October/November 2025 Paper 63 Worked Answers (A-Level Maths 9709 A2)

13 questions · 50 marks · 75 minutes

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1. Question 1a

   2 marksStrict Two-Sided Poisson Probability
   Step 1: Strict inequality $2 < X < 5$ means $X \in \{3, 4\}$. Step 2: $P(X = 3) = \dfrac{e^{-3} \cdot 3^3}{3!} = e^{-3} \cdot 4.5 = 0.22404$. Step 3: $P(X = 4) = \dfrac{e^{-3} \cdot 3^4}{4!} = e^{-3} \cdot 3.375 = 0.16803$. Step 4: Sum: $0.22404 + 0.16803 = 0.39207 \approx 0.392$ (3 s.f.).
   Method:

   Identify the included integers, compute and sum the relevant Poisson probabilities.
   Examiner tips

   • Strict inequalities exclude boundary integers.
   • Use the formula $P(X = k) = \dfrac{e^{-\lambda} \lambda^k}{k!}$.

2. Question 1b

   3 marksSum of Independent Poissons — Upper Tail
   Step 1: Sum of independent Poissons: $X + Y \sim Po(3 + 2) = Po(5)$. Step 2: $P(X + Y > 2) = 1 - P(X + Y \le 2) = 1 - [P(=0) + P(=1) + P(=2)]$. Step 3: $P(=0) = e^{-5} = 0.0067379$; $P(=1) = e^{-5} \cdot 5 = 0.0336897$; $P(=2) = \dfrac{e^{-5} \cdot 25}{2} = 0.0842243$. Step 4: Sum: $P(\le 2) = 0.1247$. So $P(X + Y > 2) = 1 - 0.1247 = 0.8753 \approx 0.875$.
   Method:

   Combine, then take $1 - P(\le 2)$.
   Examiner tips

   • Sum of independent Poissons is Poisson with summed parameters.
   • Strict 'more than $2$' means $\ge 3$, so use $1 - P(\le 2)$.

3. Question 1c

   4 marksNormal Approximation to Poisson Sum
   Step 1: Sum of independent Poissons: $T \sim Po(100 \cdot 3 + 150 \cdot 2) = Po(600)$. Approximate by $N(600, 600)$ since $\lambda$ is large. Step 2: $P(T < 560) = P(T \le 559) \approx P(N \le 559.5)$ (continuity correction). Step 3: Standardise: $z = \dfrac{559.5 - 600}{\sqrt{600}} = \dfrac{-40.5}{24.4949} = -1.6534$. Step 4: $P(N \le 559.5) = \Phi(-1.6534) = 1 - \Phi(1.6534) = 1 - 0.9508 = 0.0492$ (3 s.f.).
   Method:

   Combine, approximate, apply cc for the strict inequality, standardise, take lower tail.
   Examiner tips

   • $Po(\lambda) \approx N(\lambda, \lambda)$ for large $\lambda$.
   • Continuity correction: $P(T < 560) = P(T \le 559) \to P(N \le 559.5)$.

4. Question 2

   8 marksTwo-Tailed Hypothesis Test for Population Mean
   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.
   Method:

   Compute summary stats, form test stat, compare with two-tailed critical value, conclude.
   Examiner tips

   • Two-tailed $5\%$ critical value is $\pm 1.96$ (not $\pm 1.645$).
   • Always use unbiased variance ($n - 1$ denominator) when estimating from a sample.

5. Question 3a

   3 marksPoisson Approximation — Cumulative Probability
   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.).
   Method:

   Use $Po(np)$, sum four Poisson terms.
   Examiner tips

   • Strict $X < 4$ on integers means $X \le 3$.
   • Sum from $k = 0$ up to $3$ inclusive.

6. Question 3b

   3 marksSolving for New Poisson Parameter
   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.).
   Method:

   Form the doubling equation, take natural logs, isolate $p$.
   Examiner tips

   • Doubling a probability does not double or halve $\lambda$ in a simple way — use logs.
   • $\ln 2 \approx 0.693$.

7. Question 4a

   3 marksConfidence Level from Width Ratio
   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 $\alpha = 98$.
   Method:

   Use width-$z$ proportionality, find $z$, look up confidence level.
   Examiner tips

   • $1.414 \approx \sqrt{2}$ — note that the ratio $\sqrt{2}$ corresponds to a specific $z$ change.
   • $z = 2.326 \to 99\%$ one-tail, $98\%$ two-tail.

8. Question 4b

   1 marksConditional Probability of Nested CIs
   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: $= \dfrac{0.90}{0.98} = \dfrac{45}{49} = 0.9184 \approx 0.918$.
   Method:

   Use the subset relation and the definition of conditional probability.
   Examiner tips

   • When one event is a subset of another, $P(A \mid B) = P(A)/P(B)$.
   • Two CIs from the same sample are nested, not independent.

9. Question 5a

   4 marksType I Error from Largest Critical Region
   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.).
   Method:

   Compute cumulative probabilities at boundaries, find largest critical region, report actual significance.
   Examiner tips

   • Discrete tests achieve actual significance levels less than the nominal level.
   • Always test both boundaries of the candidate critical region.

10. Question 6a

    5 marksSum of Two Independent Normals
    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.).
    Method:

    Form sum distribution, standardise, compute upper tail.
    Examiner tips

    • Variances add for independent sums; standard deviations do not.
    • Note carefully whether the problem gives variance or standard deviation.

11. Question 6b

    5 marksLinear Combination of Normals
    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.).
    Method:

    Define $D = L - 3 S$, compute its mean and variance, standardise, find lower tail at $0$.
    Examiner tips

    • $\mathrm{Var}(aX + bY) = a^2 \mathrm{Var}(X) + b^2 \mathrm{Var}(Y)$ — both coefficients squared.
    • $P(L < kS)$ becomes $P(L - kS < 0)$.

12. Question 7a

    4 marksProbability from a Polynomial PDF
    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.).
    Method:

    Expand and integrate $f(x)$ from $4.5$ to $5$.
    Examiner tips

    • Always expand quadratic factors before integrating.
    • $5/32$ is exact — recognise the fraction.

13. Question 7c

    2 marksSymmetry of PDF for Conditional Probability
    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.).
    Method:

    Recognise symmetry, subtract twice the tail from $1$.
    Examiner tips

    • Quadratic PDF on a symmetric interval is symmetric about the midpoint.
    • Subtract twice the tail probability when both tails are equal.