9709/62 Feb/Mar 2024 — Mathematics Solved Past Paper

Q1Medium-EasySampling and Estimation

The lengths, $X\text{ cm}$ , of a sample of 100 insects of a certain type were summarised as follows.

n = 100 \qquad \Sigma x = 36.8 \qquad \Sigma x^2 = 17.34

(a)

Calculate unbiased estimates for the population mean and variance of $X$ .

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DifficultyMedium-Easy

Worked solution

Approach

The sample mean $\bar x$ is an unbiased estimate of the population mean $\mu$ . The unbiased estimate of the population variance is obtained using the divisor $n - 1$ :

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

Working

The unbiased estimate of the population mean is the sample mean:

\bar x = \frac{\sum x}{n} = \frac{36.8}{100} = 0.368

So

\text{est}(\mu) = 0.368 = \frac{46}{125}

For the variance, use the unbiased formula:

s^2 = \frac{1}{99}\left(17.34 - 100(0.368)^2\right)

First calculate $100(0.368)^2$ :

100(0.368)^2 = 100 \times 0.135424 = 13.5424

Then:

s^2 = \frac{1}{99}(17.34 - 13.5424) = \frac{3.7976}{99}

s^2 = 0.0383596\ldots = 0.0384 \quad (3\text{ s.f.})

Answer

Unbiased estimate of the population mean: $0.368$ .

Unbiased estimate of the population variance: $0.0384$ (3 s.f.).

Final answer

Mean = 0.368, variance = 0.0384 (3 s.f.)

Detailed explanation

Walkthrough

We are given summary statistics for a sample of 100 insects: $n = 100$ , $\sum x = 36.8$ and $\sum x^2 = 17.34$ . Because these are sample data, we want unbiased estimates of the population mean and variance.

The sample mean $\bar x$ is an unbiased estimator of the population mean, so we divide the total of the observations by the sample size:

\bar x = \frac{36.8}{100} = 0.368

For the population variance, we cannot simply divide by $n$ . Since we have estimated the mean from the same data, one degree of freedom is lost, so we use $n - 1$ as the divisor. This gives the unbiased estimate:

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

Substituting the given values gives $0.0384$ to 3 significant figures.

Key Takeaways

The sample mean is an unbiased estimate of the population mean.
An unbiased estimate of the population variance uses the divisor $n - 1$ , not $n$ .
Summary statistics such as $\sum x$ and $\sum x^2$ are sufficient to calculate these estimates.

Common Mistakes

Using $n$ instead of $n - 1$ in the variance formula, which gives the biased sample variance.
Forgetting to subtract $n\bar x^2$ when using $\sum x^2$ .
Using $\left(\sum x\right)^2$ instead of $\sum x^2$ .
Rounding intermediate values too early, which can change the final answer.

Things to Be Careful About

The unbiased variance estimate is $0.0384$ , not $0.037976$ ; the latter would come from dividing by $100$ instead of $99$ .
Give the variance to 3 significant figures as required by the mark scheme.
The mean has units cm, while the variance has units $\text{cm}^2$ .

Techniques used

compute the sample meanapply the unbiased variance formula with divisor n - 1substitute the given summary statistics

(b)

State a necessary condition for the estimates found in part (a) to be reliable.

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Q2Medium-EasySampling and Estimation

A random sample of 250 people living in Barapet was chosen. It was found that 78 of these people owned a BETEC phone.

(a)

Calculate an approximate 98% confidence interval for the proportion of people living in Barapet who own a BETEC phone.

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DifficultyMedium-Easy

Worked solution

Approach

The confidence interval for a population proportion is based on the sample proportion $\hat{p}$ and the standard error of $\hat{p}$ . For a 98% confidence interval, the critical value from the standard normal distribution is $z = 2.326$ . Substitute $\hat{p} = \frac{78}{250}$ into

\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

and evaluate the two endpoints.

Working

The sample proportion is

\hat{p} = \frac{78}{250} = 0.312

For a 98% confidence interval,

z = 2.326

The standard error is

\sqrt{\frac{0.312(1-0.312)}{250}} = \sqrt{\frac{0.312 \times 0.688}{250}}

So the confidence interval is

0.312 \pm 2.326 \sqrt{\frac{0.312 \times 0.688}{250}}

Evaluating the margin of error:

2.326 \sqrt{\frac{0.312 \times 0.688}{250}} \approx 0.0682

Therefore,

0.312 - 0.0682 = 0.2438

and

0.312 + 0.0682 = 0.3802

Answer

0.244 \text{ to } 0.380

Final answer

0.244 to 0.380

Detailed explanation

Walkthrough

We are estimating the unknown proportion $p$ of all Barapet residents who own a BETEC phone. From a sample of $n = 250$ people, 78 own one, so the sample proportion is $\hat{p} = 78/250 = 0.312$ . Because the sample is large, $\hat{p}$ is approximately normally distributed with standard error $\sqrt{\hat{p}(1-\hat{p})/n}$ . For a 98% confidence interval, we want the central 98% of the normal distribution, so the upper tail is 1% and the critical value is $z = 2.326$ . We then form $\hat{p} \pm z \times$ standard error. This gives the lower and upper endpoints. The interval is $(0.244, 0.380)$ , meaning we are 98% confident that the true proportion lies in this range.

Key Takeaways

The sample proportion $\hat{p}$ is the point estimate for the population proportion.
The standard error of $\hat{p}$ is $\sqrt{\hat{p}(1-\hat{p})/n}$ .
For a 98% CI, use $z = 2.326$ , not 1.96.
The interval is $\hat{p} \pm z \times$ standard error.

