May/June 2025 Paper 62 Worked Answers (A-Level Maths 9709 A2)

13 questions · 50 marks · 75 minutes

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1. Question 1b

   5 marksDifference of Two Independent Normals
   Step 1: Define $D = T_1 - T_2$. For independent normals, $D \sim N(125 - 125, \; 50 + 50) = N(0, 100)$. Step 2: 'Difference more than $12$' means $|D| > 12$, i.e. $P(D > 12) + P(D < -12) = 2 P(D > 12)$ by symmetry. Step 3: Standardise: $z = \dfrac{12 - 0}{\sqrt{100}} = 1.2$. From tables $\Phi(1.2) = 0.8849$, so $P(D > 12) = 1 - 0.8849 = 0.1151$. Step 4: $P(|D| > 12) = 2 \times 0.1151 = 0.2302 \approx 0.230$ (3 s.f.).
   Method:

   Form the difference, find its distribution, standardise, and double the upper-tail probability for the two-sided event.
   Examiner tips

   • $\mathrm{Var}(T_1 - T_2) = \mathrm{Var}(T_1) + \mathrm{Var}(T_2)$ for independent variables.
   • 'Difference more than $k$' is two-tailed: $P(|D| > k) = 2(1 - \Phi(k/\sigma_D))$.

2. Question 2

   3 marksRecovering Sum of Squares from Variance
   Step 1: Use $s^2 = \dfrac{1}{n-1}\!\left(\Sigma h^2 - n\bar{h}^2\right)$, so $\Sigma h^2 = (n-1)s^2 + n\bar{h}^2$. Step 2: $n\bar{h}^2 = 100 \times 80.2^2 = 100 \times 6432.04 = 643204$. Step 3: $(n-1)s^2 = 99 \times 15.6 = 1544.4$. Step 4: $\Sigma h^2 = 643204 + 1544.4 = 644748.4 \approx 644700$ (4 s.f.).
   Method:

   Solve the unbiased variance equation for $\Sigma h^2$, then evaluate the two terms.
   Examiner tips

   • $\Sigma h^2 = (n-1)s^2 + n\bar{h}^2$ for unbiased variance.
   • Carefully square $80.2$ before multiplying by $n$.

3. Question 3b

   3 marksEqual Adjacent Poisson Probabilities
   Step 1: Set $P(X=n) = P(X=n+1)$: $\dfrac{e^{-15} 15^n}{n!} = \dfrac{e^{-15} 15^{n+1}}{(n+1)!}$. Step 2: Cancel $e^{-15}$ from both sides and divide both sides by $15^n/n!$: $1 = \dfrac{15}{n+1}$. Step 3: Rearrange: $n + 1 = 15$, so $n = 14$. Step 4: Check: the ratio $\dfrac{P(X=n+1)}{P(X=n)} = \dfrac{\lambda}{n+1}$, equal to $1$ exactly when $n + 1 = \lambda = 15$. ✓
   Method:

   Substitute the Poisson formula, cancel common factors, and solve a one-step linear equation.
   Examiner tips

   • $P(X=n)$ peaks where $n + 1$ matches $\lambda$.
   • Use the ratio recursion to avoid handling factorials directly.

4. Question 4

   4 marksConfidence Interval Inverse Problem
   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 $a = 0.300$.
   Method:

   Equate the width to $0.1066$, solve for $a(1-a)$, then solve the resulting quadratic.
   Examiner tips

   • $90\%$ critical $z = 1.645$, $95\% \to 1.96$, $99\% \to 2.576$.
   • $a(1-a)$ is symmetric about $a = 1/2$, so roots come in pairs.

5. Question 5

   4 marksSmallest Sample Mean for Significance
   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.).
   Method:

   Identify the critical $z$, write the standardisation, set equality, and solve for the smallest $\bar{x}$.
   Examiner tips

   • Set $z$ equal to the critical value to find the boundary.
   • Always include $\sqrt{n}$ when forming the test statistic for a sample mean.

6. Question 6ai

   3 marksPoisson Approximation — Upper Tail
   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 $P(W \ge 4) = 1 - 0.5366 = 0.4634 \approx 0.463$.
   Method:

   Apply the Poisson approximation, compute $P(W \le 3)$, take complement.
   Examiner tips

   • Conditions for Poisson approximation: $n$ large and $p$ small.
   • $P(W \ge 4) = 1 - P(W \le 3)$ — careful with the boundary.

7. Question 6aii

   3 marksSum of Two Independent Approximations
   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.).
   Method:

   Combine into a single Poisson with parameter $7$, then sum probabilities at $k = 0, 1, 2$.
   Examiner tips

   • Sum of independent Poissons is Poisson with summed parameters.
   • 'Less than $3$' means $\le 2$, NOT $\le 3$.

8. Question 6b

   4 marksNormal Approximation to Poisson with Continuity Correction
   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.).
   Method:

   Approximate by normal, apply continuity correction for the strict inequality, standardise, take upper tail.
   Examiner tips

   • Continuity correction: $P(X > k) = P(X \ge k + 1) \to P(N \ge k + 0.5)$.
   • Variance of approximating normal equals the Poisson mean.

9. Question 7b

   3 marksFind Parameter from Expected Value (PDF)
   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.).
   Method:

   Set up the expectation integral, evaluate symbolically in $a$, then solve for $a$.
   Examiner tips

   • Don't confuse $E(X) = a/2$ (uniform) with $E(X) = 3a/4$ (this PDF).
   • Always include the $x$ factor when integrating $xf(x)$.

10. Question 7c

    3 marksMedian from a PDF
    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.).
    Method:

    Solve $\int_0^m f(x)\,dx = 1/2$ for $m$.
    Examiner tips

    • Median is where the CDF equals $1/2$, not the midpoint of the range.
    • For monotone-increasing $f$, the median is to the right of the centre.

11. Question 8a

    5 marksHypothesis Test Using Binomial
    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 $p < 1/6$.
    Method:

    Compute the lower-tail probability under $H_0$ and compare with the significance level.
    Examiner tips

    • Lower-tail test: small values of $X$ favour $H_1$, so compute $P(X \le \text{obs})$.
    • State the conclusion non-definitely ("insufficient evidence").

12. Question 8b

    2 marksType I Error from Critical Region
    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.).
    Method:

    Compute $P(X \in \text{critical region})$ under $H_0$.
    Examiner tips

    • For discrete distributions, the actual significance level is usually less than the nominal $\alpha$.
    • Always evaluate Type I error with $H_0$ parameters.

13. Question 8c

    3 marksType II Error When True Proportion Is Given
    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.).
    Method:

    Compute the probability of being outside the rejection region under the true distribution.
    Examiner tips

    • Type II error always uses the TRUE proportion, not the null.
    • Type I + Type II $\ne 1$; rather, Power $+ \beta = 1$.