Normal distribution problem; distribution of height

Question

The problem is: the height of children in age from 3.5 to 4 years is described by normal distribution with parameters $\mu =103$ centimetres and $ \sigma=4.5$ centimetres. What is the percent of children with height $\le$ 93 centimetres? Normally, I would count $F(93) = P(X \le 93) = \int_{-\infty}^x f(t)dt$, but here, I am puzzled.

Answer 1

Use the normal distribution by creating the z-score $$ z = \frac{X-\mu}{\sigma} = \frac{X-103}{4.5} $$ and then compute $$ P\left(z<= \frac{93-103}{4.5} = -\frac{10}{4.5}\right) $$