance estimator with divisor n-1use the computational shortcut involving sum and sum of squares

(b)

You should now assume that your estimates from part (a) are the true values of the population mean and variance of $T$ .

The times taken by another random sample of 75 students were noted, and the sample mean, $\bar{T}$ , was found.

Find the value of $a$ such that $P(\bar{T} > a) = 0.234$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the estimates from part (a) as the true population mean and variance. For a sample of size $n=75$ , the sample mean is approximately normal by the Central Limit Theorem, with mean $\mu$ and variance $\sigma^2/75$ . Convert the probability $P(\bar{T} > a) = 0.234$ into a $z$ -value and solve for $a$ .

Working

From part (a), take

\mu = \frac{230}{75} = 3.0667, \qquad \sigma^2 = \frac{337}{111} = 3.0360.

For a sample of 75,

\bar{T} \sim N\left(\mu, \frac{\sigma^2}{75}\right).

We need $P(\bar{T} > a) = 0.234$ , so

P\left(\bar{T} \le a\right) = 0.766.

Standardising,

\frac{a - \mu}{\sigma/\sqrt{75}} = \Phi^{-1}(0.766) = 0.726.

The standard deviation of $\bar{T}$ is

\frac{\sigma}{\sqrt{75}} = \sqrt{\frac{337/111}{75}} = \sqrt{\frac{337}{8325}} = 0.2012.

Therefore

a = \mu + 0.726(0.2012) = 3.0667 + 0.1461 = 3.2127.

Answer

a = 3.21 \ (3 \text{ sf})

Final answer

a = 3.21 (3 sf)

Detailed explanation

Walkthrough

Take the estimates from part (a) as the true values: $\mu = 230/75$ and $\sigma^2 = 337/111$ . For a large random sample of 75 students, the sample mean $\bar{T}$ is approximately normal with mean $\mu$ and variance $\sigma^2/75$ , by the Central Limit Theorem.

We want the value $a$ such that $P(\bar{T} > a) = 0.234$ . Since the total probability is 1, the lower-tail probability is $P(\bar{T} \le a) = 0.766$ . The corresponding standard normal quantile is $z = \Phi^{-1}(0.766) = 0.726$ .

Standardising the sample mean gives

z = \frac{a - \mu}{\sigma/\sqrt{75}}.

Set this equal to $0.726$ , substitute $\mu = 3.0667$ and $\sigma/\sqrt{75} = 0.2012$ , then solve for $a$ :

a = 3.0667 + 0.726(0.2012) = 3.2127.

Rounded to 3 significant figures, $a = 3.21$ .

Key Takeaways

For a large sample, $\bar{T}$ is approximately normal with mean $\mu$ and variance $\sigma^2/n$ .
To find a threshold from a probability, use the inverse normal function: $z = \Phi^{-1}(p)$ .
Always use the standard deviation of the sample mean, $\sigma/\sqrt{n}$ , not the population standard deviation or the variance.

Common Mistakes

Using $0.766$ as the $z$ -value instead of $\Phi^{-1}(0.766) = 0.726$ . The mark scheme explicitly rejects using $0.766$ as the $z$ -value.
Using $\sigma$ instead of $\sigma/\sqrt{75}$ in the denominator.
Using $\sigma^2/75$ as the denominator instead of its square root.
Forgetting that the probability is an upper-tail probability, so the lower-tail probability $0.766$ must be used.

Things to Be Careful About

The estimates from part (a) are assumed to be the true population values, so use the exact or unrounded values for accuracy.
The sample size is 75, so the population variance is divided by 75 before taking the square root.
The final answer must be given to 3 significant figures: $a = 3.21$ .

Techniques used

standardise the sample mean using the Central Limit Theoremfind the inverse normal quantile for a probabilitysolve the standardised equation for the threshold

Q4MediumContinuous Random Variables

A random variable $X$ has probability density function $f$ defined by

f(x) = \begin{cases} \frac{a}{x^2} - \frac{18}{x^3} & 2 \le x \le 3, \\ 0 & \text{otherwise,} \end{cases}

where $a$ is a constant.

(a)

Show that $a = \frac{27}{2}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $f$ is a probability density function, the total area under its graph over its support must equal 1. Integrate $f(x)$ from $x = 2$ to $x = 3$ and set the result equal to 1, then solve for the constant $a$ .

