A-Level Mathematics May/June 2024 Q2(a): Find the probability that a randomly chosen adult raccoon of this species has a tail le…

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  Let $X$ be the tail length of a randomly chosen adult raccoon. The distribution is given as $X \sim N(28, 3.3^2)$.

  We need to find the probability $P(23 < X < 35)$.

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

  For $x = 23$: $z_1 = \frac{x - \mu}{\sigma} = \frac{23 - 28}{3.3} = -\frac{5}{3.3} = -1.5151...$

  For $x = 35$: $z_2 = \frac{x - \mu}{\sigma} = \frac{35 - 28}{3.3} = \frac{7}{3.3} = 2.1212...$

  Now, we find the probability for the standard normal variable $Z \sim N(0, 1)$: $P(23 < X < 35) = P(-1.515... < Z < 2.121...)$

  This can be calculated using the cumulative distribution function, $\Phi(z)$: $P(-1.515 < Z < 2.121) = \Phi(2.121) - \Phi(-1.515)$

  Using the symmetry property $\Phi(-z) = 1 - \Phi(z)$: $= \Phi(2.121) - (1 - \Phi(1.515))$ $= \Phi(2.121) + \Phi(1.515) - 1$

  From statistical tables or a calculator: $= 0.98304 + 0.93512 - 1$ $= 1.91816 - 1$ $= 0.91816$

  Rounding to 3 significant figures, the probability is 0.918.

  $P(23 < X < 35) = 0.918$ (3 s.f.)

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  How the marks are awarded

  • M1 — Awarded for correctly standardising either $x=23$ or $x=35$ using the formula $(x - 28) / 3.3$. The working shows this for both values.
  • A1 — Awarded for obtaining both correct z-values, $-1.515...$ and $2.121...$. These are often rounded to $-1.52$ and $2.12$ for table use.
  • M1 — Awarded for the correct method to find the probability between two z-values, demonstrated by the expression $\Phi(2.121) - \Phi(-1.515)$ or the equivalent form $\Phi(2.121) + \Phi(1.515) - 1$.
  • A1 — Awarded for the final correct probability of 0.918, or any value that rounds to 0.918 (AWRT).

  Common mistakes

  • Using the variance ($3.3^2 = 10.89$) instead of the standard deviation (3.3) in the standardisation formula, leading to incorrect z-values.
  • Incorrectly calculating the area for the negative z-value, for example calculating $\Phi(2.121) - \Phi(1.515)$ instead of $\Phi(2.121) - (1 - \Phi(1.515))$.
  • Adding the probabilities $\Phi(2.121) + \Phi(1.515)$ without subtracting 1, which gives a probability greater than 1.
  • Rounding z-values too early (e.g., to one decimal place) which can lead to an inaccurate final answer outside the acceptable range.

  Examiner tip: Master the process of standardising values to z-scores and using the symmetry properties of the normal distribution, such as $\Phi(-z) = 1 - \Phi(z)$, to find probabilities between two points.

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