| Syllabus | Alternate Universe Cambridge 9709 |
| Subject | Mathematics (9709) A2 |
| Paper | 9709/62/O/N/25 · Paper 6, Variant 2 |
| Also listed as | 9709_w25_qp_62 · 9709_w25_ms_62 |
| Session | October/November 2025 |
| Questions | 10, multiple choice, with answers and explanations |
| Cost | Free |
9709/62/O/N/25 answer key
Preview the first questions
Question 1
Computers are donated at a Poisson rate of $2.4$ per week. What is the probability that, in a randomly chosen $4$-week period, more than $6$ but fewer than $9$ computers are donated, correct to 3 significant figures?
- $0.222$
- $0.778$
- $0.121$
- $0.323$
Show the answer and explanation
Correct answer: A
Step 1: Scale the rate: for a $4$-week period, $\lambda = 4 \times 2.4 = 9.6$. Step 2: Strict 'more than $6$ and less than $9$' means $X \in \{7, 8\}$. Step 3: $P(X = 7) = \dfrac{e^{-9.6} \cdot 9.6^7}{7!} = e^{-9.6} \cdot 1490.97 = 0.10098$. $P(X = 8) = \dfrac{e^{-9.6} \cdot 9.6^8}{8!} = e^{-9.6} \cdot 1789.16 = 0.12118$. Step 4: Sum: $0.10098 + 0.12118 = 0.22216 \approx 0.222$ (3 s.f.).
Question 2
Donations follow a Poisson rate of $2.4$ per week. Using a normal approximation with continuity correction, what is $P(X > 50)$ in a $20$-week period, correct to 3 significant figures?
- $0.359$
- $0.641$
- $0.382$
- $0.387$
Show the answer and explanation
Correct answer: A
Step 1: Scale rate: $\lambda = 20 \times 2.4 = 48$. For large $\lambda$, $X \approx N(48, 48)$. Step 2: $P(X > 50) = P(X \ge 51) \approx P(N \ge 50.5)$ (continuity correction). Step 3: Standardise: $z = \dfrac{50.5 - 48}{\sqrt{48}} = \dfrac{2.5}{6.9282} = 0.3608$. Step 4: $P(N \ge 50.5) = 1 - \Phi(0.361) = 1 - 0.6411 = 0.3589 \approx 0.359$ (3 s.f.).
Question 3
$X$ has $N(10, 3^2)$ and $Y$ is independent with $Y \sim Po(4)$. What is the standard deviation of $X + Y$, correct to 3 significant figures?
- $\sqrt{13} \approx 3.61$
- $\sqrt{5} \approx 2.24$
- $\sqrt{7} \approx 2.65$
- $5$
Show the answer and explanation
Correct answer: A
Step 1: $\mathrm{Var}(X) = 3^2 = 9$ (since sd is $3$). $\mathrm{Var}(Y) = 4$ (mean of Poisson is also its variance). Step 2: Independent variables: $\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y) = 9 + 4 = 13$. Step 3: $\mathrm{sd}(X + Y) = \sqrt{13} = 3.6056 \approx 3.61$ (3 s.f.).
Worked answers: questions 4 to 10
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Question 4 · Answer A
Step 1: $\mathrm{Var}(5X) = 5^2 \, \mathrm{Var}(X) = 25 \times 9 = 225$. Step 2: $\mathrm{Var}(Y) = 4$ (Poisson). Step 3: $\mathrm{Var}(5X - Y) = \mathrm{Var}(5X) + \mathrm{Var}(Y) = 225 + 4 = 229$ (independents; subtraction adds variances). Step 4: $\mathrm{sd} = \sqrt{229} = 15.133 \approx 15.1$ (3 s.f.).
Question 5 · Answer A
Step 1: CI midpoint: $\bar{x} = (31.02 + 33.98)/2 = 32.5$. Half-width: $33.98 - 32.5 = 1.48$. Step 2: Half-width = $1.96 \sigma/\sqrt{n}$, so $1.96 \sigma/10 = 1.48$. Step 3: $\sigma = 1.48 \times 10 / 1.96$. Step 4: $\sigma = 14.8/1.96 = 7.551 \approx 7.55$ (3 s.f.).
