A Level Mathematics 9709/61 October/November 2025 Past Paper with Answers

Question 1

If $X \sim Po(3)$, what is $P(2 < X < 5)$, correct to 3 significant figures?

1. $0.392$
2. $0.616$
3. $0.224$
4. $0.168$

Question 2

Independent random variables $X \sim Po(3)$ and $Y \sim Po(2)$. What is $P(X + Y > 2)$, correct to 3 significant figures?

1. $0.875$
2. $0.125$
3. $0.960$
4. $0.0067$

Question 3

$T$ is the total of $100$ independent values of $X \sim Po(3)$ and $150$ independent values of $Y \sim Po(2)$. Using a normal approximation with continuity correction, what is $P(T < 560)$, correct to 3 significant figures?

1. $0.0492$
2. $0.951$
3. $0.0511$
4. $0.207$

Question 4 · Answer A

Step 1: Sample mean: $\bar{x} = 29970/60 = 499.5$. Step 2: Unbiased variance: $s^2 = \dfrac{60}{59}\!\left(\dfrac{14970300}{60} - 499.5^2\right) = \dfrac{60}{59}(249505 - 249500.25) = \dfrac{60 \cdot 4.75}{59} = 4.831$. Step 3: Test statistic: $z = \dfrac{499.5 - 500}{\sqrt{4.831/60}} = \dfrac{-0.5}{\sqrt{0.08051}} = \dfrac{-0.5}{0.2838} = -1.762$. Step 4: Two-tailed $5\%$ critical value: $\pm 1.96$. Since $|-1.762| = 1.762 < 1.96$, do not reject $H_0$. There is insufficient evidence at the $5\%$ level that the population mean differs from $500$ g.

Question 5 · Answer A

Step 1: $\lambda = np = 14500 \times 0.0001 = 1.45$. With $n$ large and $p$ small, $X \approx Po(1.45)$. Step 2: $P(X < 4) = P(X \le 3) = \sum_{k=0}^{3} \dfrac{e^{-1.45} (1.45)^k}{k!}$. Step 3: Terms: $P(X=0) = e^{-1.45} = 0.23457$; $P(X=1) = 0.34013$; $P(X=2) = 0.24659$; $P(X=3) = 0.11919$. Step 4: Sum: $0.23457 + 0.34013 + 0.24659 + 0.11919 = 0.94048 \approx 0.940$ (3 s.f.).

Question 6 · Answer A

Step 1: Under the Poisson approximation, $P(X = 0) = e^{-\lambda} = e^{-14500 p}$. The original value is $e^{-1.45} = 0.23457$. Step 2: Set $e^{-14500 p_{\text{new}}} = 2 e^{-1.45}$. Step 3: Take natural logs: $-14500 p_{\text{new}} = \ln 2 + (-1.45) = 0.6931 - 1.45 = -0.7569$. Step 4: $p_{\text{new}} = \dfrac{0.7569}{14500} = 5.220 \times 10^{-5}$ (3 s.f.).

Question 7 · Answer A

Step 1: Width of a CI is proportional to the critical $z$. So $z_\alpha = 1.414 \times z_{90\%} = 1.414 \times 1.645 = 2.326$. Step 2: Find the confidence level corresponding to $z = 2.326$. From standard normal tables, $\Phi(2.326) = 0.990$. Step 3: For an $\alpha\%$ CI, the upper-tail $z$ is $z_{(1-\alpha/100)/2}$. So $1 - \alpha/100 = 2(1 - 0.990) = 0.020$, giving $\alpha = 98$. Step 4: Therefore $\alpha = 98$.

