Mathematics 9709/61 — May/June 2025

Walkthrough

We begin with the binomial random variable $X \sim \text{B}(8, \frac{3}{4})$. The first step is to compute its mean and variance using the standard binomial formulas: $E(X) = np$ and $\text{Var}(X) = np(1-p)$. With $n = 8$ and $p = \frac{3}{4}$, we get $E(X) = 6$ and $\text{Var}(X) = 1.5$.

Next, we take the sample mean $\bar{X}$ of 100 independent values of $X$. The mean of the sample mean equals the population mean: $E(\bar{X}) = \mu = 6$. The variance of the sample mean is the population variance divided by the sample size: $\text{Var}(\bar{X}) = \frac{\sigma^2}{n} = \frac{1.5}{100}$.

Because $X$ is binomial (not normal), the Central Limit Theorem is required to justify that $\bar{X}$ is approximately normal for a large sample. With $n = 100$, the CLT applies, so we treat $\bar{X} \sim N(6, \frac{1.5}{100})$.

To find $P(\bar{X} > 6.2)$, we standardise: $z = \frac{6.2 - 6}{\sqrt{1.5/100}} \approx 1.633$. Then $P(\bar{X} > 6.2) = P(Z > 1.633) = 1 - \Phi(1.633)$. Using normal tables, $\Phi(1.633) \approx 0.9488$, so the probability is $1 - 0.9488 = 0.0512$.

Key Takeaways

• The mean and variance of a binomial distribution are $np$ and $np(1-p)$.
• For a sample mean, $E(\bar{X}) = \mu$ and $\text{Var}(\bar{X}) = \frac{\sigma^2}{n}$.
• The Central Limit Theorem allows the sample mean to be treated as normal for large $n$ even when the population is not normal.
• Standardising a normal variable to $z$ enables probability tables from standard normal tables to be used.

Common Mistakes

• Forgetting to divide the variance by the sample size $n$ when computing $\text{Var}(\bar{X})$.
• Using the population standard deviation $\sigma$ instead of $\sigma/\sqrt{n}$ when standardising.
• Forgetting to subtract from 1 when the question asks for a "greater than" probability.
• Applying a continuity correction when the question explicitly says not to use one.

Things to Be Careful About

• The standard deviation of $\bar{X}$ is $\sqrt{1.5/100}$, not $\sqrt{1.5}$.
• The mean of $\bar{X}$ is 6, not $6 \times 100$.
• The standardisation must include $\sqrt{100}$ in the denominator.
• The final answer should be given to 3 significant figures: $0.0512$ or $0.0513$.

Walkthrough

The Central Limit Theorem is needed because the underlying variable $X$ follows a binomial distribution, not a normal distribution. The CLT guarantees that the sample mean $\bar{X}$ is approximately normal for a large sample regardless of the population distribution. Without it, we could not assume $\bar{X}$ was normal in part (a).

Key Takeaways

• The CLT is the justification for using the normal distribution for a sample mean when the population is not normal.
• A sample size of 100 is considered large enough for the CLT to apply.

Common Mistakes

• Saying the CLT is needed because the sample is large — this is not the reason; the reason is that the population is not normal.
• Confusing the CLT with the normal approximation to the binomial distribution.

Things to Be Careful About

• The answer should mention that the underlying distribution is not normal — it is binomial.
• The mark scheme accepts "Population ($X$) is not normally distributed" or "Population ($X$) is Binomial" or equivalent.

Walkthrough

We are given only the summary statistics from a sample of 60 days, so the best point estimate of the population mean is the sample mean. We therefore divide the sum of the times by the sample size.

For the variance, we need an unbiased estimate. The formula uses $n-1$ in the denominator:

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

This corrects for the fact that using the sample mean in place of the true population mean tends to underestimate the variance. We substitute $n=60$, $\Sigma t=2750$ and $\Sigma t^2=127000$, then simplify to obtain $s^2=16.2429$, which is $16.2$ to 3 significant figures.

Key Takeaways

• The sample mean is an unbiased estimator of the population mean.
• The unbiased estimate of the population variance uses denominator $n-1$, not $n$.
• Summary statistics $\Sigma t$ and $\Sigma t^2$ can be substituted directly into the estimator formulas.