Common Mistakes

Using $z = 1.96$ for a 95% interval or $z = 2.576$ for a 99% interval instead of $2.326$ .
Forgetting to divide by $n$ inside the square root.
Using a value such as $0.4$ instead of the sample proportion $0.312$ in the standard error.
Giving only the margin of error rather than the full interval.

Things to Be Careful About

The formula uses $\hat{p}(1-\hat{p})$ in the numerator, not $\hat{p}^2$ .
The interval must be written as two values, not just one endpoint.
The answer should be rounded to 3 significant figures: $0.244$ to $0.380$ .
This is an approximate interval, valid because the sample size is large.

Techniques used

calculate the sample proportiondetermine the critical z-value for 98% confidencecompute the standard errorconstruct the confidence interval

(b)

Manjit claims that more than 40% of the people living in Barapet own a BETEC phone.

Use your answer to part (a) to comment on this claim.

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Q3MediumThe Poisson Distribution

In a certain lottery, on average 1 in every 10 000 tickets is a prize-winning ticket. An agent sells 6000 tickets.

(a)

Use a suitable approximating distribution to find the probability that at least 3 of the tickets sold by the agent are prize-winning tickets.

3M Similar

DifficultyMedium

Worked solution

Approach

Let $X$ be the number of prize-winning tickets sold by the agent. The exact distribution is binomial, but with large $n$ and small $p$ we approximate it by a Poisson distribution with the same mean.

Working

There are $n = 6000$ tickets and the probability that one ticket wins is $p = \frac{1}{10000}$ . Therefore

\lambda = np = 6000 \times \frac{1}{10000} = 0.6

so $X \sim \mathrm{Po}(0.6)$ approximately. We need $P(X \geq 3)$ . Using the complement rule,

P(X \geq 3) = 1 - P(X \leq 2)

For a Poisson distribution $P(X = r) = \frac{e^{-\lambda} \lambda^r}{r!}$ . Hence

\begin{aligned} P(X \leq 2) &= P(X=0) + P(X=1) + P(X=2) \\ &= e^{-0.6} + 0.6e^{-0.6} + \frac{0.6^2}{2}e^{-0.6} \\ &= e^{-0.6}\left(1 + 0.6 + \frac{0.6^2}{2}\right) \end{aligned}

Numerically,

P(X \leq 2) = 0.5488 + 0.3293 + 0.0988 = 0.9769

So

P(X \geq 3) = 1 - 0.9769 = 0.0231

Answer

P(\text{at least 3 prize-winning tickets}) = 0.0231

Final answer

0.0231

Detailed explanation

Walkthrough

Let $X$ be the number of prize-winning tickets among the 6000 sold. Since each ticket either wins or does not win, $X$ has a binomial distribution $\mathrm{Bin}(6000, 1/10000)$ . Computing $P(X \geq 3)$ exactly with $n = 6000$ would involve very large numbers, so we look for an approximation. When $n$ is large and $p$ is small, the Poisson distribution is a standard approximation to the binomial. The mean of the binomial is $np = 0.6$ , so we use $X \sim \mathrm{Po}(0.6)$ .

To find $P(X \geq 3)$ , the complement is easier: $P(X \geq 3) = 1 - P(X \leq 2)$ . We calculate $P(X \leq 2)$ by adding the Poisson probabilities for $X = 0$ , $X = 1$ and $X = 2$ . The formula is $P(X = r) = e^{-\lambda} \lambda^r / r!$ . Substituting $\lambda = 0.6$ gives the three terms shown. Their sum is approximately $0.9769$ , so the required probability is $1 - 0.9769 = 0.0231$ .

Key Takeaways

This question tests the Poisson approximation to the binomial: using $\lambda = np$ when $n$ is large and $p$ is small. It also tests routine Poisson probability calculations and the complement rule for an inequality such as at least 3.

Common Mistakes

Using the exact binomial distribution and attempting a lengthy calculation instead of the approximation.
Forgetting to include $P(X = 2)$ when computing $P(X \leq 2)$ .
Computing $P(X < 3)$ incorrectly; the complement requires $P(X \leq 2)$ .
Omitting the expression in part (a). The mark scheme requires the method expression to be seen for the method mark; a bare answer without working can lose marks.
Confusing $\lambda = 0.6$ with a probability; $\lambda$ is the mean, not the per-ticket probability.

Things to Be Careful About

The final numerical value is $0.0231$ to 3 significant figures. Use $\lambda = np = 0.6$ , not $\lambda = 6000$ . When using the complement, subtract from 1 after summing exactly $P(X=0), P(X=1), P(X=2)$ . The mark scheme accepts the algebraic expression $1 - e^{-0.6}(1 + 0.6 + 0.6^2/2)$ ; showing that expression is a clear way to secure the method mark.

Techniques used

identify the binomial settingapproximate a binomial distribution by a Poisson distributioncompute Poisson probabilities using the Poisson formulaapply the complement rule

(b)

Justify the use of your approximating distribution in this context.

1M Similar

DifficultyEasy

Worked solution

Approach

The underlying distribution of the number of prize-winning tickets is binomial. It is appropriate to approximate this binomial distribution by a Poisson distribution when $n$ is large and either $p$ or $np$ is small. We check both conditions explicitly.

Working

There are $n = 6000$ tickets, so

n = 6000 > 50

The probability of a prize-winning ticket is

p = \frac{1}{10000} = 0.0001 < 0.1

Equivalently,

np = 6000 \times \frac{1}{10000} = 0.6 < 5

Both conditions are satisfied, so the Poisson approximation with $\lambda = np = 0.6$ is justified.

