A-Level Mathematics May/June 2025 Q4(c): 80 competitors who took part in this Saturday's event are selected at random. Use a sui…

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  Let $X$ be the number of competitors with times less than 36.0 minutes. From the context of the problem, the probability $p$ of a single competitor having a time less than 36.0 minutes is 0.6. The number of competitors selected is $n=80$. Thus, $X$ follows a binomial distribution: $X \sim B(80, 0.6)$.

  We need to find the probability $P(X > 50)$. Since $n$ is large, we can use a normal approximation. The conditions for approximation are met as $np = 80 \times 0.6 = 48 > 5$ and $n(1-p) = 80 \times 0.4 = 32 > 5$.

  Calculate the mean and variance for the approximation: Mean, $\mu = np = 80 \times 0.6 = 48$ Variance, $\sigma^2 = np(1-p) = 80 \times 0.6 \times 0.4 = 19.2$

  So, $X$ is approximated by the normal distribution $Y \sim N(48, 19.2)$.

  We need to find $P(X > 50)$. Using a continuity correction, this becomes $P(Y > 50.5)$.

  Standardise the value: $Z = \frac{Y - \mu}{\sigma} = \frac{50.5 - 48}{\sqrt{19.2}}$ $Z = \frac{2.5}{4.38178...} \approx 0.5705$

  Now, find the probability from the standard normal distribution: $P(Z > 0.5705) = 1 - P(Z < 0.5705)$ $= 1 - \Phi(0.5705)$ From the tables, $\Phi(0.5705) \approx 0.7158$. $P(Z > 0.5705) = 1 - 0.7158 = 0.2842$

  The probability that more than 50 of these competitors had times less than 36.0 minutes is approximately 0.284 (to 3 s.f.).

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  How the marks are awarded

  • B1 — Correctly stating or using the mean $\mu = 48$ and variance $\sigma^2 = 19.2$ for the normal approximation.
  • M1 — Substituting the calculated mean (48) and variance (19.2) into the standardisation formula, e.g., $\frac{x - 48}{\sqrt{19.2}}$.
  • M1 — Applying the correct continuity correction for $P(X > 50)$ by using the value 50.5 in the standardisation formula.
  • M1 — Finding the correct probability area for a 'greater than' question with a positive z-value, i.e., calculating $1 - \Phi(z)$. This results in a probability less than 0.5.
  • A1 — Obtaining the final correct answer, which rounds to 0.284.

  Common mistakes

  • Forgetting to apply a continuity correction, or using the wrong one (e.g., using 50 or 49.5 instead of 50.5).
  • Using the variance (19.2) instead of the standard deviation ($\sqrt{19.2}$) in the denominator of the standardisation formula.
  • Calculating the wrong tail probability, for example finding $\Phi(0.5705)$ instead of $1 - \Phi(0.5705)$, leading to an answer greater than 0.5.
  • Assuming $p=0.5$ instead of using the value of $p=0.6$ established in the earlier parts of the question.

  Examiner tip: When approximating a discrete distribution (like Binomial) with a continuous one (like Normal), always apply a continuity correction to the value of interest before standardising.

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