Common Mistakes

• Using denominator $n$ instead of $n-1$ gives a biased estimate, which the mark scheme explicitly disallows.
• Confusing $\Sigma t^2$ with $(\Sigma t)^2$.
• Forgetting to square the sample mean inside the variance formula.

Things to Be Careful About

• The variance formula can be written in two equivalent forms; either is acceptable.
• Keep enough precision in the variance for part (b) to avoid rounding errors.
• The mean estimate is $45.8$ to 3 significant figures, and the variance estimate is $16.2$ to 3 significant figures.

Walkthrough

We want to test whether the mean time is greater than $45$ minutes. The null hypothesis $H_0$ represents the company's claim, so $H_0: \mu = 45$. The passenger's belief is the alternative hypothesis, $H_1: \mu > 45$, making this a one-tailed test.

Because the population is normal and the sample size is $60$, the sample mean is normally distributed. Under $H_0$, the mean of the sample mean is $45$ and its variance is $\sigma^2/n$. Since $\sigma^2$ is unknown, we use the unbiased estimate $s^2 = 16.2429$ from part (a).

The standard error is

$$\sqrt{\frac{s^2}{n}} = \sqrt{\frac{16.2429}{60}} = 0.520303$$

We standardise the observed sample mean:

$$z = \frac{45.8333 - 45}{0.520303} = 1.602$$

For a one-tailed $5\%$ test, the critical value is $1.645$. Since $1.602 < 1.645$, the observed result is not unusual enough to reject $H_0$. Therefore we conclude that there is insufficient evidence to support the passenger's belief.

Key Takeaways

• State $H_0$ and $H_1$ clearly before performing the test.
• For a normal population with an estimated variance, use the standardised test statistic $z = (\bar{x} - \mu)/\sqrt{s^2/n}$.
• Compare the test statistic with the one-tailed critical value.
• The conclusion must be in context and should not claim a definite effect when the result is not significant.

Common Mistakes

• Using the two-tailed critical value $1.96$ instead of the one-tailed value $1.645$.
• Forgetting the $\sqrt{n}$ in the denominator when standardising.
• Using a biased variance estimate from part (a) when the question requires the unbiased estimate.
• Concluding "the mean is not greater than 45" rather than "there is insufficient evidence that it is greater than 45".
• Stating the hypotheses without defining the parameter or without using the correct inequality direction.

Things to Be Careful About

• Use the unbiased variance estimate from part (a), not a prematurely rounded value.
• The test statistic is approximately $1.602$; values between $1.595$ and $1.602$ are acceptable depending on rounding.
• The critical value for a one-tailed $5\%$ test is $1.645$.
• A critical-value method is also acceptable: the critical sample mean is $45 + 1.645\sqrt{16.2429/60} = 45.856$, and since $45.833 < 45.856$, the conclusion is the same.
• The final conclusion must mention the company's claim or the passenger's belief in context.

Walkthrough

Let $X$ be the number of minutes. The amount paid is $C = 0.4X$, so we need the mean and variance of $0.4X$.

For the mean, scaling a variable by a constant scales its mean by the same constant:

$$E(C) = 0.4E(X) = 0.4 \times 15 = 6.$$

For the variance, the constant is squared when it is taken out:

$$\mathrm{Var}(C) = 0.4^2 \mathrm{Var}(X) = 0.16 \times 9 = 1.44.$$

So the mean amount paid is 6 and the variance is 1.44.

Key Takeaways

This question tests the linear transformation of a random variable:

$$E(aX) = aE(X), \quad \mathrm{Var}(aX) = a^2\mathrm{Var}(X).$$

A common pattern is to multiply the mean by the constant but square the constant for the variance.

Common Mistakes

• Multiplying the variance by $0.4$ instead of $0.4^2$.
• Forgetting that variance is in squared units.
• Using the standard deviation $3$ instead of the variance $9$ in the calculation.

Things to Be Careful About

The variance is $9$, not the standard deviation $3$. If units are used, the variance is in squared monetary units, but units are not required here.

Walkthrough

From part (a), each person's payment has mean $6$ and variance $1.44$. Let $C_i$ be the payment of the $i$-th person and let $T = C_1 + C_2 + \cdots + C_{35}$ be the total.

Because the 35 people are independent, the mean of the total is the sum of the individual means:

$$E(T) = 35 \times 6 = 210.$$

The variance of the total is the sum of the individual variances:

$$\mathrm{Var}(T) = 35 \times 1.44 = 50.4.$$

The key point is that when adding independent variables, variances add, not standard deviations, and the number of variables is multiplied once, not squared.

