A-Level Mathematics May/June 2024 Q2: The random variable X has the distribution N(31.2, 10.4²). Two independent random value…

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  Let the random variable $X$ have the distribution $X \sim N(31.2, 10.4^2)$. We are given two independent random values, $X_1$ and $X_2$. We need to find the probability $P(X_1 > 3X_2)$.

  This inequality can be rewritten as $P(X_1 - 3X_2 > 0)$.

  Let a new random variable $Y = X_1 - 3X_2$. Since $X_1$ and $X_2$ are independent and normally distributed, $Y$ is also normally distributed.

  We need to find the mean and variance of $Y$.

  Mean of $Y$: $E(Y) = E(X_1 - 3X_2) = E(X_1) - 3E(X_2)$ $E(Y) = 31.2 - 3(31.2) = -62.4$

  Variance of $Y$: Since $X_1$ and $X_2$ are independent: $Var(Y) = Var(X_1 - 3X_2) = Var(X_1) + (-3)^2Var(X_2)$ $Var(Y) = Var(X_1) + 9Var(X_2)$ $Var(Y) = 10.4^2 + 9(10.4^2) = 10(10.4^2)$ $Var(Y) = 10 \times 108.16 = 1081.6$

  So, the distribution of $Y$ is $Y \sim N(-62.4, 1081.6)$.

  We want to find $P(Y > 0)$. We standardise this value: $Z = \frac{Y - \mu_Y}{\sigma_Y}$ $P(Y > 0) = P\left(Z > \frac{0 - (-62.4)}{\sqrt{1081.6}}\right)$ $P(Y > 0) = P\left(Z > \frac{62.4}{32.88768...}\right)$ $P(Y > 0) = P(Z > 1.8973...)$

  Using the standard normal distribution tables: $P(Z > 1.897) = 1 - P(Z < 1.897)$ $P(Z > 1.897) = 1 - \Phi(1.897)$ $P(Z > 1.897) = 1 - 0.9711$ $P(Z > 1.897) = 0.0289$

  Therefore, $P(X_1 > 3X_2) = 0.0289$ (to 3 s.f.).

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  How the marks are awarded

  • B1 — Correctly calculating the expectation of the combined variable: E(X₁ - 3X₂) = 31.2 - 3(31.2) = -62.4.
  • B1 — Correctly calculating the variance of the combined variable: Var(X₁ - 3X₂) = 10.4² + 9(10.4²) = 1081.6. This includes squaring the coefficient '3' and adding the variances.
  • M1 — Correctly standardising the value 0 using the calculated mean and variance: (0 - (-62.4)) / √1081.6.
  • M1 — Finding the correct upper tail probability for their Z-value. In the model answer, this is finding 1 - Φ(1.897).
  • A1 — Obtaining the final correct answer of 0.0289, correctly rounded to 3 significant figures.

  Common mistakes

  • Incorrectly calculating the variance by subtracting instead of adding, e.g., Var(X₁) - 9Var(X₂).
  • Forgetting to square the coefficient '3' when calculating the variance, e.g., Var(X₁) + 3Var(X₂).
  • Finding the wrong area from the normal distribution table, for example calculating P(Z < 1.897) instead of P(Z > 1.897), resulting in an answer of 0.9711.
  • Rearranging the inequality incorrectly, or attempting to standardise X₁ and X₂ separately before combining them.

  Examiner tip: Master the rules for combining independent random variables: E(aX ± bY) = aE(X) ± bE(Y) and Var(aX ± bY) = a²Var(X) + b²Var(Y).

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