Answer

Poisson approximation is justified because $n = 6000 > 50$ and $np = 0.6 < 5$ (equivalently $p = 0.0001 < 0.1$ ).

Final answer

n = 6000 > 50 and np = 0.6 < 5 (equivalently p = 0.0001 < 0.1)

Detailed explanation

Walkthrough

The exact distribution is binomial: there are $n = 6000$ tickets and each has success probability $p = 1/10000$ . The standard rule for a Poisson approximation is that $n$ should be large and $p$ should be small, with $np$ moderate. The mark scheme wants the values to be stated. Here $n = 6000$ , which is greater than 50, so $n$ is large. Also $p = 1/10000 = 0.0001 < 0.1$ , or equivalently $np = 0.6 < 5$ . Since these conditions hold, the Poisson model with mean $0.6$ is appropriate.

Key Takeaways

The justification for the Poisson approximation is not just large $n$ , small $p$ . The examiner requires explicit numerical checks: $n > 50$ and either $p < 0.1$ or $np < 5$ . These conditions make the approximation reliable and show that you have understood why the Poisson model is suitable.

Common Mistakes

Writing only since $n$ is large and $p$ is small is insufficient for the mark.
Stating $np = 0.6$ but not also checking $n > 50$ , or vice versa.
Using $p = 1/10000 = 0.0001$ but writing it incorrectly.
Claiming the distribution becomes Poisson exactly; it is only an approximation to the binomial distribution.

Things to Be Careful About

The two acceptable justifications are $n = 6000 > 50$ and $np = 0.6 < 5$ , or $n = 6000 > 50$ and $p = 0.0001 < 0.1$ . State both values clearly. Note that $np$ is also the parameter $\lambda$ used in the approximating Poisson distribution.

Techniques used

state the large n conditionstate the small p or small np conditionverify the approximation conditions

Q4MediumLinear Combinations of Random Variables

Each year a transport firm uses $X$ litres of gasoline and $Y$ litres of diesel fuel, where $X$ and $Y$ have the independent distributions $X \sim \text{N}(10\,700, 950^2)$ and $Y \sim \text{N}(13\,400, 1210^2)$ .

(a)

Find the probability that in a randomly chosen year the firm uses more gasoline than diesel fuel.

5M Similar

DifficultyMedium

Worked solution

Approach

Let $D = X - Y$ . Since $X$ and $Y$ are independent normal variables, $D$ is also normal. The firm uses more gasoline than diesel when $D > 0$ , i.e. $X - Y > 0$ . Compute the mean and variance of $D$ , standardise, and use the normal distribution tables.

Working

Since $X$ and $Y$ are independent,

E(D) = E(X - Y) = 10\,700 - 13\,400 = -2700

and

\text{Var}(D) = \text{Var}(X - Y) = 950^2 + 1210^2 = 2\,366\,600.

Thus $D \sim \text{N}(-2700, 2\,366\,600)$ .

The required probability is

P(X > Y) = P(X - Y > 0) = P(D > 0).

Standardise:

z = \frac{0 - (-2700)}{\sqrt{2\,366\,600}} = \frac{2700}{\sqrt{2\,366\,600}} \approx 1.755.

Therefore

P(D > 0) = 1 - \Phi(1.755) \approx 1 - 0.9604 = 0.0396.

Answer

P(X > Y) \approx 0.0396

Final answer

0.0396

Detailed explanation

Walkthrough

We are asked for the probability that gasoline use exceeds diesel use, i.e. $X > Y$ . It is easier to work with one variable: define $D = X - Y$ . Because $X$ and $Y$ are independent and both normal, $D$ is normal, so we only need its mean and variance.

For the mean, expectation is linear: $E(X-Y)=E(X)-E(Y)$ . For independent variables, the variance of a difference is the sum of the variances: $\text{Var}(X-Y)=\text{Var}(X)+\text{Var}(Y)$ . This is a common point to remember: subtracting does not subtract variances.

Then $D$ has mean $-2700$ and variance $2\,366\,600$ . We need $P(D>0)$ . Convert to the standard normal by subtracting the mean and dividing by the standard deviation. Here $0$ is $1.755$ standard deviations above the mean, so $P(D>0)=P(Z>1.755)=1-\Phi(1.755)$ . Reading the normal tables gives $\Phi(1.755)\approx 0.9604$ , so the probability is about $0.0396$ .

Key Takeaways

A linear combination of independent normal variables is normal.
$E(aX+bY)=aE(X)+bE(Y)$ .
For independent $X$ and $Y$ , $\text{Var}(aX+bY)=a^2\text{Var}(X)+b^2\text{Var}(Y)$ , so $\text{Var}(X-Y)=\text{Var}(X)+\text{Var}(Y)$ .
To find $P(X>Y)$ , rewrite it as $P(X-Y>0)$ and standardise.

Common Mistakes

Writing $\text{Var}(X-Y)=\text{Var}(X)-\text{Var}(Y)$ ; variances always add for independent variables.
Finding $P(X-Y>0)$ but forgetting to use the correct sign of the mean; the $z$ -score uses $0 - (-2700)$ .
Using $P(Z<1.755)$ instead of the upper tail $1-\Phi(1.755)$ .
Not showing the standardisation step; the mark scheme requires a method mark for standardising with the correct mean and variance.

Things to Be Careful About

The question asks for "more gasoline than diesel", so the event is $X>Y$ , not $X<Y$ .
If you instead use $Y-X$ , the event becomes $Y-X<0$ ; the final probability is the same.
Use the variance, not the standard deviation, in the formula $\text{Var}(X-Y)$ before taking the square root.
Give the final answer to 3 significant figures as requested.

