| Syllabus | Alternate Universe Cambridge 9709 |
| Subject | Mathematics (9709) A2 |
| Paper | 9709/63/M/J/25 · Paper 6, Variant 3 |
| Also listed as | 9709_s25_qp_63 · 9709_s25_ms_63 |
| Session | May/June 2025 |
| Questions | 9, multiple choice, with answers and explanations |
| Cost | Free |
9709/63/M/J/25 answer key
Preview the first questions
Question 1
Customers arrive at a constant rate of $23.4$ per hour. What is the probability that at least $2$ customers arrive in a randomly chosen $1$-minute period, correct to 3 significant figures?
- $0.0589$
- $0.941$
- $0.264$
- $0.323$
Show the answer and explanation
Correct answer: A
Step 1: Convert hourly rate to per-minute: $\lambda = 23.4/60 = 0.39$. So $X \sim Po(0.39)$. Step 2: $P(X \ge 2) = 1 - P(X \le 1) = 1 - [P(X = 0) + P(X = 1)]$. Step 3: $P(X = 0) = e^{-0.39} = 0.67706$ and $P(X = 1) = e^{-0.39} \cdot 0.39 = 0.26405$. Step 4: $P(X \ge 2) = 1 - (0.67706 + 0.26405) = 1 - 0.94111 = 0.0589$ (3 s.f.).
Question 2
The number $X$ of customers in a $1$-hour period is approximated by $N(23.4, 23.4)$. Using continuity correction, what is $P(20 < X < 30)$, correct to 3 significant figures?
- $0.622$
- $0.378$
- $0.673$
- $0.547$
Show the answer and explanation
Correct answer: A
Step 1: $X \approx N(23.4, 23.4)$, so $\mathrm{sd}(X) = \sqrt{23.4} = 4.8374$. Step 2: Apply continuity correction so that $P(20 < X < 30) = P(21 \le X \le 29) \approx P(20.5 \le N \le 29.5)$. Step 3: Standardise: $z_1 = (20.5 - 23.4)/4.8374 = -0.5995$, $z_2 = (29.5 - 23.4)/4.8374 = 1.2610$. Step 4: $\Phi(1.2610) - \Phi(-0.5995) = 0.8964 - (1 - 0.7257) = 0.8964 - 0.2743 = 0.6221 \approx 0.622$ (3 s.f.).
Question 3
A hypothesis test gives test statistic $z = 1.995$ from sample mean $\bar{x} = 10.03$, $\mu_0 = 10$, $n = 50$. What is the population standard deviation $\sigma$, correct to 3 significant figures?
- $\sigma = 0.106$
- $\sigma = 0.0150$
- $\sigma = 0.752$
- $\sigma = 0.0598$
Show the answer and explanation
Correct answer: A
Step 1: Test statistic: $z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$. Step 2: Substitute: $1.995 = \dfrac{10.03 - 10}{\sigma/\sqrt{50}} = \dfrac{0.03 \sqrt{50}}{\sigma}$. Step 3: Rearrange: $\sigma = \dfrac{0.03 \sqrt{50}}{1.995} = \dfrac{0.03 \times 7.0711}{1.995} = \dfrac{0.21213}{1.995}$. Step 4: $\sigma = 0.10634 \approx 0.106$ (3 s.f.).
Worked answers: questions 4 to 9
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Question 4 · Answer A
Step 1: Upper-tail $2.5\%$ critical value: $z_{0.025} = 1.96$. Step 2: Compare: $1.995 > 1.96$, so the test statistic lies in the rejection region. Step 3: Equivalently, $p = 1 - \Phi(1.995) \approx 0.023$, which is less than $0.025$. Step 4: Reject $H_0$. There is sufficient evidence at the $2.5\%$ significance level that the mean length is greater than $10$ cm.
Question 5 · Answer A
Step 1: Sample mean: $\bar{v} = 46350/150 = 309$. Step 2: Compute $\dfrac{(\Sigma v)^2}{n} = \dfrac{46350^2}{150} = \dfrac{2148322500}{150} = 14322150$. Step 3: $\Sigma v^2 - \dfrac{(\Sigma v)^2}{n} = 14410800 - 14322150 = 88650$. Step 4: Unbiased variance: $s^2 = \dfrac{88650}{n - 1} = \dfrac{88650}{149} = 594.97 \approx 595$ (3 s.f.).
Question 6 · Answer A
Step 1: Standard error: $\mathrm{SE} = \sqrt{s^2/n} = \sqrt{595/150} = \sqrt{3.9667} = 1.9917$. Step 2: Critical value: $z_{0.025} = 1.96$ (95%). Step 3: Margin of error: $1.96 \times 1.9917 = 3.904$. Step 4: CI: $309 \pm 3.904 = (305.10, \; 312.90)$, so $(305, \; 313)$ to 3 s.f.
Question 7 · Answer A
Step 1: Total mass $T_n \sim N(70.3 n, \; 5.9^2 n)$ (sum of $n$ independent normals). Step 2: Set up the boundary $P(T_n > 1500) = 0.01$, with critical $z_{0.01} = 2.326$ (upper tail): $\dfrac{1500 - 70.3 n}{5.9 \sqrt{n}} = 2.326$. Step 3: Substitute $u = \sqrt{n}$: $1500 - 70.3 u^2 = 2.326 \cdot 5.9 \, u = 13.7234 \, u$, giving $70.3 u^2 + 13.7234 u - 1500 = 0$. Step 4: Quadratic formula: $u = \dfrac{-13.7234 + \sqrt{188.33 + 421800}}{140.6} = \dfrac{-13.7234 + 649.61}{140.6} = 4.5226$. Then $n = u^2 = 20.45$. Since $P(T > 1500)$ increases with $n$ and we need it $< 0.01$, the maximum integer $n$ is $20$.
Question 8 · Answer A
Step 1: For a lower-tail test, we reject $H_0$ if $P(X \le r) < 0.04$. Find the largest $r$ satisfying this. Step 2: Compute $P(X \le 3) = (0.75)^{35} + 35(0.75)^{34}(0.25) + \binom{35}{2}(0.75)^{33}(0.25)^2 + \binom{35}{3}(0.75)^{32}(0.25)^3 = 0.0000424 + 0.000494 + 0.00280 + 0.01027 = 0.01361$. Step 3: $0.01361 < 0.04$, so $r = 3$ is in the rejection region. Now check $r = 4$: $P(X \le 4) = 0.01361 + \binom{35}{4}(0.75)^{31}(0.25)^4 = 0.01361 + 0.02740 = 0.04101$. Step 4: $0.04101 > 0.04$, so $r = 4$ is NOT in the rejection region. Therefore the largest $r$ is $3$.
Question 9 · Answer A
Step 1: Type II error: retain $H_0$ when $H_0$ is false. Retention region is $X \ge 4$. Step 2: Under true $p = 0.05$, $X \sim B(35, 0.05)$. Compute $P(X \le 3) = \sum_{k=0}^{3} \binom{35}{k}(0.05)^k (0.95)^{35-k}$. Step 3: Term-by-term: $P(X=0) = 0.95^{35} = 0.16608$; $P(X=1) = 35 \cdot 0.95^{34} \cdot 0.05 = 0.30594$; $P(X=2) = \binom{35}{2}(0.95)^{33}(0.05)^2 = 0.27374$; $P(X=3) = \binom{35}{3}(0.95)^{32}(0.05)^3 = 0.15848$. Sum = $0.90425$. Step 4: $P(\text{Type II}) = 1 - 0.90425 = 0.09575 \approx 0.0958$ (3 s.f.).