A-Level Mathematics October/November 2024 Q5(c): Use approximating distributions to find the probability that the total number of lemon…

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  Let $L$ be the total number of lemon sweets in 10 bags, and $O$ be the total number of orange sweets in 10 bags. We assume the number of sweets of each flavour in a bag follows a Poisson distribution, and from previous parts of the question, the mean for a single bag is 3.7 for lemon and 2.6 for orange.

  For 10 bags, the total number of sweets of each flavour also follows a Poisson distribution: $L_{total} \sim \text{Po}(10 \times 3.7) \Rightarrow L_{total} \sim \text{Po}(37)$ $O_{total} \sim \text{Po}(10 \times 2.6) \Rightarrow O_{total} \sim \text{Po}(26)$

  Since the parameters ($\lambda$) are both greater than 15, we can use a Normal approximation for each distribution: $L \sim \text{N}(37, 37)$ $O \sim \text{N}(26, 26)$

  We want to find the probability that the total number of lemon sweets is less than the total number of orange sweets, which is $P(L < O)$. This is equivalent to $P(O - L > 0)$.

  Let the random variable $X = O - L$. We find the distribution of $X$. $E(X) = E(O) - E(L) = 26 - 37 = -11$ $\text{Var}(X) = \text{Var}(O) + \text{Var}(L) = 26 + 37 = 63$

  So, $X \sim \text{N}(-11, 63)$.

  Now we find $P(X > 0)$ by standardising: $P(X > 0) = P\left(Z > \frac{0 - (-11)}{\sqrt{63}}\right)$ $= P\left(Z > \frac{11}{\sqrt{63}}\right)$ $= P(Z > 1.3856...)$

  Using standard Normal distribution tables: $= 1 - \Phi(1.386)$ $= 1 - 0.9171$ $= 0.0829$

  The probability that the total number of lemon sweets is less than the total number of orange sweets is 0.0829 (3 s.f.).

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  How the marks are awarded

  • B1 — Stating the correct Normal approximations for the total number of sweets in 10 bags, L ~ N(37, 37) and O ~ N(26, 26), seen or implied by later working.
  • B1 — Stating the correct distribution for the difference O - L, specifically N(-11, 63), seen or implied by the mean and variance values.
  • M1 — Correctly calculating the variance for the difference of the two independent variables by adding their individual variances: Var(X) = 26 + 37 = 63.
  • M1 — Correctly standardising to find a Z-value, using their mean and variance. The calculation (0 - (-11)) / √63 earns this mark. A continuity correction is not required.
  • M1 — Finding the correct probability area consistent with their working. For a positive Z-score from P(X > 0), this involves calculating 1 - Φ(Z).
  • A1 — Obtaining the final correct answer of 0.0829 or 0.0828. (The alternative answer 0.0737 or 0.0736 from using a continuity correction would also be accepted).

  Common mistakes

  • Incorrectly calculating the variance of the difference, e.g., Var(O - L) = Var(O) - Var(L) = 26 - 37 = -11. Variances of independent variables are always added.
  • Using the parameters for a single bag (e.g., N(3.7, 3.7)) instead of the total for 10 bags (N(37, 37)).
  • Applying a continuity correction incorrectly. For P(O - L > 0), which is P(X > 0), the correct correction would be to find P(X ≥ 1) and then standardise 0.5. A common error is to standardise -0.5 or 0.5 without justification.
  • Finding the wrong area from the Normal distribution tables, for example calculating Φ(1.386) instead of 1 - Φ(1.386).

  Examiner tip: Remember that for any two independent random variables X and Y, the variance of their sum or difference is always the sum of their variances: Var(X ± Y) = Var(X) + Var(Y).

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