Working

\int_2^3 f(x)\,dx = 1

\int_2^3 \left(\frac{a}{x^2} - \frac{18}{x^3}\right) dx = 1

Integrating term by term:

\left[-\frac{a}{x} + \frac{9}{x^2}\right]_2^3 = 1

Substituting the limits:

\left(-\frac{a}{3} + \frac{9}{9}\right) - \left(-\frac{a}{2} + \frac{9}{4}\right) = 1

-\frac{a}{3} + 1 + \frac{a}{2} - \frac{9}{4} = 1

\frac{a}{6} - \frac{5}{4} = 1

\frac{a}{6} = \frac{9}{4}

a = \frac{54}{4} = \frac{27}{2}

Answer

a = \frac{27}{2}

Final answer

a = 27/2

Detailed explanation

Walkthrough

A probability density function has the defining property that the total area under its graph over the entire support is exactly 1. Since $f(x) = 0$ outside $x \in [2,3]$ , we only need to integrate from $x = 2$ to $x = 3$ and set the result equal to 1.

We integrate each term separately using the power rule for negative powers. The integral of $\frac{a}{x^2}$ is $-\frac{a}{x}$ , and the integral of $-\frac{18}{x^3}$ is $+\frac{9}{x^2}$ (because $\int x^{-3}\,dx = -\frac{x^{-2}}{2}$ , and multiplying by $-18$ gives $+9x^{-2}$ ).

We then evaluate the antiderivative at the upper limit $x = 3$ and the lower limit $x = 2$ , subtracting the latter from the former. This produces a linear equation in $a$ :

\frac{a}{6} - \frac{5}{4} = 1 \implies \frac{a}{6} = \frac{9}{4} \implies a = \frac{27}{2}.

Key Takeaways

The defining property of a PDF — $\int f(x)\,dx = 1$ over the support — is what determines the constant in any density function. This question also reinforces the power rule for integrating negative powers and the fact that the constant of integration cancels in a definite integral.

Common Mistakes

Forgetting to set the integral equal to 1.
Sign errors when integrating $-\frac{18}{x^3}$ : the antiderivative is $+\frac{9}{x^2}$ , not $-\frac{9}{x^2}$ .
Arithmetic errors when combining the fractions at the two limits (e.g. mishandling $1 - \frac{9}{4}$ ).

Things to Be Careful About

The limits of integration must be 2 and 3 — the support of the distribution — not the whole real line. The constant $a$ must be positive for $f$ to be a valid density, and indeed $\frac{27}{2} > 0$ .

Techniques used

integrate the PDF over its supportset the total integral equal to 1solve the resulting equation for the constant

(b)

Show that $E(X) = \frac{27}{2} \ln \frac{3}{2} - 3$ .

3M Similar

DifficultyMedium

Worked solution

Approach

Use the formula $E(X) = \int x\,f(x)\,dx$ over the support of the distribution. Substitute $a = \frac{27}{2}$ from part (a), integrate $x\,f(x)$ from $x = 2$ to $x = 3$ , then simplify using the laws of logarithms.

Working

E(X) = \int_2^3 x \cdot f(x)\,dx

= \int_2^3 x\left(\frac{27}{2x^2} - \frac{18}{x^3}\right) dx

= \int_2^3 \left(\frac{27}{2x} - \frac{18}{x^2}\right) dx

Integrating:

\left[\frac{27}{2}\ln x + \frac{18}{x}\right]_2^3

Substituting the limits:

\left(\frac{27}{2}\ln 3 + \frac{18}{3}\right) - \left(\frac{27}{2}\ln 2 + \frac{18}{2}\right)

= \frac{27}{2}\ln 3 + 6 - \frac{27}{2}\ln 2 - 9

= \frac{27}{2}(\ln 3 - \ln 2) - 3

= \frac{27}{2}\ln\frac{3}{2} - 3

Answer

E(X) = \frac{27}{2}\ln\frac{3}{2} - 3

Final answer

E(X) = (27/2) ln(3/2) - 3

Detailed explanation

Walkthrough

The expectation of a continuous random variable is defined as $E(X) = \int x\,f(x)\,dx$ over its support. We substitute $a = \frac{27}{2}$ into the PDF and multiply by $x$ before integrating.

Multiplying out: $x \cdot \frac{27}{2x^2} = \frac{27}{2x}$ and $x \cdot \frac{18}{x^3} = \frac{18}{x^2}$ , so the integrand becomes $\frac{27}{2x} - \frac{18}{x^2}$ .

Integrating term by term: $\int \frac{27}{2x}\,dx = \frac{27}{2}\ln x$ (since $\int \frac{1}{x}\,dx = \ln x$ ) and $\int -\frac{18}{x^2}\,dx = \frac{18}{x}$ (since $\int x^{-2}\,dx = -x^{-1}$ , multiplied by $-18$ gives $\frac{18}{x}$ ).

Evaluating at the limits: at $x = 3$ we get $\frac{27}{2}\ln 3 + 6$ ; at $x = 2$ we get $\frac{27}{2}\ln 2 + 9$ . Subtracting gives constants $6 - 9 = -3$ , and the logarithms combine using the law $\ln 3 - \ln 2 = \ln\frac{3}{2}$ .