Techniques used

form the difference of two independent normal variablescompute the mean and variance of a linear combinationstandardise a normal variableuse the standard normal distribution table

(b)

The costs per litre of gasoline and diesel fuel are $0.80 and $0.85 respectively.

Find the probability that the total cost of gasoline and diesel fuel in a randomly chosen year is between $20,000 and $22,000.

5M Similar

DifficultyMedium

Worked solution

Approach

Let the total cost be $C = 0.8X + 0.85Y$ . Since $X$ and $Y$ are independent normal variables, $C$ is normal. Find $E(C)$ and $\text{Var}(C)$ , standardise both boundary values $20,000 and $22,000, and subtract the cumulative probabilities.

Working

E(C) = 0.8(10\,700) + 0.85(13\,400) = 8560 + 11\,390 = 19\,950.

Since $X$ and $Y$ are independent,

\begin{aligned} \text{Var}(C) &= 0.8^2(950^2) + 0.85^2(1210^2) \\ &= 0.64(902\,500) + 0.7225(1\,464\,100) \\ &= 1\,635\,412.25. \end{aligned}

So $C \sim \text{N}(19\,950, 1\,635\,412.25)$ .

Standardise the upper limit:

z_{U} = \frac{22\,000 - 19\,950}{\sqrt{1\,635\,412.25}} \approx \frac{2050}{1278.83} \approx 1.603.

Standardise the lower limit:

z_{L} = \frac{20\,000 - 19\,950}{\sqrt{1\,635\,412.25}} \approx \frac{50}{1278.83} \approx 0.0391.

Hence

P(20\,000 < C < 22\,000) = \Phi(1.603) - \Phi(0.0391) \approx 0.9455 - 0.5156 = 0.4299.

Answer

P(20\,000 < C < 22\,000) \approx 0.430

Final answer

0.430

Detailed explanation

Walkthrough

The total cost is $C = 0.8X + 0.85Y$ . Since $X$ and $Y$ are independent normal variables, $C$ is also normal. We need $P(20\,000 < C < 22\,000)$ .

First find the mean: $E(C)=0.8(10\,700)+0.85(13\,400)=8560+11\,390=19\,950$ . Then find the variance. The coefficients are squared when they multiply variances: $\text{Var}(C)=0.8^2(950^2)+0.85^2(1210^2)=577\,600+1\,057\,812.25=1\,635\,412.25$ .

Now standardise the two boundaries. The upper boundary $22\,000$ gives $z_U=(22\,000-19\,950)/\sqrt{1\,635\,412.25}\approx 1.603$ . The lower boundary $20\,000$ gives $z_L=(20\,000-19\,950)/\sqrt{1\,635\,412.25}\approx 0.0391$ . Therefore the probability is $\Phi(1.603)-\Phi(0.0391)\approx 0.9455-0.5156=0.4299\approx 0.430$ .

Key Takeaways

For a weighted sum $aX+bY$ , the mean uses the coefficients directly, but the variance uses their squares.
A linear combination of independent normal variables is normal, so standard normal tables can be used.
For a probability between two values, standardise both and subtract cumulative probabilities: $P(a<C<b)=\Phi(z_b)-\Phi(z_a)$ .

Common Mistakes

Forgetting to square the cost coefficients when computing variance, e.g. writing $0.8(950^2)+0.85(1210^2)$ .
Using $E(C)=E(X)E(Y)$ or mixing up addition and multiplication.
Subtracting the cumulative probabilities in the wrong order, giving a negative probability.
Not using the variance in the denominator; the denominator must be $\sqrt{\text{Var}(C)}$ .

Things to Be Careful About

The total cost is measured in dollars; the boundaries are $20,000 and $22,000.
Keep enough precision in the $z$ -scores; small rounding can change the final 3 significant figures.
The mark scheme allows the final answer $0.430$ ; do not give an unsupported answer.
When standardising, the mean is $19,950, which is below both boundaries, so both $z$ -scores are positive.

Techniques used

form a weighted sum of independent normal variablescompute the mean and variance of a weighted sumstandardise both boundariessubtract cumulative normal probabilities

Q5MediumThe Poisson DistributionLinear Combinations of Random VariablesHypothesis Tests

A teacher models the numbers of girls and boys who arrive late for her class on any day by the independent random variables $G \sim \text{Po}(0.10)$ and $B \sim \text{Po}(0.15)$ respectively.

(a)

Find the probability that during a randomly chosen 2-day period no girls arrive late.

1M Similar

(b)

Find the probability that during a randomly chosen 5-day period the total number of students who arrive late is less than 3.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Combine the girls and boys rates per day, then scale to 5 days. The total number of late students is Poisson, so sum the probabilities for 0, 1 and 2 late students.

Working

Per day, the total rate is

\lambda_{\text{day}} = 0.10 + 0.15 = 0.25

Over 5 days,

\lambda = 5 \times 0.25 = 1.25

P(T < 3) = P(T = 0) + P(T = 1) + P(T = 2)

= e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2}\right)

= 0.2865 + 0.3581 + 0.2238 = 0.8684

Answer

The probability is $0.868$ (3 s.f.).

Final answer

0.868 (3 s.f.)

Detailed explanation

Walkthrough

On any one day, the total number of students late is $G + B$ . Because $G$ and $B$ are independent Poisson variables, $G+B$ is Poisson with mean $0.10 + 0.15 = 0.25$ . Over 5 independent days, the total $T$ is Poisson with mean $5 \times 0.25 = 1.25$ . The event 'total less than 3' means $T = 0$ , $T = 1$ or $T = 2$ . Using $P(T=r)=e^{-\lambda}\lambda^r/r!$ , we sum these three probabilities:

P(T<3)=e^{-1.25}\left(1+1.25+\frac{1.25^2}{2}\right)

This equals $0.2865 + 0.3581 + 0.2238 = 0.8684$ , so the required probability is 0.868 to 3 significant figures.

