A-Level Mathematics May/June 2025 Q4(a): How many competitors would you expect to have times within 1.2 minutes of the mean time?

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  Let $X$ be the time taken by a competitor in minutes. We are given that the times are normally distributed with a mean $\mu = 32.4$ and a standard deviation $\sigma = 2.5$. So, $X \sim N(32.4, 2.5^2)$. The total number of competitors is 150.

  We want to find the number of competitors with times 'within 1.2 minutes of the mean'. This corresponds to the interval:

  $P(32.4 - 1.2 < X < 32.4 + 1.2) = P(31.2 < X < 33.6)$

  First, we standardise the values to find the corresponding z-scores:

  $Z = \frac{X - \mu}{\sigma}$

  $P(\frac{31.2 - 32.4}{2.5} < Z < \frac{33.6 - 32.4}{2.5})$

  $= P(\frac{-1.2}{2.5} < Z < \frac{1.2}{2.5})$

  $= P(-0.48 < Z < 0.48)$

  To find this probability, we use the standard normal distribution table or a calculator:

  $P(-0.48 < Z < 0.48) = \Phi(0.48) - \Phi(-0.48)$

  Using the symmetry property $\Phi(-z) = 1 - \Phi(z)$:

  $= \Phi(0.48) - (1 - \Phi(0.48))$ $= 2\Phi(0.48) - 1$

  From the tables, $\Phi(0.48) = 0.6844$.

  Probability $= 2 \times 0.6844 - 1 = 1.3688 - 1 = 0.3688$

  The question asks for the expected number of competitors. We multiply the probability by the total number of competitors (150):

  Expected number $= 150 \times 0.3688 = 55.32$

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

  Expected number of competitors = 55.

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  How the marks are awarded

  • M1 — Correctly applying the standardisation formula with the given values, seen in the step $P(\frac{-1.2}{2.5} < Z < \frac{1.2}{2.5})$. Using $\pm 1.2$ and the standard deviation 2.5 is required.
  • M1 — Calculating the correct probability area for the symmetric interval. This is demonstrated by the method $2\Phi(0.48) - 1$ or $\Phi(0.48) - \Phi(-0.48)$.
  • A1 — Obtaining the correct probability of 0.3688, accurate to 4 decimal places.
  • B1 — Multiplying the calculated probability by the total number of competitors (150) and giving the final answer as a single integer, 55.

  Common mistakes

  • Using the variance ($2.5^2$) or the square root of the standard deviation ($\sqrt{2.5}$) in the standardisation formula instead of the standard deviation (2.5).
  • Calculating the wrong probability area, such as $P(Z > 0.48)$ or just finding $\Phi(0.48)$, instead of the area between -0.48 and 0.48.
  • Leaving the answer as a probability (0.3688) or a decimal number (55.32) instead of rounding to the nearest integer as required for a 'number of competitors'.
  • Incorrectly applying a continuity correction, which is not appropriate for a continuous variable that has not been approximated from a discrete one.

  Examiner tip: Master translating phrases like 'within a certain value of the mean' into a symmetric probability calculation, $P(-z < Z < z)$, which simplifies to $2\Phi(z) - 1$.

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