Key Takeaways

The formula for expectation of a continuous random variable: $E(X) = \int x\,f(x)\,dx$ . The integral of $\frac{1}{x}$ is $\ln x$ . The logarithm law $\ln a - \ln b = \ln\frac{a}{b}$ is essential to match the given form of the answer.

Common Mistakes

Forgetting to multiply $f(x)$ by $x$ before integrating.
Sign error on the $\frac{18}{x^2}$ term (it integrates to $+\frac{18}{x}$ , not $-\frac{18}{x}$ ).
Substituting the limits incorrectly — both the logarithm and the $\frac{18}{x}$ term must be evaluated at 2 and 3.
Failing to combine the logarithms at the end, leaving the answer in an unsimplified form.

Things to Be Careful About

The value of $E(X)$ must be positive since $X \geq 2 > 0$ . The final simplification $\ln 3 - \ln 2 = \ln\frac{3}{2}$ is required to obtain the exact expression given in the question.

Techniques used

multiply the PDF by xintegrate to find the expectationapply the laws of logarithms to simplify

Q5MediumHypothesis Tests

The lengths, in centimetres, of worms of a certain kind are normally distributed with mean $\mu$ and standard deviation 2.3. An article in a magazine states that the value of $\mu$ is 12.7. A scientist wishes to test whether this value is correct. He measures the lengths, $x\text{ cm}$ , of a random sample of 50 worms of this kind and finds that $\sum x = 597.1$ . He plans to carry out a test, at the 1% significance level, of whether the true value of $\mu$ is different from 12.7.

(a)

State, with a reason, whether he should use a one-tailed or a two-tailed test.

1M Similar

DifficultyEasy

Worked solution

Approach

For the test, the scientist wants to test whether the mean worm length is different from 12.7. This means the alternative hypothesis is $H_1: \mu \neq 12.7$ , covering values both above and below 12.7.

Working

Because the alternative hypothesis is two-sided, the rejection region must be split between the two tails of the normal distribution. A one-tailed test would only be used if the scientist were looking for a mean greater than 12.7 or less than 12.7, but the word "different" indicates no direction is specified.

Answer

Use a two-tailed test, because the scientist is looking for a difference in either direction.

Final answer

Two-tailed test, because the scientist is looking for a difference in either direction.

Detailed explanation

Walkthrough

The claim being tested is that the mean worm length is 12.7 cm. The scientist wants to test whether the true mean is different from 12.7. The phrase "different from" means the true mean could be either greater than 12.7 or less than 12.7.

A one-tailed test is used when the alternative hypothesis is directional, such as "greater than" or "less than". Here no direction is given, so both tails of the distribution must be considered. Therefore the test is two-tailed.

Key Takeaways

The tail type is determined by the alternative hypothesis, not by the sample data.
"Different" or "not equal" implies a two-tailed test.
"Greater than" or "less than" implies a one-tailed test.

Common Mistakes

Choosing a one-tailed test because the sample mean is less than 12.7. The sample result should not determine the test type.
Stating "two-tailed" without giving the required reason.

Things to Be Careful About

Read the wording carefully: "whether the value is different from 12.7" is two-sided.
Do not change to a one-tailed test after seeing the sample mean.

Techniques used

identify the test direction from the wordingchoose the number of tails based on the alternative hypothesis

(b)

Carry out the test.

5M Similar

DifficultyMedium

Worked solution

Approach

Set up the null and alternative hypotheses, calculate the sample mean, find the standard error, compute the standardised test statistic, and compare it with the two-tailed critical value at the 1% significance level.

Working

State the hypotheses:

H_0: \mu = 12.7, \quad H_1: \mu \neq 12.7

Calculate the sample mean:

\bar{x} = \frac{597.1}{50} = 11.942

The standard error is:

\frac{\sigma}{\sqrt{n}} = \frac{2.3}{\sqrt{50}} = 0.3253

The test statistic is:

z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} = \frac{11.942 - 12.7}{0.3253} = -2.330

At the 1% significance level, the two-tailed critical values are:

z^* = \pm 2.576

Since $-2.330 > -2.576$ , the test statistic is not in the critical region.

Answer

Do not reject $H_0$ . There is insufficient evidence to conclude that the true mean length of worms is different from 12.7 cm.

Final answer

Do not reject H0; insufficient evidence that the true mean worm length is different from 12.7 cm.

Detailed explanation

Walkthrough

First write the hypotheses. The null hypothesis is the magazine's claim: $H_0: \mu = 12.7$ . Since part (a) established a two-tailed test, the alternative hypothesis is $H_1: \mu \neq 12.7$ .

