The heights of an adult female population are normally distributed with mean 162 cm and standard deviation 7.5 cm - Edexcel - A-Level Maths Statistics - Question 7 - 2012 - Paper 2

For a normally distributed population, the height at the 60th percentile can be calculated using the mean and standard deviation:

The z-score for the 60th percentile is approximately 0.253. Now, using the inverse z-score formula for height:

$X = \mu + z \cdot \sigma$

Where:

• $\mu = 162$ cm (mean height)
• $\sigma = 7.5$ cm (standard deviation)
• $z \approx 0.253$

Calculating Sarah's estimated height:

$X = 162 + 0.253 \cdot 7.5 \approx 162 + 1.8975 \approx 163.9$

Thus, Sarah's estimated height as an adult is approximately 163.9 cm.

Given that 90% of adult males are taller than the mean height of adult females, we can say:

$P(X > \text{mean height of adult females}) = 0.90$

This means the mean height of adult females corresponds to the 10th percentile of the male height distribution. Using the z-score corresponding to the 10%:

The z-score for the 10th percentile is approximately -1.28. Hence, using the standard deviation for males ($\sigma = 9.0$ cm):

Let $M$ be the mean height of adult males, then we can express:

$z = \frac{X - M}{\sigma} \rightarrow -1.28 = \frac{\text{mean height of adult females} - M}{9}$

We already know from part (a) that the mean height of adult females is 162 cm:

$-1.28 = \frac{162 - M}{9}$

Solving for $M$:

$-1.28 \cdot 9 = 162 - M \rightarrow -11.52 = 162 - M \rightarrow M = 162 + 11.52 = 173.52$

Thus, the mean height of an adult male is approximately 173.52 cm.