May/June 2025 Paper 61 Worked Answers (A-Level Maths 9709 A2)

11 questions · 50 marks · 75 minutes

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1. Question 1

   3 marksPoisson Approximation to Binomial
   Step 1: $X \sim B(400, 0.01)$. Since $n$ is large and $p$ small (with $np = 4$), use the Poisson approximation $X \approx Po(4)$. Step 2: $P(X \le 3) = e^{-4}\!\left(\dfrac{4^0}{0!} + \dfrac{4^1}{1!} + \dfrac{4^2}{2!} + \dfrac{4^3}{3!}\right) = e^{-4}\!\left(1 + 4 + 8 + \dfrac{32}{3}\right)$. Step 3: Sum inside the bracket: $1 + 4 + 8 + 10.667 = 23.667$. Step 4: $e^{-4} = 0.018316$, so $P(X \le 3) = 0.018316 \times 23.667 = 0.433$ (3 s.f.).
   Method:

   Identify the Poisson approximation conditions, compute $\lambda = np$, then sum the discrete Poisson probabilities up to the required value.
   Examiner tips

   • Conditions for Poisson approximation: $n$ large and $p$ small, e.g. $n > 50$ and $np < 5$.
   • Sum from $k = 0$ up to and including the upper bound for $P(X \le 3)$.

2. Question 2a

   6 marksDistribution of Sample Mean
   Step 1: For $X \sim B(8, \tfrac{3}{4})$: $E(X) = np = 6$ and $\mathrm{Var}(X) = np(1-p) = 8 \cdot \tfrac{3}{4} \cdot \tfrac{1}{4} = 1.5$. Step 2: By the Central Limit Theorem, $\bar{X} \sim N\!\left(6, \dfrac{1.5}{100}\right) = N(6, 0.015)$, so $\mathrm{sd}(\bar{X}) = \sqrt{0.015} = 0.1225$. Step 3: Standardise: $z = \dfrac{6.2 - 6}{0.1225} = \dfrac{0.2}{0.1225} = 1.633$. Step 4: $P(\bar{X} \ge 6.2) = 1 - \Phi(1.633) = 1 - 0.9488 = 0.0512$ (3 s.f.).
   Method:

   Find $\mu$ and $\sigma^2$ for $X$, then use $\bar{X} \sim N(\mu, \sigma^2/n)$ via the CLT and standardise to get the upper-tail probability.
   Examiner tips

   • $\mathrm{Var}(\bar{X}) = \sigma^2/n$, not $\sigma^2$
   • For an upper tail probability, take $1 - \Phi(z)$

3. Question 3a

   3 marksUnbiased Estimates from Summary Data
   Step 1: Sample mean: $\bar{t} = \dfrac{\Sigma t}{n} = \dfrac{2750}{60} = 45.833 \approx 45.8$. Step 2: Unbiased variance: $s^2 = \dfrac{1}{n-1}\!\left(\Sigma t^2 - \dfrac{(\Sigma t)^2}{n}\right) = \dfrac{1}{59}\!\left(127000 - \dfrac{2750^2}{60}\right)$. Step 3: $\dfrac{2750^2}{60} = \dfrac{7562500}{60} = 126041.67$, so the bracket is $127000 - 126041.67 = 958.33$. Step 4: $s^2 = \dfrac{958.33}{59} = 16.243 \approx 16.2$ (3 s.f.).
   Method:

   Compute $\bar{t} = \Sigma t/n$, then apply the unbiased variance formula $s^2 = \dfrac{1}{n-1}\!\left(\Sigma t^2 - \dfrac{(\Sigma t)^2}{n}\right)$.
   Examiner tips

   • Always divide by $n - 1$ for an unbiased variance estimate.
   • Compute the bracket $\Sigma t^2 - (\Sigma t)^2/n$ first, then divide once.