Next, calculate the sample mean from the given sum: $\bar{x} = 597.1/50 = 11.942$ . The population standard deviation is known to be 2.3, so we use the normal distribution. The standard error is $\sigma/\sqrt{n} = 2.3/\sqrt{50} = 0.3253$ .

The test statistic measures how far the sample mean is from the claimed mean in standard errors:

z = \frac{11.942 - 12.7}{0.3253} = -2.330

At the 1% significance level, a two-tailed test has 0.005 in each tail. The critical values are approximately $\pm 2.576$ . Since $-2.330$ is greater than $-2.576$ , it lies inside the acceptance region. Equivalently, the one-sided p-value is about 0.0099, which is greater than 0.005, so the two-sided p-value is about 0.0198, greater than 0.01.

Therefore we do not reject $H_0$ . There is insufficient evidence that the true mean worm length is different from 12.7 cm.

Key Takeaways

When the population standard deviation is known, use the normal distribution for the test statistic.
The test statistic is $z = (\bar{x} - \mu)/(\sigma/\sqrt{n})$ .
A two-tailed 1% test uses critical values $\pm 2.576$ .
Failing to reject $H_0$ does not prove $H_0$ is true; it only means the evidence is not strong enough.

Common Mistakes

Using the sample standard deviation instead of the known population standard deviation.
Comparing the test statistic with the wrong critical value for a two-tailed test.
Saying that $\mu = 12.7$ in the conclusion; we can only say there is insufficient evidence that it is different.
Using a one-tailed test in part (b) despite part (a).

Things to Be Careful About

The test statistic is negative because the sample mean is below the claimed mean; compare its magnitude with the critical value.
The conclusion should be stated in context: the length of worms.
If using p-values, compare the two-sided p-value with 0.01, or the one-sided p-value with 0.005.

Techniques used

state null and alternative hypothesescalculate the sample meancompute the standardised test statisticcompare the test statistic with the critical valuedraw a conclusion in context

Q6MediumThe Poisson DistributionLinear Combinations of Random Variables

The numbers of customers arriving at service desks $A$ and $B$ during a 10-minute period have the independent distributions $Po(1.8)$ and $Po(2.1)$ respectively.

(a)

Find the probability that during a randomly chosen 15-minute period more than 2 customers will arrive at desk $A$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

For a 15-minute period, scale the 10-minute mean for desk $A$ . Since we need more than 2 arrivals, work out $1-P(X\le 2)$ using the Poisson probability formula.

Working

The number of customers arriving at desk $A$ in 15 minutes is

X \sim \mathrm{Po}\left(1.8 \times \frac{15}{10}\right) = \mathrm{Po}(2.7).

Now

\begin{aligned} P(X\le 2) &= e^{-2.7}\left(1 + 2.7 + \frac{2.7^2}{2!}\right)\\ &= e^{-2.7}(1 + 2.7 + 3.645)\\ &= 0.06721 + 0.1815 + 0.2450\\ &= 0.4937. \end{aligned}

Therefore

P(X>2) = 1 - 0.4937 = 0.5063 \approx 0.506.

Answer

P(\text{more than 2 at desk }A) = 0.506 \text{ (3 sf)}

Final answer

0.506 (3 sf)

Detailed explanation

Walkthrough

In a 10-minute period the mean number of arrivals at desk $A$ is $1.8$ . A 15-minute period is $1.5$ times as long, so the Poisson parameter scales to $1.8 \times 1.5 = 2.7$ . This rescaling property is essential: the mean is proportional to the length of the observation interval.

The event “more than 2 customers” means $X=3,4,5,\dots$ . It is easier to use the complement, so calculate $P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$ . The expression

e^{-2.7}\left(1+2.7+\frac{2.7^2}{2!}\right)

is the full Poisson cumulative probability for 0, 1 and 2 arrivals. Subtracting it from 1 gives the required tail probability.

Key Takeaways

The Poisson mean is proportional to the time interval: if a 10-minute interval has mean $\lambda$ , then a $k \times 10$ -minute interval has mean $k\lambda$ .
For a “more than” probability, use the complement: $P(X>r)=1-P(X\le r)$ .
Always show the Poisson formula when a mark scheme awards method marks; an unsupported numerical answer may not receive full credit.

Common Mistakes

Using $\lambda=1.8$ for the 15-minute period instead of $2.7$ .
Confusing $P(X>2)$ with $P(X\ge 2)$ . The complement of $X>2$ is $X\le 2$ , which includes 0, 1 and 2, not just 0 and 1.
Omitting the expression and writing only $0.506$ ; this can lose method marks.

