A-Level Mathematics May/June 2024 Q6(a): A random sample of 110 residents of Mahjing is chosen. Use a suitable approximation to…

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  Let $X$ be the random variable representing the number of residents who classified their bus service as good. Given the sample size $n=110$ and the probability of classifying the service as good $p=0.25$ (from the question context), the distribution is: $X \sim B(110, 0.25)$

  Since $n$ is large, we can use a Normal approximation. We first find the mean and variance: Mean, $\mu = np = 110 \times 0.25 = 27.5$ Variance, $\sigma^2 = np(1-p) = 110 \times 0.25 \times 0.75 = 20.625$

  So, the approximating distribution is $X \approx N(27.5, 20.625)$.

  We want to find the probability that fewer than 22 residents classified the service as good, which is $P(X < 22)$. For a discrete distribution, this is equivalent to $P(X \le 21)$.

  Applying the continuity correction: $P(X \le 21) \approx P(Y < 21.5)$, where $Y \sim N(27.5, 20.625)$.

  Now, we standardise the value to find the Z-score: $Z = \frac{x - \mu}{\sigma} = \frac{21.5 - 27.5}{\sqrt{20.625}}$ $Z = \frac{-6}{4.541475...} \approx -1.32116$

  Finally, we find the probability from the standard normal distribution: $P(Z < -1.32116) = \Phi(-1.32116)$ By symmetry of the normal distribution, this is $1 - \Phi(1.32116)$. $P(Z < -1.32116) = 1 - 0.90678 = 0.09322$

  So, the probability is $0.0932$ (3 s.f.).

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  How the marks are awarded

  • B1 — Correctly calculating and stating the Mean = 27.5 and Variance = 20.625, as shown in the first step of the working.
  • M1 — Substituting the calculated mean (27.5) and variance (20.625) into the standardisation formula, specifically in the form (value - 27.5) / sqrt(20.625).
  • M1 — Applying the correct continuity correction. The discrete value 'fewer than 22' (X ≤ 21) becomes the continuous value 21.5 in the standardisation formula.
  • M1 — Finding the correct probability area for a negative Z-value. The calculation P(Z < -1.32116) correctly seeks the left tail of the distribution, resulting in a probability less than 0.5.
  • A1 — Obtaining the final correct answer of 0.0932, which falls within the required range of 0.0932 to 0.09325.

  Common mistakes

  • Forgetting to apply a continuity correction and using 22 or 21 in the standardisation formula instead of 21.5.
  • Using an incorrect continuity correction, such as 22.5, by misinterpreting 'fewer than 22'.
  • Using the variance (20.625) instead of the standard deviation (√20.625) in the denominator of the standardisation formula.
  • Calculating the wrong tail probability, for example finding P(Z > -1.321) instead of P(Z < -1.321), leading to an answer of 1 - 0.0932 = 0.9068.

  Examiner tip: When using a normal approximation for a discrete distribution, always convert the discrete inequality (e.g., X < 22) into its integer equivalent (X ≤ 21) before applying the continuity correction (Y < 21.5).

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