Key Takeaways

For independent variables $X_1, \ldots, X_n$, the sum $S = X_1 + \cdots + X_n$ has

$$E(S) = \sum E(X_i), \quad \mathrm{Var}(S) = \sum \mathrm{Var}(X_i).$$

This is why the total variance is $35$ times the individual variance, not $35^2$ times.

Common Mistakes

• Multiplying the variance by $35^2$ instead of $35$.
• Multiplying standard deviations instead of variances.
• Forgetting that the variance addition formula requires independence.

Things to Be Careful About

• Use the answers from part (a) for the mean and variance of one person's payment.
• The variance is in squared monetary units, but units are not required here.
• The independence of the 35 people is given in the question, so no covariance terms are needed.

Walkthrough

Start with the given equation:

$$\frac{5}{2}\mathrm{P}(Y=3) + \mathrm{P}(Y=4) = \mathrm{P}(Y=5)$$

For $Y \sim \text{Po}(\lambda)$, $\mathrm{P}(Y=r)=e^{-\lambda}\frac{\lambda^r}{r!}$. Substitute this into the equation:

$$\frac{5}{2}e^{-\lambda}\frac{\lambda^3}{3!} + e^{-\lambda}\frac{\lambda^4}{4!} = e^{-\lambda}\frac{\lambda^5}{5!}$$

Since $e^{-\lambda}$ is never zero, it can be cancelled from every term:

$$\frac{5}{2}\frac{\lambda^3}{6} + \frac{\lambda^4}{24} = \frac{\lambda^5}{120}$$

Multiply through by $120$:

$$50\lambda^3 + 5\lambda^4 = \lambda^5$$

Since $\lambda > 0$, divide by $\lambda^3$:

$$\lambda^2 - 5\lambda - 50 = 0$$

Factorise:

$$(\lambda - 10)(\lambda + 5) = 0$$

So $\lambda = 10$ or $\lambda = -5$. Since $\lambda > 0$, the only valid solution is $\lambda = 10$.

Key Takeaways

• The Poisson probability formula is $\mathrm{P}(Y=r)=e^{-\lambda}\frac{\lambda^r}{r!}$.
• When an equation contains the same exponential factor in every term, it can be cancelled.
• Always check the domain of the parameter when solving; here $\lambda > 0$ removes the negative root.

Common Mistakes

• Forgetting to cancel $e^{-\lambda}$ and making the algebra more complicated.
• Making factorial errors when simplifying $3!$, $4!$ and $5!$.
• Dividing by $\lambda^3$ without noting that $\lambda > 0$, so the division is valid.
• Giving $\lambda = -5$ as an answer without rejecting it because $\lambda > 0$.

Things to Be Careful About

• The mark scheme accepts the equation with or without the $e^{-\lambda}$ factors, but the probabilities must be written correctly.
• After multiplying by $120$, rearrange carefully to get the quadratic in the standard form.
• The final answer must be the positive root only.

Walkthrough

The manufacturer's claim is that the population proportion is exactly 25%, so this is the null hypothesis. Jeyeraj thinks the proportion is less, so the alternative hypothesis must be one-sided in the downward direction. Therefore $H_0: p = 0.25$ and $H_1: p < 0.25$.

Key Takeaways

Hypotheses are statements about the population proportion, not about the sample. The null hypothesis always contains the equality, and the alternative expresses the direction of suspicion.

Common Mistakes

• Writing $H_1: p \neq 0.25$ when the test is specifically one-sided.
• Stating hypotheses using the sample proportion instead of the population proportion.
• Omitting the inequality direction in $H_1$.

Things to Be Careful About

Use $p$ for the population proportion. Keep $H_0$ as $p = 0.25$ and make sure $H_1$ is $p < 0.25$ because Jeyeraj believes the proportion is less.

Walkthrough

The rejection rule is to reject the manufacturer's claim if fewer than 5 students own a Pumpkin phone. A Type II error is failing to reject the null hypothesis when the alternative is true. Since the true proportion is 10%, the number of owners in 30 students follows $B(30, 0.1)$. The test fails to reject when $X \geq 5$, so we need $P(X \geq 5)$. It is easier to calculate $1 - P(X \leq 4)$ using the binomial probability formula for $X = 0, 1, 2, 3, 4$. Substituting the probabilities gives $0.175491$, which rounds to $0.175$.