Things to Be Careful About

Poisson probabilities are discrete, so “more than 2” means 3 or more; do not include 2 in the required probability.
Give the final answer to 3 significant figures as usual, here $0.506$ .
Check the direction of the complement: $P(X>2)=1-P(X\le 2)$ , not $1-P(X<2)$ .

Techniques used

scale a Poisson mean to a new time intervaluse the complement rule for a tail probabilityevaluate the required Poisson cumulative probability terms

(b)

Find the probability that during a randomly chosen 5-minute period the total number of customers arriving at both desks is less than 4.

3M Similar

DifficultyMedium

Worked solution

Approach

Use the fact that the sum of independent Poisson variables is again Poisson. First scale both means to a 5-minute period, then add them and compute $P(T\le 3)$ , since fewer than 4 means 0, 1, 2 or 3 customers.

Working

For a 5-minute period:

\lambda_A = \frac{1.8}{2} = 0.9, \qquad \lambda_B = \frac{2.1}{2} = 1.05.

Let $T=X+Y$ , where $X$ is the number at desk $A$ and $Y$ the number at desk $B$ . Since they are independent,

T \sim \mathrm{Po}(0.9+1.05) = \mathrm{Po}(1.95).

The probability that the total is less than 4 is

\begin{aligned} P(T<4)=P(T\le 3) &= e^{-1.95}\left(1+1.95+\frac{1.95^2}{2!}+\frac{1.95^3}{3!}\right)\\ &= e^{-1.95}(1+1.95+1.90125+1.2358)\\ &= 0.1423+0.2774+0.2705+0.1758\\ &= 0.8660. \end{aligned}

Answer

P(\text{fewer than 4 customers in total}) = 0.866 \text{ (3 sf)}

Final answer

0.866 (3 sf)

Detailed explanation

Walkthrough

First convert the 10-minute rates to a 5-minute period. Since 5 minutes is half of 10, desk $A$ has mean $1.8/2=0.9$ and desk $B$ has mean $2.1/2=1.05$ .

Because the arrivals at the two desks are independent and each follows a Poisson distribution, their total $T$ also follows a Poisson distribution. The mean of the total is the sum of the individual means:

T \sim \mathrm{Po}(0.9+1.05)=\mathrm{Po}(1.95).

“Less than 4” means $T=0,1,2,3$ , so we need $P(T\le 3)$ . The formula is

e^{-1.95}\left(1+1.95+\frac{1.95^2}{2!}+\frac{1.95^3}{3!}\right).

Work through the terms separately and sum them. The probability is approximately $0.866$ .

Key Takeaways

The sum of independent Poisson random variables is Poisson, and its parameter is the sum of the parameters.
When an observation interval changes, rescale every Poisson mean before adding them.
For an integer-valued random variable, “less than 4” is the same as “less than or equal to 3”.

Common Mistakes

Adding the original 10-minute means $1.8+2.1=3.9$ instead of first halving them.
Treating “less than 4” as $P(T\le 4)$ , which would include 4 customers.
Forgetting that $T$ must be handled as a single Poisson variable rather than calculating the convolution of two Poisson distributions term by term.

Things to Be Careful About

“Less than 4” includes $T=3$ ; since $T$ is discrete, $T<4$ and $T\le 3$ are identical.
Show the full formula for method marks; an unsupported $0.866$ may receive only reduced credit according to the mark scheme.
Keep 3 significant figures in the final answer.

Techniques used

halve Poisson means for a shorter time intervaladd independent Poisson meanscompute a Poisson cumulative probability

(c)

An inspector waits at desk $B$ . She wants to wait long enough to be 90% certain of seeing at least one customer arrive at the desk.

Find the minimum time for which she should wait, giving your answer correct to the nearest minute.

4M Similar

DifficultyMedium

Worked solution

Approach

If the inspector waits $x$ 10-minute periods, the number of arrivals at desk $B$ is $\mathrm{Po}(2.1x)$ . Set the probability of at least one arrival at 0.90 and solve the resulting inequality for $x$ , then convert to minutes.

Working

Let $x$ be the number of 10-minute periods she waits. Then the number of arrivals is

Y \sim \mathrm{Po}(2.1x).

She needs

P(Y\ge 1)=1-P(Y=0)=1-e^{-2.1x} \ge 0.90.

So

e^{-2.1x} \le 0.1.

Taking natural logarithms:

-2.1x \le \ln(0.1) = -2.3026.

Dividing by $-2.1$ , which reverses the inequality:

x \ge \frac{2.3026}{2.1} = 1.0965.

This $x$ is in units of 10-minute periods, so the waiting time is

10x \ge 10.965 \text{ minutes}.

To the nearest minute, she must wait for at least 11 minutes.

