| Syllabus | Alternate Universe Cambridge 9709 |
| Subject | Mathematics (9709) A2 |
| Paper | 9709/61/M/J/25 · Paper 6, Variant 1 |
| Also listed as | 9709_s25_qp_61 · 9709_s25_ms_61 |
| Session | May/June 2025 |
| Questions | 11, multiple choice, with answers and explanations |
| Cost | Free |
9709/61/M/J/25 answer key
Preview the first questions
Question 1
It is known that 1% of houses in a certain area have a wind turbine. A random sample of 400 houses is chosen and the number of houses in the sample with a wind turbine is denoted by $X$. Using a suitable approximating distribution, what is $P(X \le 3)$, correct to 3 significant figures?
- $0.433$
- $0.567$
- $0.238$
- $0.195$
Show the answer and explanation
Correct answer: A
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.).
Question 2
The random variable $X$ has distribution $B(8, \tfrac{3}{4})$. A random sample of 100 values of $X$ is taken and the sample mean $\bar{X}$ is computed. Without using a continuity correction, what is $P(\bar{X} \ge 6.2)$, correct to 3 significant figures?
- $0.0512$
- $0.949$
- $0.435$
- $0.500$
Show the answer and explanation
Correct answer: A
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.).
Question 3
A random sample of $60$ daily times $t$ minutes for a bus journey is summarised by $\Sigma t = 2750$ and $\Sigma t^2 = 127000$. Calculate unbiased estimates of the population mean $\mu$ and population variance $\sigma^2$, correct to 3 significant figures.
- $\bar{t} = 45.8, \; s^2 = 16.2$
- $\bar{t} = 45.8, \; s^2 = 16.0$
- $\bar{t} = 45.8, \; s^2 = 2117$
- $\bar{t} = 0.764, \; s^2 = 16.2$
Show the answer and explanation
Correct answer: A
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.).
Worked answers: questions 4 to 11
Why each answer is right. The full questions are waiting in every Paperverse world, with your mark at the end.
Question 4 · Answer A
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 $\mu > 45$.
Question 5 · Answer A
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: $\mathrm{Var}(C) = 1.44$.
Question 6 · Answer A
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 $= 50.4$.
Question 7 · Answer A
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.).
Question 8 · Answer A
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.).
Question 9 · Answer A
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 $\mu > 0$, $\mu = 10$.
Question 10 · Answer A
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.).
Question 11 · Answer A
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: $0.125 \pm 0.1025 = (0.0225, \; 0.2275) \approx (0.0225, \; 0.227)$.