A-Level Mathematics October/November 2024 Q6(a): The heights of the female students at Breven college are normally distributed: • 90% of…

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  Let $H$ be the height of a female student. We are given that $H \sim N(\mu, \sigma^2)$.

  We are given two pieces of information:

  1. $P(H < 182.7) = 0.90$
  2. $P(H < 162.5) = 0.40$

  We standardise these using $Z = \frac{H - \mu}{\sigma}$, where $Z \sim N(0, 1)$.

  For the first piece of information: $P(Z < \frac{182.7 - \mu}{\sigma}) = 0.90$ We find the z-value corresponding to a probability of 0.90. Let this be $z_1$. $z_1 = \Phi^{-1}(0.90) = 1.282$ So, we have the equation: $\frac{182.7 - \mu}{\sigma} = 1.282 \implies 182.7 - \mu = 1.282\sigma$ (Equation 1)

  For the second piece of information: $P(Z < \frac{162.5 - \mu}{\sigma}) = 0.40$ Since the probability is less than 0.5, the z-value will be negative. Let this be $z_2$. $z_2 = \Phi^{-1}(0.40) = -\Phi^{-1}(1-0.40) = -\Phi^{-1}(0.60)$ From the tables, $\Phi^{-1}(0.60) = 0.2533$. So, $z_2 = -0.2533$. So, we have the equation: $\frac{162.5 - \mu}{\sigma} = -0.2533 \implies 162.5 - \mu = -0.2533\sigma$ (Equation 2)

  Now we solve the two linear equations simultaneously. From (1): $\mu = 182.7 - 1.282\sigma$ From (2): $\mu = 162.5 + 0.2533\sigma$

  Set the expressions for $\mu$ equal: $182.7 - 1.282\sigma = 162.5 + 0.2533\sigma$ $182.7 - 162.5 = 1.282\sigma + 0.2533\sigma$ $20.2 = 1.5353\sigma$ $\sigma = \frac{20.2}{1.5353} = 13.157...$

  Substitute the value of $\sigma$ back into Equation 2 to find $\mu$: $\mu = 162.5 + 0.2533(13.157...)$ $\mu = 162.5 + 3.3326...$ $\mu = 165.8326...$

  Rounding the answers to an appropriate degree of accuracy (3 significant figures or 1 decimal place as per context): Mean, $\mu = 165.8$ cm Standard deviation, $\sigma = 13.2$ cm

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  How the marks are awarded

  • B1 — Awarded for correctly identifying the z-value for a probability of 0.90. In the model answer, this is shown by stating '$z_1 = \Phi^{-1}(0.90) = 1.282$'.
  • B1 — Awarded for correctly identifying the negative z-value for a probability of 0.40. In the model answer, this is shown by finding '$z_2 = -0.2533$', which is within the required range.
  • M1 — Awarded for correctly setting up at least one standardisation equation, linking a height, the mean, the standard deviation, and a z-value. For example, '$\frac{182.7 - \mu}{\sigma} = 1.282$'.
  • M1 — Awarded for a correct method to solve the two simultaneous equations. The model answer shows this by eliminating $\mu$ to form a single equation in $\sigma$: '$20.2 = 1.5353\sigma$'.
  • A1 — Awarded for obtaining the correct final values for both the mean and standard deviation, rounded to at least one decimal place. The model answer provides '$\mu = 165.8$' and '$\sigma = 13.2$'.

  Common mistakes

  • Using a positive z-value for the probability of 0.40 (i.e., using 0.2533 instead of -0.2533), which ignores that the height is below the mean.
  • Incorrectly using the probability values (0.90 and 0.40) in the standardisation formula instead of the corresponding z-values.
  • Making an algebraic error when solving the simultaneous equations, such as incorrectly handling the subtraction of a negative number (e.g., $1.282 - (-0.2533)$).
  • Rounding z-values too early or to too few significant figures, leading to a final answer outside the acceptable range for accuracy marks.

  Examiner tip: When given two probability points for a normal distribution, the standard approach is to convert them into two standardised equations and solve them simultaneously for the mean and standard deviation.

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