Answer

\text{Minimum waiting time} = 11 \text{ minutes}

Final answer

11 minutes

Detailed explanation

Walkthrough

At desk $B$ , the mean number of arrivals in 10 minutes is $2.1$ . If the inspector waits for $x$ of these 10-minute periods, the mean becomes $2.1x$ , so the number of arrivals in that time is $\mathrm{Po}(2.1x)$ .

The probability of seeing at least one customer is the complement of seeing none:

P(Y\ge 1)=1-P(Y=0)=1-e^{-2.1x}.

She wants this to be at least $0.90$ , so

1-e^{-2.1x}\ge 0.90.

Rearranging gives $e^{-2.1x}\le 0.1$ . Taking logs removes the exponential. Since $\ln(0.1)=-2.3026$ , we get

-2.1x\le -2.3026.

Multiplying or dividing by the negative number $-2.1$ reverses the inequality, giving $x\ge 1.0965$ . Because $x$ was measured in 10-minute periods, the real waiting time is $10x\ge 10.965$ minutes.

A time of $10.965$ minutes is closer to 11 minutes than to 10 minutes, and to be certain of meeting the 90% condition the inspector should wait 11 minutes. There is no integer-minute time of 10 minutes that satisfies the inequality, since 10 minutes would give $\lambda=2.1$ and probability $1-e^{-2.1}=0.8775<0.90$ .

Key Takeaways

The Poisson mean is proportional to the length of the interval, so a waiting time of $x$ periods has parameter $2.1x$ .
The probability of “at least one” is $1-e^{-\lambda}$ for a Poisson variable.
Inverses of exponential inequalities are solved with natural logarithms; remember to reverse the inequality when dividing by a negative number.
A hidden unit conversion may be required: the Poisson parameter may be per 10 minutes, while the answer is requested in minutes.

Common Mistakes

Forgetting to convert from 10-minute periods to minutes, leaving the answer as $1.096$ periods rather than $11$ minutes.
Not reversing the inequality when dividing by $-2.1$ .
Writing $P(Y\ge 1)=P(Y=1)$ instead of $1-P(Y=0)$ ; this is wrong because at least one means any positive number, not exactly one.
Rounding $10.965$ to $10$ minutes because it is “nearest 10”, or writing $10.97$ minutes instead of the final integer minute.
Omitting the expression, making it impossible to gain the method marks.

Things to Be Careful About

The minimum time required is just over 10.96 minutes; since the question asks for the nearest whole minute and she must satisfy the 90% condition, the answer is 11 minutes.
The mark scheme accepts the intermediate value $1.096$ or $10.96$ , then requires 11 minutes.
In the rearrangement, taking logs of the correct side is important: from $e^{-2.1x}\le 0.1$ , the logarithm gives $-2.1x\le \ln 0.1$ .

Techniques used

express the probability of at least one arrival using the Poisson zero-probability termset up an inequality for a probability conditionsolve an exponential inequality using natural logarithmsconvert the Poisson time period to minutes and round up to the requested unit

Q7MediumHypothesis TestsThe Poisson Distribution

The number of accidents per year on a certain road has the distribution $Po(\lambda)$ . In the past the value of $\lambda$ was 3.3. Recently, a new speed limit was imposed and the council wishes to test whether the value of $\lambda$ has decreased. The council notes the total number, $X$ , of accidents during two randomly chosen years after the speed limit was introduced and it carries out a test at the 5% significance level.

(a)

Calculate the probability of a Type I error.

4M Similar

DifficultyMedium

Worked solution

Approach

The two-year total is $X \sim Po(6.6)$ under $H_0$ . The Type I error is the probability of rejecting $H_0$ when it is true, which is the size of the critical region. We find the critical region by comparing Poisson probabilities with the 5% significance level.

Working

Since two years are used, the total under $H_0$ is

X \sim Po(2 \times 3.3) = Po(6.6)

For a one-tailed test at the 5% level, the critical region consists of the smallest values of $X$ until the cumulative probability exceeds 0.05.

P(X \leq 2) = e^{-6.6}\left(1 + 6.6 + \frac{6.6^2}{2}\right)

= e^{-6.6}(1 + 6.6 + 21.78) = e^{-6.6} \times 29.38

= 0.001360 + 0.008978 + 0.02963 = 0.0400

Next,

P(X \leq 3) = P(X \leq 2) + e^{-6.6} \times \frac{6.6^3}{3!}

= 0.0400 + 0.001360 \times 47.916 = 0.0400 + 0.0652 = 0.105

Since $0.0400 < 0.05$ and $0.105 > 0.05$ , the critical region is $X \leq 2$ .