Key Takeaways

• A Type II error is the probability of not rejecting $H_0$ when $H_0$ is false.
• The true proportion under the alternative is used to set up the binomial distribution.
• Tail probabilities are often easier computed as $1$ minus the lower tail.

Common Mistakes

• Using $p = 0.25$ instead of the true proportion $p = 0.1$.
• Finding $P(X < 5)$ instead of $P(X \geq 5)$.
• Forgetting to subtract from 1 after summing the lower-tail probabilities.
• Giving an unsupported answer; the mark scheme requires the binomial expression and terms to be shown.

Things to Be Careful About

The rejection region is $X < 5$, so the non-rejection region is $X \geq 5$. Use the true proportion $0.1$ in the binomial model. Show the full binomial sum and the individual terms to earn the method marks. Round the final answer to 3 significant figures.

Walkthrough

We have a random sample of 40 students, of whom 5 own a Pumpkin phone. The sample proportion is $\hat{p} = 5/40 = 0.125$. For a 95% confidence interval for a proportion, we use $\hat{p} \pm z \sqrt{\hat{p}(1-\hat{p})/n}$, where $z = 1.96$. The standard error is approximately $0.05229$. Multiplying by $1.96$ gives the margin of error $0.1025$. Adding and subtracting this margin from $0.125$ gives the interval from $0.0225$ to $0.2275$, which rounds to $0.0225$ to $0.227$ at 3 significant figures.

Key Takeaways

The confidence interval for a proportion is centered at the sample proportion and uses a normal approximation. The critical value for 95% confidence is $1.96$. The interval is written as a pair of lower and upper bounds.

Common Mistakes

• Using $z = 1.645$, which is for a one-sided 95% interval, instead of $1.96$.
• Using $n - 1$ in the denominator of the standard error.
• Forgetting to give both endpoints of the interval.
• Rounding intermediate values too early, which can change the final 3 significant figures.

Things to Be Careful About

The mark scheme requires a $z$-value; any $z$ used must be a standard normal critical value. Only calculating one side of the interval can lose marks, so always show both the lower and upper bounds. Use $n = 40$, not $n-1$, and give the final interval to 3 significant figures.

Walkthrough

We need the probability that $X$ is less than $\frac{1}{2}$. For a continuous random variable, this is the integral of the probability density function over the interval $0 \le x \le \frac{1}{2}$. The PDF is zero outside $[0,1]$, and $\frac{1}{2}$ lies inside that interval, so we integrate $f(x) = 1 + \cos\pi x$ from $0$ to $\frac{1}{2}$.

The integral splits into two pieces: $\int 1 \,dx = x$ and $\int \cos\pi x\,dx = \frac{\sin\pi x}{\pi}$. So the antiderivative is $x + \frac{\sin\pi x}{\pi}$.

We evaluate this antiderivative at the upper limit $\frac{1}{2}$ and the lower limit $0$, subtracting the lower value from the upper. At $x = \frac{1}{2}$, $\sin\frac{\pi}{2} = 1$, so the first term is $\frac{1}{2} + \frac{1}{\pi}$. At $x = 0$, $\sin 0 = 0$, so the lower value is $0$. The result is $\frac{1}{2} + \frac{1}{\pi}$.

Key Takeaways

• For a continuous random variable, the probability of falling in an interval is the integral of its PDF over that interval.
• Integrating $\cos(ax)$ gives $\frac{\sin(ax)}{a}$; the factor $a$ in the denominator is essential.

Common Mistakes

• Forgetting the $\frac{1}{\pi}$ factor when integrating $\cos\pi x$.
• Using the wrong limits (e.g., starting at $-1$); the PDF is zero outside $[0,1]$, so the written working should start at $0$.
• Not showing the intermediate antiderivative $\left[x + \frac{\sin\pi x}{\pi}\right]$. Since the answer is given, a bare final number is not sufficient.

Things to Be Careful About

• This is an "answer given" (AG) question: the working must be convincing and free of errors. The mark scheme awards the marks only if the intermediate step is shown and no errors are seen.
• Evaluate $\sin\frac{\pi}{2}$ and $\sin 0$ correctly; both are standard values.