A Level Mathematics 9709/62 May/June 2025 Past Paper with Answers

Question 1

Times $T$ taken by students to complete a test are independently $N(125, 50)$. Two students are chosen at random. What is the probability that the difference between their times is more than $12$ minutes, correct to 3 significant figures?

1. $0.230$
2. $0.115$
3. $0.885$
4. $0.420$

Question 2

From a sample of $n = 100$ plants, $\bar{h} = 80.2$ and an unbiased estimate of $\sigma^2$ is $s^2 = 15.6$. What is $\Sigma h^2$, correct to 4 significant figures?

1. $644700$
2. $643200$
3. $1544$
4. $644800$

Question 3

If $X \sim Po(15)$, what is the value of $n$ for which $P(X = n) = P(X = n + 1)$?

1. $n = 14$
2. $n = 15$
3. $n = 16$
4. $n = 1$

Question 4 · Answer A

Step 1: Width = $2 z \sqrt{\dfrac{a(1-a)}{n}}$ with $z = 1.645$ (90%), so $2 \times 1.645 \sqrt{\dfrac{a(1-a)}{200}} = 0.1066$. Step 2: Divide: $\sqrt{\dfrac{a(1-a)}{200}} = \dfrac{0.1066}{3.29} = 0.03240$. Step 3: Square: $\dfrac{a(1-a)}{200} = 0.001050$, so $a(1-a) = 0.20997$. Step 4: Solve $a^2 - a + 0.20997 = 0$ via the quadratic formula: $a = \dfrac{1 \pm \sqrt{1 - 0.83988}}{2} = \dfrac{1 \pm 0.4}{2}$, giving $a = 0.700$ or $a = 0.300$.

Question 5 · Answer A

Step 1: Critical region: reject $H_0$ when $z \ge 1.645$ (one-tailed at $5\%$). Step 2: Test statistic: $z = \dfrac{\bar{x} - 10.5}{3.8/\sqrt{10}}$. Step 3: Set $z = 1.645$ and solve: $\bar{x} = 10.5 + 1.645 \cdot \dfrac{3.8}{\sqrt{10}} = 10.5 + 1.645 \cdot 1.2017 = 10.5 + 1.9768$. Step 4: $\bar{x}_{\min} = 12.477 \approx 12.5$ (3 s.f.).

Question 6 · Answer A

Step 1: $\lambda = np = 700 \times 0.005 = 3.5$. Since $n$ is large and $p$ small, $W \approx Po(3.5)$. Step 2: $P(W \ge 4) = 1 - P(W \le 3)$. Step 3: $P(W \le 3) = e^{-3.5}\left(1 + 3.5 + \dfrac{3.5^2}{2!} + \dfrac{3.5^3}{3!}\right) = e^{-3.5}(1 + 3.5 + 6.125 + 7.1458) = e^{-3.5} \times 17.7708$. Step 4: $e^{-3.5} = 0.030197$, so $P(W \le 3) = 0.030197 \times 17.7708 = 0.5366$. Hence $P(W \ge 4) = 1 - 0.5366 = 0.4634 \approx 0.463$.

Question 7 · Answer A

Step 1: Each $W \approx Po(3.5)$, and the sum of two independent Poissons is Poisson: $W_1 + W_2 \approx Po(7)$. Step 2: $P(W_1 + W_2 < 3) = P(W_1 + W_2 \le 2) = P(=0) + P(=1) + P(=2)$. Step 3: $e^{-7}\left(1 + 7 + \dfrac{7^2}{2!}\right) = e^{-7}(1 + 7 + 24.5) = e^{-7} \times 32.5$. Step 4: $e^{-7} = 0.0009119$, so probability $= 0.0009119 \times 32.5 = 0.02964 \approx 0.0296$ (3 s.f.).

Question 8 · Answer A

Step 1: For large $\lambda$, $X \approx N(\lambda, \lambda) = N(200, 200)$. Step 2: $P(X > 205) = P(X \ge 206) \approx P(N \ge 205.5)$ (continuity correction). Step 3: Standardise: $z = \dfrac{205.5 - 200}{\sqrt{200}} = \dfrac{5.5}{14.142} = 0.3889$. Step 4: $P(X > 205) = 1 - \Phi(0.389) = 1 - 0.6515 = 0.3485 \approx 0.349$ (3 s.f.).

Question 9 · Answer A

Step 1: $E(X) = \int_0^a x \cdot \dfrac{3x^2}{a^3}\,dx = \dfrac{3}{a^3}\int_0^a x^3\,dx$. Step 2: Evaluate: $\dfrac{3}{a^3}\!\left[\dfrac{x^4}{4}\right]_0^a = \dfrac{3}{a^3} \cdot \dfrac{a^4}{4} = \dfrac{3a}{4}$. Step 3: Set equal to $1$: $\dfrac{3a}{4} = 1$. Step 4: Solve: $a = \dfrac{4}{3} = 1.333\ldots \approx 1.33$ (3 s.f.).

Question 10 · Answer A

Step 1: Median: $\int_0^m f(x)\,dx = 0.5$, i.e. $\int_0^m \dfrac{3x^2}{a^3}\,dx = 0.5$. Step 2: With $a = 4/3$, $a^3 = 64/27$, and the integrand becomes $\dfrac{3x^2}{64/27} = \dfrac{81 x^2}{64}$. Step 3: Integrate: $\dfrac{81}{64} \cdot \dfrac{m^3}{3} = \dfrac{27 m^3}{64} = 0.5$, so $m^3 = \dfrac{32}{27}$. Step 4: $m = \sqrt[3]{32/27} = \sqrt[3]{1.1852} = 1.0583 \approx 1.06$ (3 s.f.).

Question 11 · Answer A

Step 1: Test stat: under $H_0$, $X \sim B(30, 1/6)$. For a lower-tail test, compute $P(X \le 2 \mid H_0)$. Step 2: $P(X = 0) = (5/6)^{30} = 0.00421$; $P(X = 1) = 30(5/6)^{29}(1/6) = 0.02528$; $P(X = 2) = \binom{30}{2}(5/6)^{28}(1/6)^2 = 0.07330$. Step 3: Sum: $P(X \le 2) = 0.00421 + 0.02528 + 0.07330 = 0.1028 \approx 0.103$. Step 4: Compare with significance level: $0.103 > 0.05$, so the result is not in the rejection region. Do not reject $H_0$ — insufficient evidence that $p < 1/6$.

Question 12 · Answer A

Step 1: Type I error = $P(\text{reject } H_0 \mid H_0 \text{ true}) = P(X \le 1 \mid X \sim B(30, 1/6))$. Step 2: $P(X = 0) = (5/6)^{30} = 0.00421$. Step 3: $P(X = 1) = 30(5/6)^{29}(1/6) = 0.02528$. Step 4: Sum: $P(\text{Type I}) = 0.00421 + 0.02528 = 0.02949 \approx 0.0295$ (3 s.f.).

Question 13 · Answer A

Step 1: Type II error: $H_0$ retained when $H_0$ false. Retention region is $X \ge 2$, so $P(\text{Type II}) = P(X \ge 2 \mid p = 0.02)$. Step 2: Under $p = 0.02$ on $n = 30$: $P(X = 0) = 0.98^{30} = 0.54548$ and $P(X = 1) = 30 \cdot 0.98^{29} \cdot 0.02 = 0.33397$. Step 3: $P(X \le 1) = 0.54548 + 0.33397 = 0.87945$. Step 4: $P(\text{Type II}) = 1 - 0.87945 = 0.12055 \approx 0.121$ (3 s.f.).