The probability of a Type I error is the probability of falling in the critical region when $H_0$ is true:

P(\text{Type I error}) = P(X \leq 2 \mid \lambda = 6.6) = 0.0400

Answer

0.0400

Final answer

0.0400

Detailed explanation

Walkthrough

The question states that the number of accidents per year is $Po(\lambda)$ with $\lambda = 3.3$ in the past. The council records the total number of accidents in two randomly chosen years. Because the sum of two independent Poisson variables with parameters $\lambda_1$ and $\lambda_2$ is Poisson with parameter $\lambda_1 + \lambda_2$ , the total $X$ under the null hypothesis is $Po(6.6)$ .

The test is one-tailed at the 5% significance level. The council suspects the mean has decreased, so the alternative hypothesis is $\lambda < 6.6$ . The critical region is therefore of the form $X \leq c$ .

To find the critical region, compute the cumulative probabilities $P(X \leq c)$ for increasing $c$ until the probability exceeds 0.05. We find $P(X \leq 2) = 0.0400 < 0.05$ , so $X = 0, 1, 2$ are in the critical region. We then check $P(X \leq 3) = 0.105 > 0.05$ , so $X = 3$ is not in the critical region. Hence the critical region is $X \leq 2$ .

The Type I error is the probability of rejecting $H_0$ when it is true, i.e. $P(X \leq 2)$ when $\lambda = 6.6$ , which is 0.0400.

Key Takeaways

The sum of two independent Poisson variables is Poisson with parameter equal to the sum of the parameters. The Type I error is the size of the critical region, i.e. the probability of rejecting a true null hypothesis. The critical region is found by taking the smallest possible values of the test statistic until the cumulative probability exceeds the significance level.

Common Mistakes

Using $\lambda = 3.3$ instead of $6.6$ for the two-year total is a common error. The mark scheme allows the correct use of 3.3 in the M1 expression, but the final answer must use $\lambda = 6.6$ . Another common error is computing $P(X = 2)$ instead of $P(X \leq 2)$ for the critical region.

Things to Be Careful About

The critical region is cumulative: always add the probabilities for $X = 0, 1, 2$ . The final answer is $P(X \leq 2) = 0.0400$ (3 significant figures), and the mark scheme also accepts 0.040 or 0.04.

Techniques used

determine the critical region for a Poisson testcalculate Poisson probabilitiesidentify the Type I error as the size of the critical region

(b)

Given that $X = 2$ , carry out the test.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

State the null and alternative hypotheses, compare the observed value $X = 2$ with the critical region found in part (a), and conclude in context.

Working

The hypotheses are:

H_0: \lambda = 6.6, \quad H_1: \lambda < 6.6

The critical region from part (a) is $X \leq 2$ , and the observed value is $X = 2$ , which lies in the critical region.

From part (a), $P(X \leq 2) = 0.0400 < 0.05$ .

Since $0.0400 < 0.05$ , we reject $H_0$ .

Answer

There is evidence to suggest that the mean number of accidents has decreased.

Final answer

Reject H0; there is evidence that the mean number of accidents has decreased.

Detailed explanation

Walkthrough

The hypotheses are $H_0: \lambda = 6.6$ and $H_1: \lambda < 6.6$ . The observed value is $X = 2$ . From part (a), the critical region is $X \leq 2$ , so the observed value lies in the critical region. Equivalently, $P(X \leq 2) = 0.0400 < 0.05$ , so the result is significant at the 5% level.

We therefore reject $H_0$ and conclude that there is evidence that the mean number of accidents has decreased.

Key Takeaways

A hypothesis test compares the observed test statistic with the critical region. If the observed value falls in the critical region, we reject $H_0$ . The conclusion must be stated in the context of the question.

Common Mistakes

Not writing the hypotheses, or writing them with the wrong parameter value. A common error is stating the conclusion too definitely (e.g., "the mean has decreased") without the phrase "evidence to suggest" — the mark scheme requires the softer wording.

Things to Be Careful About

The conclusion must be in context (about accidents), not just "reject H_0". The mark scheme requires the comparison $0.0400 < 0.05$ to be shown.

Techniques used

state null and alternative hypothesescompare the test statistic with the critical regiondraw a conclusion in context

(c)

The council decides to carry out another similar test at the 5% significance level using the same hypotheses and two different randomly chosen years.

Given that the true value of $\lambda$ is 0.6, calculate the probability of a Type II error.

3M Similar

DifficultyMedium

Worked solution

Approach

A Type II error is accepting $H_0$ when $H_1$ is true. The acceptance region is the complement of the critical region, $X \geq 3$ . Under $H_1$ the true value of $\lambda$ is $0.6$ per year, so for two years $\lambda = 1.2$ . We compute $P(X \geq 3 \mid \lambda = 1.2)$ .