Question 8 · Answer A

Step 1: The $90\%$ CI is contained inside the $98\%$ CI (both centred at $\bar{x}$ with the wider one corresponding to higher confidence). Step 2: $P(90\%\text{ contains } \mu) = 0.90$, $P(98\%\text{ contains } \mu) = 0.98$, and 'inner contains' implies 'outer contains'. Step 3: $P(90\% \mid 98\%) = \dfrac{P(90\% \cap 98\%)}{P(98\%)} = \dfrac{P(90\%)}{P(98\%)}$ (since $90\% \subset 98\%$). Step 4: $= \dfrac{0.90}{0.98} = \dfrac{45}{49} = 0.9184 \approx 0.918$.

Question 9 · Answer A

Step 1: Find the largest critical region $\{X \le r\}$ with $P(X \le r \mid p = 0.2) < 0.025$. Step 2: $P(X \le 2) = 0.8^{40} + 40 \cdot 0.8^{39} \cdot 0.2 + \binom{40}{2} \cdot 0.8^{38} \cdot 0.2^2 = 0.000133 + 0.001329 + 0.006480 = 0.00794$. Step 3: Check $P(X \le 3) = 0.00794 + \binom{40}{3}(0.8)^{37}(0.2)^3 = 0.00794 + 0.02052 = 0.02852 > 0.025$. Step 4: So critical region is $X \le 2$, and $P(\text{Type I}) = 0.00794$ (3 s.f.).

Question 10 · Answer A

Step 1: $L + S \sim N(2.5 + 0.8, \; 0.05 + 0.02) = N(3.3, 0.07)$ (sum of independent normals). Step 2: $\mathrm{sd}(L + S) = \sqrt{0.07} = 0.2646$. Step 3: Standardise: $z = \dfrac{3.55 - 3.3}{0.2646} = \dfrac{0.25}{0.2646} = 0.9449$. Step 4: $P(L + S > 3.55) = 1 - \Phi(0.9449) = 1 - 0.8276 = 0.1724 \approx 0.172$ (3 s.f.).

Question 11 · Answer A

Step 1: Define $D = L - 3 S$. $E(D) = 2.5 - 3 \times 0.8 = 0.1$. Step 2: $\mathrm{Var}(D) = \mathrm{Var}(L) + 3^2 \, \mathrm{Var}(S) = 0.05 + 9 \times 0.02 = 0.05 + 0.18 = 0.23$. Step 3: $P(L < 3 S) = P(D < 0)$. Standardise: $z = \dfrac{0 - 0.1}{\sqrt{0.23}} = \dfrac{-0.1}{0.4796} = -0.2085$. Step 4: $P(D < 0) = \Phi(-0.2085) = 1 - \Phi(0.2085) = 1 - 0.5826 = 0.4174 \approx 0.417$ (3 s.f.).

Question 12 · Answer A

Step 1: Expand: $f(x) = -\dfrac{3}{4}(x^2 - 8x + 15)$. Step 2: $P(X > 4.5) = \int_{4.5}^{5} f(x)\,dx = -\dfrac{3}{4}\!\left[\dfrac{x^3}{3} - 4 x^2 + 15 x\right]_{4.5}^{5}$. Step 3: At $x = 5$: $\dfrac{125}{3} - 100 + 75 = 41.667 - 25 = 16.667$. At $x = 4.5$: $\dfrac{91.125}{3} - 81 + 67.5 = 30.375 - 13.5 = 16.875$. Step 4: Difference: $16.667 - 16.875 = -0.2083$. Multiply by $-3/4$: $-3/4 \times -0.2083 = 0.15625 = 5/32 \approx 0.156$ (3 s.f.).

Question 13 · Answer A

Step 1: $f(x)$ is symmetric about $x = 4$ (midpoint of $[3, 5]$). Step 2: By symmetry, $P(X < 3.5) = P(X > 4.5) = 5/32$. Step 3: $P(3.5 < X < 4.5) = 1 - P(X < 3.5) - P(X > 4.5) = 1 - 2 \times 5/32$. Step 4: $= 1 - 10/32 = 22/32 = 11/16 = 0.6875 \approx 0.688$ (3 s.f.).