Key Takeaways

Independent Poisson counts can be combined by adding their means, both across different sources and across repeated periods. Cumulative Poisson probabilities are found by summing individual probabilities.

Common Mistakes

A frequent error is using 0.25 as the mean for the 5-day period instead of 1.25. Another is omitting $P(T=0)$ or $P(T=2)$ when finding $P(T<3)$ . The mark scheme allows one end error for the method mark, but the final answer must be correct for full marks.

Things to Be Careful About

'Less than 3' includes 0, 1 and 2, not 3. Use $\lambda = 1.25$ for the 5-day total. Give the final probability to 3 significant figures.

Techniques used

combine independent Poisson ratesscale rate over multiple periodssum Poisson probabilities for a range of values

(c)

It is given that the values of $\text{P}(G = r)$ and $\text{P}(B = r)$ for $r \geqslant 3$ are very small and can be ignored.

Find the probability that on a randomly chosen day more girls arrive late than boys.

3M Similar

DifficultyMedium

Worked solution

Approach

We need $P(G > B)$ . Since values with $r \geq 3$ can be ignored, only the pairs $(1,0)$ , $(2,0)$ and $(2,1)$ contribute. Use independence to multiply Poisson probabilities, then add the mutually exclusive cases.

Working

P(B = 0) = e^{-0.15} = 0.8607, \quad P(B = 1) = e^{-0.15}(0.15) = 0.1291

P(G = 1) = e^{-0.1}(0.1) = 0.09048, \quad P(G = 2) = e^{-0.1}\frac{0.1^2}{2} = 0.004524

P(G > B) = P(B = 0)P(G = 1) + P(B = 0)P(G = 2) + P(B = 1)P(G = 2)

= e^{-0.15}e^{-0.1}(0.1) + e^{-0.15}e^{-0.1}\left(\frac{0.1^2}{2}\right) + e^{-0.15}(0.15)e^{-0.1}\left(\frac{0.1^2}{2}\right)

= 0.077879 + 0.003894 + 0.0005841 = 0.0824

Answer

The probability is $0.0824$ (3 s.f.).

Final answer

0.0824 (3 s.f.)

Detailed explanation

Walkthrough

We need $P(G > B)$ . Since $r \geq 3$ can be ignored, the only possible values of $G$ and $B$ are 0, 1 and 2. For $G > B$ , the favourable pairs are $(G,B) = (1,0)$ , $(2,0)$ and $(2,1)$ . Because $G$ and $B$ are independent, the probability of each pair is the product of the individual Poisson probabilities. We compute:

P(B=0)=e^{-0.15}=0.8607,\quad P(B=1)=e^{-0.15}(0.15)=0.1291

P(G=1)=e^{-0.1}(0.1)=0.09048,\quad P(G=2)=e^{-0.1}\frac{0.1^2}{2}=0.004524

Then add the three products:

P(G>B)=0.8607(0.09048)+0.8607(0.004524)+0.1291(0.004524)=0.0824

Key Takeaways

Discrete probability questions often require enumerating all favourable outcomes. Independence means multiplying probabilities, and mutually exclusive cases are added. The instruction to ignore very small tail probabilities reduces the number of cases.

Common Mistakes

Missing the $(2,1)$ case is common. Another common error is using $P(G \geq B)$ instead of $P(G > B)$ . Students sometimes add the two Poisson means or probabilities instead of multiplying independent probabilities. The mark scheme gives a method mark for one correct product and a second method mark for adding all three terms.

Things to Be Careful About

'More girls than boys' is a strict inequality. Only three cases contribute because $r \geq 3$ is negligible. Use the exact Poisson probabilities and keep enough decimal places before rounding the final answer to 3 significant figures.

Techniques used

enumerate favourable independent Poisson outcomesmultiply independent Poisson probabilitiessum mutually exclusive cases

(d)

Following a timetable change the teacher claims that on average more students arrive late than before the change. During a randomly chosen 5-day period a total of 4 students are late.

Test the teacher's claim at the 5% significance level.

5M Similar

DifficultyMedium

Worked solution

Approach

Set up a one-tailed hypothesis test for the Poisson mean over a 5-day period. Calculate the probability of observing 4 or more late students under the null hypothesis, compare it with the 5% significance level, and conclude in context.

Working

Before the change, the mean number of late students per day is

0.10 + 0.15 = 0.25

so over 5 days

\lambda = 5 \times 0.25 = 1.25

Let $X$ be the total number of late students in 5 days. The hypotheses are

H_0: \lambda = 1.25, \quad H_1: \lambda > 1.25

Under $H_0$ , $X \sim \text{Po}(1.25)$ . The probability of 4 or more is

P(X \geq 4) = 1 - P(X \leq 3)

= 1 - e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2} + \frac{1.25^3}{6}\right)

= 1 - (0.2865 + 0.3581 + 0.2238 + 0.09326)

= 0.0383

Since $0.0383 < 0.05$ , reject $H_0$ at the 5% significance level.

Answer

There is sufficient evidence to support the teacher's claim that more students arrive late on average after the timetable change.

Final answer

Reject H0 at the 5% level; there is sufficient evidence to support the teacher's claim.

