The measure of intelligence, IQ, of a group of students is assumed to be Normally distributed with mean 100 and standard deviation 15 - Edexcel - A-Level Maths Statistics - Question 7 - 2007 - Paper 1

We know that:

$P(X > 100 + k) = 0.2090$

This can be rewritten as:

$P(X < 100 + k) = 1 - 0.2090 = 0.7910$

Standardizing this probability, we use the Z-score formula again:

$Z = \frac{X - \mu}{\sigma}$

Let’s denote $X$ as $100 + k$:

$P(X < 100 + k) = P\left(Z < \frac{100 + k - 100}{15}\right) = P\left(Z < \frac{k}{15}\right)$

From the Z-table, we need to find $Z$ such that:

$P(Z < z) = 0.7910$

This gives us $z \approx 0.81$.

Equating it we find:

$\frac{k}{15} = 0.81 \implies k = 15 \times 0.81 = 12.15$

Rounding to the nearest integer, we conclude:

$k = 12$.