Question 6 · Answer A
Step 1: Each CI independently contains $\mu$ with probability $0.95$. So all $r$ contain $\mu$ with probability $0.95^r$. Step 2: Need $0.95^r > 0.5$. Take logs (and remember $\ln 0.95 < 0$, so the inequality flips): $r \ln 0.95 > \ln 0.5 \Rightarrow r < \dfrac{\ln 0.5}{\ln 0.95}$. Step 3: $\dfrac{\ln 0.5}{\ln 0.95} = \dfrac{-0.6931}{-0.0513} = 13.513$. Step 4: So $r < 13.513$. The largest integer satisfying this is $r = 13$. Check: $0.95^{13} = 0.5133 > 0.5$ ✓ and $0.95^{14} = 0.4877 < 0.5$ ✓.
Question 7 · Answer A
Step 1: Under $H_0$, $X \sim B(40, 0.18)$. Lower-tailed test, so compute $P(X \le 3 \mid H_0)$. Step 2: Term-by-term: $P(X = 0) = 0.82^{40} = 0.000357$; $P(X = 1) = 40 \cdot 0.82^{39} \cdot 0.18 = 0.003134$; $P(X = 2) = \binom{40}{2} \cdot 0.82^{38} \cdot 0.18^2 = 0.013414$; $P(X = 3) = \binom{40}{3} \cdot 0.82^{37} \cdot 0.18^3 = 0.037298$. Step 3: Sum: $P(X \le 3) = 0.000357 + 0.003134 + 0.013414 + 0.037298 = 0.05420$. Step 4: Compare with $\alpha = 0.05$: $0.0542 > 0.05$, so do not reject $H_0$. Insufficient evidence at $5\%$ to support the owner's claim that $p < 0.18$.
Question 8 · Answer A
Step 1: $E(X) = \dfrac{3}{4}\!\int_0^2 (2x^3 - x^4)\,dx = \dfrac{3}{4}\!\left[\dfrac{x^4}{2} - \dfrac{x^5}{5}\right]_0^2 = \dfrac{3}{4}\!\left(8 - \dfrac{32}{5}\right) = \dfrac{3}{4} \cdot \dfrac{8}{5} = \dfrac{6}{5} = 1.2$. Step 2: $P(X \le 1.2) = \dfrac{3}{4}\!\int_0^{1.2} (2x^2 - x^3)\,dx = \dfrac{3}{4}\!\left[\dfrac{2x^3}{3} - \dfrac{x^4}{4}\right]_0^{1.2}$. Step 3: Evaluate at $1.2$: $\dfrac{2 \cdot 1.728}{3} - \dfrac{1.2^4}{4} = 1.152 - \dfrac{2.0736}{4} = 1.152 - 0.5184 = 0.6336$. So $P(X \le 1.2) = \dfrac{3}{4} \cdot 0.6336 = 0.4752$. Step 4: Therefore $P(E(X) \le X \le m) = P(X \le m) - P(X \le E(X)) = 0.5 - 0.4752 = 0.0248$.
Question 9 · Answer A
Step 1: Hypotheses: $H_0: \mu = 736$, $H_1: \mu < 736$. Step 2: Test statistic: $z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{725 - 736}{26/\sqrt{35}} = \dfrac{-11}{4.3949} = -2.503$. Step 3: Critical value at $2\%$ lower-tail: $-2.054$ (since $\Phi(-2.054) = 0.02$). Step 4: Compare: $-2.503 < -2.054$, so the test statistic is in the rejection region. Reject $H_0$. There is sufficient evidence at the $2\%$ level that the mean weekly profit has decreased.
Question 10 · Answer A
Step 1: Find critical $\bar{x}$: $\dfrac{a - 736}{26/\sqrt{35}} = -2.054$, so $a = 736 - 2.054 \cdot 4.3949 = 736 - 9.027 = 726.973$. Step 2: Reject $H_0$ if $\bar{x} < 726.973$; retain $H_0$ if $\bar{x} \ge 726.973$. Step 3: Type II error: $P(\text{retain } H_0 \mid \mu = 718) = P(\bar{x} \ge 726.973 \mid \mu = 718)$. Standardise: $z = \dfrac{726.973 - 718}{26/\sqrt{35}} = \dfrac{8.973}{4.3949} = 2.042$. Step 4: $P(\text{Type II}) = 1 - \Phi(2.042) = 1 - 0.9794 = 0.0206$ (3 s.f.).