Detailed explanation

Walkthrough

Before the change, the mean number of late students per day is $0.10 + 0.15 = 0.25$ , so over 5 days the mean is $5 \times 0.25 = 1.25$ . The teacher claims the mean has increased, so this is a one-tailed test:

H_0: \lambda = 1.25, \quad H_1: \lambda > 1.25

Under $H_0$ , the total number late in 5 days is $X \sim \text{Po}(1.25)$ . The observed value is 4, so we calculate the probability of 4 or more:

P(X \geq 4) = 1 - P(X \leq 3)

= 1 - e^{-1.25}\left(1 + 1.25 + \frac{1.25^2}{2} + \frac{1.25^3}{6}\right)

= 1 - (0.2865 + 0.3581 + 0.2238 + 0.09326) = 0.0383

Since $0.0383 < 0.05$ , we reject $H_0$ at the 5% significance level. There is sufficient evidence to support the teacher's claim that more students arrive late on average after the timetable change.

Key Takeaways

A hypothesis test for a Poisson mean uses the Poisson distribution to find the probability of the observed result or more extreme. The comparison of this probability with the significance level determines whether to reject the null hypothesis. The conclusion must be in context and should not claim certainty.

Common Mistakes

Using the per-day mean 0.25 instead of the 5-day mean 1.25 is a common error. Another is testing two-tailed when the claim is clearly one-sided ('more'). Some students calculate $P(X = 4)$ instead of $P(X \geq 4)$ . The mark scheme notes that 0.0383 with no working scores only B1, so the full expression must be shown.

Things to Be Careful About

State both hypotheses clearly. Use the 5-day mean $\lambda = 1.25$ . Compare the tail probability with 0.05 and reject $H_0$ because $0.0383 < 0.05$ . The final conclusion should be in context, for example 'there is sufficient evidence to suggest that the teacher's claim is true', not 'the claim is correct' or 'more students are late' as a definite statement.

Techniques used

state null and alternative hypothesescalculate upper-tail Poisson probabilitycompare p-value with significance leveldraw conclusion in context

Q6MediumContinuous Random Variables

The graph of the probability density function $\text{f}$ of a random variable $X$ is symmetrical about the line $x = 2$ . It is given that $\text{P}(2 < X < 5) = \frac{117}{256}$ .

(a)

Using only this information show that $\text{P}(X > -1) = \frac{245}{256}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Approach

The PDF is symmetric about $x=2$ , so the probability of being within 3 units to the right of 2 equals the probability of being within 3 units to the left of 2. Use this to find the left tail $P(X < -1)$ , then use the complement rule.

Working

Since the graph is symmetric about $x=2$ ,

P(-1 < X < 2) = P(2 < X < 5) = \frac{117}{256}.

The total probability is 1, and by symmetry the two outer tails are equal:

P(X < -1) = P(X > 5).

Therefore,

2P(X < -1) + 2\left(\frac{117}{256}\right) = 1.

So

P(X < -1) = \frac{1 - \frac{234}{256}}{2} = \frac{22}{256} \cdot \frac{1}{2} = \frac{11}{256}.

Hence

P(X > -1) = 1 - P(X < -1) = 1 - \frac{11}{256} = \frac{245}{256}.

Answer

P(X > -1) = \frac{245}{256}

Final answer

P(X > -1) = 245/256

Detailed explanation

Walkthrough

The key idea is symmetry. The PDF is symmetric about $x=2$ , so the probability of $X$ being between $-1$ and $2$ (three units to the left of $2$ ) equals the probability of $X$ being between $2$ and $5$ (three units to the right of $2$ ). We are told the latter is $\frac{117}{256}$ . The total probability under any PDF is 1. By symmetry, the two outer tails $P(X < -1)$ and $P(X > 5)$ are equal. So the total probability is

P(X < -1) + P(-1 < X < 2) + P(2 < X < 5) + P(X > 5) = 1,

which becomes

2P(X < -1) + 2\left(\frac{117}{256}\right) = 1.

Solving gives $P(X < -1) = \frac{11}{256}$ . Then $P(X > -1)$ is the complement of $P(X < -1)$ , so

P(X > -1) = 1 - \frac{11}{256} = \frac{245}{256}.

Key Takeaways

Symmetry of a PDF about a line means equal probabilities for equal distances on either side. The complement rule is useful for finding the probability of the opposite event. For a continuous random variable, single-point probabilities are zero, so strict and non-strict inequalities give the same probability.

Common Mistakes

Forgetting that there are two outer tails; subtracting $\frac{117}{256}$ from 1 without accounting for both tails gives the wrong result.
Not using the symmetry of the outer tails $P(X < -1) = P(X > 5)$ .
For an answer given (AG) question, not showing enough numerical working to reach $\frac{245}{256}$ .

Things to Be Careful About

The interval $(-1,2)$ mirrors $(2,5)$ because both are three units from $x=2$ .
Since $X$ is continuous, $P(X = -1)=0$ , so $P(X > -1)=P(X \geq -1)$ .
The question says to use only the symmetry information; do not introduce a formula for $\text{f}(x)$ .

Techniques used

apply symmetry of the PDFuse complement ruleequate tail probabilities

(b)

It is now given that, for $x$ in a suitable domain,

\text{f}(x) = k(12 + 4x - x^2)

where $k$ is a constant.

Find the value of $k$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since $P(2 < X < 5) = \frac{117}{256}$ , integrate the given PDF over the interval $2$ to $5$ , set the result equal to $\frac{117}{256}$ , and solve for $k$ .

Working

For a continuous random variable, probability is the integral of the PDF:

k\int_2^5 (12+4x-x^2)\,dx = \frac{117}{256}.

Integrate term by term:

\int (12+4x-x^2)\,dx = 12x + 2x^2 - \frac{x^3}{3}.

