The heights of a population of women are normally distributed with mean $\mu$ cm and standard deviation $\sigma$ cm - Edexcel - A-Level Maths Statistics - Question 7 - 2010 - Paper 1

Using the information provided, we can derive the equation:
Given $P(X < 154) = 0.05$, we get:
$z = \frac{154 - \mu}{\sigma} = -1.6449$
This implies that:
$154 - \mu = -1.6449\sigma$
Rearranging, we find:
$\mu = 154 + 1.6449\sigma$

We know that 30% of the women are taller than 172 cm, thus $P(X > 172) = 0.30$, which implies $P(X < 172) = 0.70$.
By using the z-score for this percentile:
$z = \frac{172 - \mu}{\sigma} = 0.524 \text{ (approximately)}$
This leads us to the equation:
$172 - \mu = 0.524\sigma$
We now have a system of two equations:

1. $\mu = 154 + 1.6449\sigma$
2. $172 - \mu = 0.524\sigma$
   Substituting the first equation into the second allows us to solve for $\mu$ and $\sigma$. After solving, we find: $\mu \approx 167.65 \text{ and } \sigma \approx 8.30$

To find the probability that a woman chosen at random is taller than 160 cm, we find:
$P(Taller \ than\ 160) = P(Z > \frac{160 - \mu}{\sigma})$
Calculating the z-score: $z = \frac{160 - 167.65}{8.30} \approx -0.91$
Utilizing standard normal distribution tables, we find: $P(Z > -0.91) = 0.8186 \text{ (approximately)}$
Thus, the probability that a randomly chosen woman is taller than 160 cm is approximately 0.82.