4. Question 3b

   5 marksHypothesis Test for Population Mean
   Step 1: Hypotheses: $H_0: \mu = 45$ vs $H_1: \mu > 45$ (one-tailed, upper). Step 2: Test statistic: $z = \dfrac{\bar{t} - \mu_0}{\sqrt{s^2/n}} = \dfrac{45.833 - 45}{\sqrt{16.243/60}} = \dfrac{0.833}{\sqrt{0.2707}} = \dfrac{0.833}{0.5203} = 1.602$. Step 3: Critical value at $5\%$ for upper tail: $z_{\text{crit}} = 1.645$. Step 4: Compare: $1.602 < 1.645$, so the test statistic is not in the rejection region. Do not reject $H_0$. Step 5: Conclusion: there is insufficient evidence at the $5\%$ level to support the passenger's belief that $\mu > 45$.
   Method:

   State hypotheses, compute the $z$ test statistic, compare with the appropriate critical value, and conclude in context.
   Examiner tips

   • For an upper one-tailed test at $5\%$, the critical value is $1.645$.
   • State the conclusion in context — non-definite language ('insufficient evidence').

5. Question 4a

   3 marksMean and Variance of a Linear Function
   Step 1: $C = 0.4T$ is a linear function of $T$ with $a = 0.4$ and $b = 0$. Step 2: $E(C) = E(0.4T) = 0.4 \, E(T) = 0.4 \times 15 = 6$ dollars. Step 3: $\mathrm{Var}(C) = \mathrm{Var}(0.4T) = 0.4^2 \, \mathrm{Var}(T) = 0.16 \times 9$. Step 4: $\mathrm{Var}(C) = 1.44$.
   Method:

   Apply $E(aT+b) = aE(T)+b$ and $\mathrm{Var}(aT+b) = a^2 \mathrm{Var}(T)$ to the linear cost function.
   Examiner tips

   • Variance of $aT$ scales by $a^2$, not by $a$.
   • Adding a constant does not change variance.

6. Question 4b

   3 marksSum of Independent Variables
   Step 1: $T_{35} = \sum_{i=1}^{35} C_i$ where the $C_i$ are independent and identically distributed with $E(C_i) = 6$ and $\mathrm{Var}(C_i) = 1.44$. Step 2: Expectation: $E(T_{35}) = 35 \, E(C_i) = 35 \times 6 = 210$. Step 3: Variance (of a sum of independents): $\mathrm{Var}(T_{35}) = 35 \, \mathrm{Var}(C_i) = 35 \times 1.44 = 50.4$. Step 4: So mean $= 210$ dollars, variance $= 50.4$.
   Method:

   Apply linearity of expectation and variance-of-independent-sum to compute total mean and total variance.
   Examiner tips

   • $\mathrm{Var}(X_1 + X_2 + \ldots + X_n) = n\,\mathrm{Var}(X)$ when the $X_i$ are i.i.d.
   • Be careful: $\mathrm{Var}(nX) = n^2 \mathrm{Var}(X)$ — different scenario.

7. Question 5ai

   2 marksSum of Two Poisson Distributions
   Step 1: For independent Poissons, $W + X \sim Po(\lambda_W + \lambda_X) = Po(3.5)$. Step 2: Required probability = $P(Y = 3) + P(Y = 4) + P(Y = 5)$ with $Y \sim Po(3.5)$. Step 3: Compute each term: $P(Y=3) = \dfrac{e^{-3.5} \cdot 3.5^3}{3!} = 0.21580$; $P(Y=4) = \dfrac{e^{-3.5} \cdot 3.5^4}{4!} = 0.18881$; $P(Y=5) = \dfrac{e^{-3.5} \cdot 3.5^5}{5!} = 0.13217$. Step 4: Sum: $0.21580 + 0.18881 + 0.13217 = 0.5368 \approx 0.537$ (3 s.f.).
   Method:

   Combine $W$ and $X$ into a single Poisson, then sum probabilities at $k = 3, 4, 5$.
   Examiner tips

   • Sum of independent Poissons is Poisson with parameter equal to the sum of parameters.
   • $P(a \le Y \le b) = \sum_{k=a}^{b} P(Y = k)$ — list all values.

