A-Level Mathematics October/November 2024 Q4(a): How many of these trees would you expect to have height less than 18.2 metres?

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  Let $H$ be the random variable representing the height of a tree. We are given that the heights are normally distributed with a mean $\mu = 19.8$ and a standard deviation $\sigma = 2.4$. So, $H \sim N(19.8, 2.4^2)$.

  We want to find the number of trees out of 450 with a height less than 18.2 metres. First, we find the probability $P(H < 18.2)$.

  Standardise the value $H = 18.2$: $Z = \frac{H - \mu}{\sigma}$ $z = \frac{18.2 - 19.8}{2.4} = \frac{-1.6}{2.4} = -\frac{2}{3} \approx -0.6667$

  Now, we find the probability $P(Z < -0.6667)$. Using the properties of the standard normal distribution: $P(Z < -0.6667) = \Phi(-0.6667) = 1 - \Phi(0.6667)$

  Using a calculator, the probability is: $P(H < 18.2) = P(Z < -0.6667) = 0.25249...$

  The total number of trees is 450. The expected number of trees with height less than 18.2 m is: Expected Number = $P(H < 18.2) \times 450$ Expected Number = $0.25249... \times 450 = 113.620...$

  Since the number of trees must be an integer, we round to the nearest whole number.

  Expected number of trees = 114

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  How the marks are awarded

  • M1 — Correctly substituting the values 18.2, 19.8, and 2.4 into the standardisation formula, $z = \frac{18.2 - 19.8}{2.4}$.
  • M1 — Finding the probability corresponding to the calculated z-score. This involves finding the area in the left tail of the normal distribution, $P(Z < -0.6667)$, which must be a value less than 0.5.
  • A1 — Obtaining the correct probability of approximately 0.252 or 0.2525. This value is seen in the working as 0.25249...
  • B1FT — Multiplying the probability (from the A1 mark, or a follow-through from an earlier attempt) by the total number of trees (450) and giving a final answer as a single integer. The calculation $0.25249 \times 450$ is correctly performed and rounded to 114.

  Common mistakes

  • Using a continuity correction (e.g., standardising 18.15 or 18.25), which is incorrect as height is a continuous variable.
  • Calculating the wrong probability area, such as $P(H > 18.2)$ or $1 - P(H < 18.2)$, often by mishandling the negative z-score, leading to a probability of ~0.748.
  • Incorrectly standardising by reversing the mean and the value, e.g., $\frac{19.8 - 18.2}{2.4}$, which results in a positive z-score and a probability greater than 0.5.
  • Leaving the final answer as a decimal (e.g., 113.6) instead of rounding to the nearest integer for the expected number of trees.

  Examiner tip: Always draw a quick sketch of the Normal distribution curve to visualise the probability area you need to find, especially when dealing with 'less than' or 'greater than' a value below the mean.

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