Working

Under $H_1$ , for two years:

\lambda = 2 \times 0.6 = 1.2

The acceptance region is $X \geq 3$ , so:

P(\text{Type II error}) = P(X \geq 3 \mid \lambda = 1.2)

= 1 - P(X \leq 2 \mid \lambda = 1.2)

= 1 - e^{-1.2}\left(1 + 1.2 + \frac{1.2^2}{2}\right)

= 1 - e^{-1.2}(1 + 1.2 + 0.72)

= 1 - e^{-1.2} \times 2.92

= 1 - 0.3012 \times 2.92

= 1 - 0.8795

= 0.1205

Answer

0.121 \text{ (3 sf)}

Final answer

0.121

Detailed explanation

Walkthrough

A Type II error is the probability of accepting $H_0$ when $H_1$ is true. The acceptance region is the complement of the critical region: since the critical region is $X \leq 2$ , the acceptance region is $X \geq 3$ .

Under $H_1$ , the true value of $\lambda$ is 0.6 per year, so for two years the total follows $Po(1.2)$ . We need $P(X \geq 3 \mid \lambda = 1.2)$ , which is $1 - P(X \leq 2 \mid \lambda = 1.2)$ .

Compute $P(X \leq 2) = e^{-1.2}(1 + 1.2 + 1.2^2/2) = 0.8795$ . Therefore the Type II error is $1 - 0.8795 = 0.1205 \approx 0.121$ .

Key Takeaways

The Type II error is the probability of accepting $H_0$ when it is false. It is computed using the true value of the parameter (the $H_1$ value). The acceptance region is the complement of the critical region.

Common Mistakes

Using $\lambda = 6.6$ or $\lambda = 0.6$ instead of $\lambda = 1.2$ for the two-year calculation. Also, computing $P(X \leq 2)$ rather than $1 - P(X \leq 2)$ — the Type II error is the probability of the acceptance region, so the complement is needed.

Things to Be Careful About

The Type II error uses the true value $\lambda = 0.6$ per year, but since the test uses two years, $\lambda = 1.2$ . The answer is 0.121 (3 sf) or 0.120.

Techniques used

identify the acceptance regioncalculate the probability of accepting H0 when H1 is trueuse Poisson probabilities with a new value of λ

(d)

Using $\lambda = 0.6$ and a suitable approximating distribution, find the probability that there will be more than 10 accidents in 30 years.

4M Similar

DifficultyMedium

Worked solution

Approach

For 30 years, the total number of accidents follows $Po(30 \times 0.6) = Po(18)$ . Since $\lambda = 18$ is large, we use the normal approximation $N(18, 18)$ . Apply the continuity correction and find $P(X > 10.5)$ .

Working

For 30 years:

\lambda = 30 \times 0.6 = 18

So $X \sim Po(18) \approx N(18, 18)$ .

Applying the continuity correction:

P(X > 10) = P(X > 10.5)

Standardise:

z = \frac{10.5 - 18}{\sqrt{18}} = \frac{-7.5}{4.243} = -1.768

Since the normal distribution is symmetric about the mean,

P(X > 10.5) = P(Z > -1.768) = \Phi(1.768)

From the standard normal tables:

\Phi(1.768) = 0.9615 \approx 0.961

Answer

0.961 \text{ (3 sf)}

Final answer

0.961

Detailed explanation

Walkthrough

For 30 years, the total number of accidents follows the Poisson distribution $Po(30 \times 0.6) = Po(18)$ . Since $\lambda = 18$ is large (greater than 10), we use the normal approximation $N(18, 18)$ .

To find $P(X > 10)$ , apply the continuity correction: $P(X > 10.5)$ . Standardise:

z = \frac{10.5 - 18}{\sqrt{18}} = -1.768

By symmetry of the normal distribution, $P(Z > -1.768) = \Phi(1.768) \approx 0.961$ .

Key Takeaways

The Poisson distribution can be approximated by a normal distribution when $\lambda$ is large. The mean and variance of the approximating normal are both $\lambda$ . The continuity correction is essential when approximating a discrete distribution by a continuous one.

Common Mistakes

Forgetting the continuity correction, or applying it in the wrong direction (e.g., using $P(X \geq 10.5)$ instead of $P(X > 10.5)$ ). Also, using the variance $18$ as the standard deviation — the correct standard deviation is $\sqrt{18}$ .

Things to Be Careful About

The continuity correction for $P(X > 10)$ is $P(X > 10.5)$ . The answer is 0.961 or 0.962 (3 sf).

Techniques used

approximate a Poisson distribution by a normal distributionapply the continuity correctionstandardise and use the standard normal distribution

Mathematics 9709/63 — October/November 2024

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