8. Question 5aii

   6 marksNormal Approximation to Poisson
   Step 1: Sum of independent Poissons: $S \sim Po(100 \cdot 1.2 + 200 \cdot 2.3) = Po(580)$. With $\lambda$ large, approximate by $N(580, 580)$. Step 2: $P(S > 600) = P(S \ge 601)$. With continuity correction, this is $P(N \ge 600.5)$. Step 3: Standardise: $z = \dfrac{600.5 - 580}{\sqrt{580}} = \dfrac{20.5}{24.083} = 0.851$. Step 4: $P(N \ge 600.5) = 1 - \Phi(0.851) = 1 - 0.8027 = 0.1973 \approx 0.197$ (3 s.f.).
   Method:

   Combine the Poissons, approximate by $N(\lambda, \lambda)$, apply continuity correction for the strict inequality, standardise and use the upper tail.
   Examiner tips

   • For large $\lambda$, $Po(\lambda) \approx N(\lambda, \lambda)$.
   • Continuity correction: $P(X > 600) = P(X \ge 601) \approx P(N \ge 600.5)$.

9. Question 5b

   3 marksSolving for Poisson Parameter
   Step 1: Substitute the Poisson p.m.f.: $\dfrac{5}{2} \cdot \dfrac{e^{-\mu}\mu^3}{3!} + \dfrac{e^{-\mu}\mu^4}{4!} = \dfrac{e^{-\mu}\mu^5}{5!}$. Step 2: Multiply through by $\dfrac{5!}{e^{-\mu}\mu^3}$ to remove $e^{-\mu}$ and $\mu^3$: $\dfrac{5}{2} \cdot \dfrac{120}{6} + \dfrac{120}{24}\mu = \dfrac{120}{120}\mu^2$, which simplifies to $50 + 5\mu = \mu^2$. Step 3: Rearrange: $\mu^2 - 5\mu - 50 = 0$. Step 4: Factorise: $(\mu - 10)(\mu + 5) = 0$, so $\mu = 10$ or $\mu = -5$. Since $\mu > 0$, $\mu = 10$.
   Method:

   Substitute the Poisson formula, cancel common factors, and solve the quadratic for $\mu$.
   Examiner tips

   • Factorise out $e^{-\mu}\mu^k$ for the smallest $k$ to simplify the equation.
   • Reject the negative root because $\mu$ is a Poisson parameter.

10. Question 6b

    3 marksType II Error Probability (Binomial)
    Step 1: Type II error occurs when $H_0$ is not rejected even though the true $p \ne 0.25$. Here, with rejection region $X < 5$, $H_0$ is not rejected when $X \ge 5$. Step 2: Under the true proportion $p = 0.10$, $X \sim B(30, 0.1)$, so $\beta = P(X \ge 5)$. Step 3: Compute $P(X \le 4)$: $\binom{30}{0}(0.9)^{30} = 0.04239$; $\binom{30}{1}(0.1)(0.9)^{29} = 0.14130$; $\binom{30}{2}(0.1)^{2}(0.9)^{28} = 0.22765$; $\binom{30}{3}(0.1)^{3}(0.9)^{27} = 0.23609$; $\binom{30}{4}(0.1)^{4}(0.9)^{26} = 0.17707$. Sum $= 0.82451$. Step 4: $\beta = 1 - 0.82451 = 0.1755 \approx 0.175$ (3 s.f.).
    Method:

    Compute the probability of NOT rejecting $H_0$ under the true proportion using the binomial cumulative distribution.
    Examiner tips

    • Type II error uses the TRUE distribution, not the null distribution.
    • Read the rejection region carefully: $X < 5$ means reject for $X \le 4$, retain for $X \ge 5$.

11. Question 6c

    3 marksConfidence Interval for a Proportion
    Step 1: Sample proportion: $\hat{p} = \dfrac{5}{40} = \dfrac{1}{8} = 0.125$. Step 2: Standard error: $\mathrm{SE} = \sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\dfrac{0.125 \times 0.875}{40}} = \sqrt{0.002734} = 0.05229$. Step 3: Margin of error: $1.96 \times 0.05229 = 0.10250$. Step 4: CI: $0.125 \pm 0.1025 = (0.0225, \; 0.2275) \approx (0.0225, \; 0.227)$.
    Method:

    Compute $\hat{p}$, then form the symmetric interval $\hat{p} \pm 1.96\sqrt{\hat{p}(1-\hat{p})/n}$.
    Examiner tips

    • $95\%$ CI uses $z = 1.96$, not $1.645$ ($90\%$) or $2.576$ ($99\%$).
    • Always divide $\hat{p}(1 - \hat{p})$ by $n$ before taking the square root.