Evaluate from $2$ to $5$ :

\left[12x + 2x^2 - \frac{x^3}{3}\right]_2^5 = \left(60+50-\frac{125}{3}\right) - \left(24+8-\frac{8}{3}\right) = \frac{205}{3} - \frac{88}{3} = \frac{117}{3} = 39.

So

39k = \frac{117}{256}.

Thus

k = \frac{117}{256 \times 39} = \frac{3}{256}.

Answer

k = \frac{3}{256}

Final answer

k = 3/256

Detailed explanation

Walkthrough

We know $P(2 < X < 5) = \frac{117}{256}$ . For a continuous random variable, this probability is the integral of the PDF over that interval. So integrate $k(12+4x-x^2)$ from $2$ to $5$ and set it equal to $\frac{117}{256}$ . First integrate term by term: $12x + 2x^2 - \frac{x^3}{3}$ . Evaluate at $5$ and $2$ and subtract. The result is $39$ . Then $39k = \frac{117}{256}$ , so $k = \frac{3}{256}$ . This uses the given probability on the interval rather than the total area over the whole domain.

Key Takeaways

A PDF integrates to 1 over its whole domain, and the probability of an interval is the integral of the PDF over that interval. A known probability can be used to determine the normalising constant $k$ .

Common Mistakes

Forgetting to include the factor $k$ when integrating.
Making arithmetic errors when evaluating the definite integral, especially with fractions.
Equating the integral to 1 instead of to $\frac{117}{256}$ .

Things to Be Careful About

Use the correct limits $2$ and $5$ .
Simplify the fraction carefully: $\frac{117}{256 \times 39} = \frac{3}{256}$ .
The mark scheme also accepts alternative limits if the total area is used, but with the information given, integrating over $2$ to $5$ is the direct method.

Techniques used

integrate the PDF over the given intervalequate the integral to a known probabilitysolve for the normalising constant

(c)

A different random variable $X$ has probability density function $\text{g}(x) = \frac{2}{9}(2 + x - x^2)$ . The domain of $X$ is all values of $x$ for which $\text{g}(x) \geqslant 0$ .

Find $\text{Var}(X)$ .

5M Similar

DifficultyMedium

Worked solution

Approach

First find the domain by solving $g(x) \geq 0$ . Then use symmetry to find the mean, integrate $x^2g(x)$ to find $E(X^2)$ , and use $\text{Var}(X)=E(X^2)-[E(X)]^2$ .

Working

Solve $2+x-x^2=0$ :

x^2-x-2=0 \quad \Rightarrow \quad (x-2)(x+1)=0,

so $x=-1$ or $x=2$ . Since $g(x) \geq 0$ between these roots, the domain is $-1 \leq x \leq 2$ .

The function $2+x-x^2$ is symmetric about $x=\frac{1}{2}$ , so the mean is

E(X)=\frac{1}{2}.

Now

E(X^2)=\int_{-1}^{2} x^2 g(x)\,dx = \frac{2}{9}\int_{-1}^{2} (2x^2+x^3-x^4)\,dx.

Integrate:

\frac{2}{9}\left[\frac{2}{3}x^3+\frac{x^4}{4}-\frac{x^5}{5}\right]_{-1}^{2}.

At $x=2$ :

\frac{2}{3}(8)+4-\frac{32}{5} = \frac{16}{3}+4-\frac{32}{5} = \frac{44}{15}.

At $x=-1$ :

-\frac{2}{3}+\frac{1}{4}+\frac{1}{5} = -\frac{13}{60}.

Therefore

E(X^2)=\frac{2}{9}\left(\frac{44}{15}-\left(-\frac{13}{60}\right)\right) = \frac{2}{9}\left(\frac{176}{60}+\frac{13}{60}\right) = \frac{2}{9}\cdot\frac{189}{60} = \frac{7}{10}.

Then

\text{Var}(X)=E(X^2)-[E(X)]^2 = \frac{7}{10}-\left(\frac{1}{2}\right)^2 = \frac{7}{10}-\frac{1}{4} = \frac{14}{20}-\frac{5}{20} = \frac{9}{20}.

Answer

\text{Var}(X)=\frac{9}{20}=0.45

Final answer

Var(X) = 9/20 = 0.45

Detailed explanation

Walkthrough

First find where $g(x) \geq 0$ . Since $g(x)=\frac{2}{9}(2+x-x^2)$ is a quadratic with a negative leading coefficient, it is nonnegative between its roots. Solve $2+x-x^2=0$ to get $x=-1$ and $x=2$ , so the domain is $-1 \leq x \leq 2$ .

Next, note that $2+x-x^2$ is symmetric about $x=\frac{1}{2}$ , and the domain is also symmetric about $\frac{1}{2}$ . Therefore the mean is $E(X)=\frac{1}{2}$ ; no integration is needed for the mean.

To find $E(X^2)$ , integrate $x^2g(x)$ over the domain:

E(X^2)=\frac{2}{9}\int_{-1}^{2}(2x^2+x^3-x^4)\,dx.

Integrate term by term and evaluate at $2$ and $-1$ . At $x=2$ the antiderivative value is $\frac{44}{15}$ ; at $x=-1$ it is $-\frac{13}{60}$ . The difference is $\frac{189}{60}$ , and multiplying by $\frac{2}{9}$ gives $\frac{7}{10}$ .

Finally use the variance formula:

\text{Var}(X)=E(X^2)-[E(X)]^2=\frac{7}{10}-\left(\frac{1}{2}\right)^2=\frac{9}{20}.

Key Takeaways

The domain of a PDF is where the density is nonnegative. Symmetry of a PDF and its domain can give the mean without integration. The variance is $E(X^2)-[E(X)]^2$ , not $E(X^2)$ alone.

