A-Level Mathematics October/November 2024 Q7(d): Use a suitable approximation to find the probability that Leno scores more than 28 goal…

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  Let $X$ be the number of goals scored. The question requires a suitable approximation, which is the Normal distribution. From previous parts of the question, we have the parameters for the approximation.

  Let $Y \sim N(\mu, \sigma^2)$.

  Parameters: Mean, $\mu = 30$ Variance, $\sigma^2 = 18$ Standard deviation, $\sigma = \sqrt{18} \approx 4.2426$

  We want to find the probability that Leno scores more than 28 goals but fewer than 35 goals, which is $P(28 < X < 35)$. Since $X$ is a discrete variable, this is equivalent to $P(29 \le X \le 34)$.

  Applying continuity correction for the continuous Normal distribution $Y$: $P(28.5 < Y < 34.5)$

  Now, we standardise the boundaries to find the $z$-scores: $Z = \frac{Y - \mu}{\sigma}$

  $z_1 = \frac{28.5 - 30}{\sqrt{18}} = \frac{-1.5}{4.2426...} = -0.35355...$

  $z_2 = \frac{34.5 - 30}{\sqrt{18}} = \frac{4.5}{4.2426...} = 1.06066...$

  So we need to find $P(-0.3536 < Z < 1.0607)$.

  This is calculated as $\Phi(1.0607) - \Phi(-0.3536)$. Using the property $\Phi(-z) = 1 - \Phi(z)$: $= \Phi(1.0607) - (1 - \Phi(0.3536))$ $= \Phi(1.0607) + \Phi(0.3536) - 1$

  From the standard normal distribution tables: $\Phi(1.0607) \approx 0.8556$ $\Phi(0.3536) \approx 0.6382$

  Probability $= 0.8556 + 0.6382 - 1$ $= 1.4938 - 1$ $= 0.4938$

  Therefore, the probability is $0.494$ (to 3 s.f.).

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  How the marks are awarded

  • B1 — Stating the correct Mean = 30 and Variance = 18. This is shown in the 'Parameters' section of the answer.
  • M1 — Substituting their mean and a positive standard deviation into a standardisation formula. This is demonstrated by writing expressions like (28.5 - 30)/√18.
  • M1 — Using the correct continuity corrections. The probability P(28 < X < 35) is correctly converted to the continuous interval P(28.5 < Y < 34.5).
  • M1 — Calculating the appropriate area under the curve from their z-scores. The expression Φ(1.0607) - (1 - Φ(0.3536)) shows the correct final process to find the probability between the two values.
  • A1 — Obtaining the final correct answer of 0.494 or anything that rounds to it (AWRT).

  Common mistakes

  • Forgetting to apply continuity correction and instead calculating P(28 < Y < 35), which leads to incorrect z-scores and loses marks.
  • Applying the continuity correction incorrectly, for example using P(28.5 < Y < 35.5) or P(27.5 < Y < 34.5), by misinterpreting the strict inequalities '>' and '<'.
  • Using the variance (18) or an incorrect root (e.g. 18²) in the standardisation formula instead of the standard deviation (√18).
  • Incorrectly handling the probability for a negative z-score, for example by evaluating Φ(-0.3536) as -Φ(0.3536) instead of the correct 1 - Φ(0.3536).

  Examiner tip: When approximating a discrete distribution with a Normal distribution, always apply a continuity correction of ±0.5, carefully adjusting the boundary based on whether the inequality is strict (>, <) or inclusive (≥, ≤).

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