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

We know that 5% is below 154 cm. Therefore, using the z-score formula:

$P(X < 154) = 0.05$

Using the z-score for 0.05 (which is approximately -1.6449):

$z = \frac{X - μ}{σ}$ implies that:

$154 - μ = -1.6449σ$

Rearranging gives:

$μ = 154 + 1.6449σ.$

This establishes the equation.

For the 172 cm measurement, we know that:

$P(X > 172) = 0.30$

Thus,

$P(X < 172) = 0.70$.

Using the z-score for 0.70, we find it to be approximately 0.524:

$P(X < 172) = 0.70$ gives:

$z = \frac{172 - μ}{σ}$

This results in:

$172 - μ = 0.524σ,$ which simplifies to:

$μ = 172 - 0.524σ.$

Now we have two equations:

1. $μ = 154 + 1.6449σ$
2. $μ = 172 - 0.524σ$

Setting these equal to each other:

$154 + 1.6449σ = 172 - 0.524σ$

Solving gives:

$2.1684σ = 18 \\ σ ≈ 8.30$

Substituting back to find μ:

$μ = 154 + 1.6449(8.30) ≈ 167.65.$

To find the probability that a woman is taller than 160 cm, we use:

$P(X > 160) = P(Z > \frac{160 - μ}{σ})$

Substituting our values:

$Z = \frac{160 - 167.65}{8.30}$

Now calculate Z:

$Z ≈ -0.91$

We then find:

$P(Z > -0.91) = 1 - P(Z < -0.91)$

Using standard normal distribution tables, we find:

$P(Z < -0.91) ≈ 0.1815$

Thus,

$P(X > 160) = 1 - 0.1815 = 0.8185.$
This gives a probability of approximately 0.82.