Common Mistakes

Choosing the wrong domain, such as $x \leq -1$ or $x \geq 2$ instead of $-1 \leq x \leq 2$ .
Forgetting the factor $\frac{2}{9}$ when integrating $x^2g(x)$ .
Using $E(X^2)$ as the variance without subtracting the square of the mean.
Sign errors when evaluating at $x=-1$ , particularly for $-\frac{x^5}{5}$ .

Things to Be Careful About

At $x=-1$ , $x^5=-1$ , so $-\frac{x^5}{5}=+\frac{1}{5}$ .
The mean is $0.5$ , not $0$ .
The mark scheme requires both $E(X^2)$ and the mean squared to be numerical before subtracting.
The final answer may be written as $\frac{9}{20}$ or $0.45$ .

Techniques used

find the domain by solving g(x) >= 0use symmetry to identify the meancompute E(X^2) by integrationapply the variance formula

Q7Medium-HardHypothesis Tests

The heights, in centimetres, of adult females in Litania have mean $\mu$ and standard deviation $\sigma$ . It is known that in 2004 the values of $\mu$ and $\sigma$ were 163.21 and 6.95 respectively. The government claims that the value of $\mu$ this year is greater than it was in 2004. In order to test this claim a researcher plans to carry out a hypothesis test at the 1% significance level. He records the heights of a random sample of 300 adult females in Litania this year and finds the value of the sample mean.

(a)

State the probability of a Type I error.

1M Similar

(b)

You should assume that the value of $\sigma$ after 2004 remains at 6.95 .

Given that the value of $\mu$ this year is actually 164.91, find the probability of a Type II error.

5M Similar

DifficultyMedium-Hard

Worked solution

Approach

Let $\bar H$ be the sample mean height. Since $\sigma$ is known and the sample is large,

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

Under the null hypothesis $H_0: \mu = 163.21$ and the alternative $H_1: \mu > 163.21$ . A Type II error is failing to reject $H_0$ when in fact $\mu = 164.91$ , so it is the probability that $\bar H$ falls below the critical value of the rejection region.

Working

Find the critical value $c$ for the rejection region $\bar H > c$ . At the 1% significance level,

2.326 = \frac{c - 163.21}{6.95/\sqrt{300}}

Therefore

c = 163.21 + 2.326 \times \frac{6.95}{\sqrt{300}} = 164.14

The rejection region is $\bar H > 164.14$ .

If the true mean is $\mu = 164.91$ , the probability of a Type II error is

P(\text{Type II}) = P(\bar H < 164.14 \mid \mu = 164.91).

Standardising using $\sigma/\sqrt{300}$ :

Z = \frac{164.14 - 164.91}{6.95/\sqrt{300}} = -1.919

Thus

P(\text{Type II}) = \Phi(-1.919) = 1 - \Phi(1.919) = 0.0275

Answer

P(\text{Type II}) = 0.0275

Final answer

0.0275

Detailed explanation

Walkthrough

To test the claim that the population mean is greater than in 2004, set

H_0: \mu = 163.21, \quad H_1: \mu > 163.21.

Since $\sigma$ is known, the sample mean is normally distributed with mean $\mu$ and standard deviation $\sigma/\sqrt{300}$ .

A Type II error occurs when the test fails to reject a false null hypothesis. Here we are told that the true mean is $\mu = 164.91$ , so we need the probability that the sample mean falls in the acceptance region of the test.

First find the boundary of the rejection region. For a one-tailed test at the 1% significance level, the critical $z$ -value is $2.326$ . Equating this to the standardised critical sample mean gives

2.326 = \frac{c - 163.21}{6.95/\sqrt{300}}

and hence

c = 163.21 + 2.326 \times \frac{6.95}{\sqrt{300}} = 164.14.

The rejection region is therefore $\bar H > 164.14$ , and the acceptance region is $\bar H \leq 164.14$ .

Now, if the true mean is $164.91$ , the Type II error probability is

P(\bar H < 164.14 \mid \mu = 164.91).

Standardise this boundary using the true mean $164.91$ :

Z = \frac{164.14 - 164.91}{6.95/\sqrt{300}} = -1.919.

Thus

P(\text{Type II}) = \Phi(-1.919) = 1 - \Phi(1.919) = 0.0275.

Key Takeaways

A Type I error has probability exactly equal to the significance level.
The critical region is calculated using the null hypothesis mean.
A Type II error is calculated under the true/alternative mean, not the null mean.
For a sample mean test with known variance, the standard error is $\sigma/\sqrt{n}$ .

Common Mistakes

Using the two-tailed critical value, such as $2.576$ or $1.96$ , instead of the one-tailed value $2.326$ .
Finding the critical value using $\mu = 164.91$ instead of $\mu = 163.21$ .
Finding the Type II probability using $\mu = 163.21$ ; it must be found using the given true value $164.91$ .
Forgetting to divide $\sigma$ by $\sqrt{300}$ when standardising.

Things to Be Careful About

The rejection region is $\bar H > 164.14$ ; the Type II probability is the probability of the acceptance region $\bar H < 164.14$ , because the test fails to reject $H_0$ there.
One-tailed 1% significance corresponds to $z = 2.326$ , not $2.576$ .
Since the boundary was rounded to $164.14$ , the standardised value is approximately $-1.919$ ; the final probability may be quoted as $0.0275$ to $0.0280$ depending on rounding.

Techniques used

compute the critical sample mean for a one-tailed normal teststandardise the critical value under the specified alternative meanfind the normal tail probability using symmetry

Mathematics 9709/62 — February/March 2024

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