Question

A normally distributed population has a mean of 25.6 and a standard deviation of 3.3. Find the probability that a single randomly selected element X of the population exceeds 30.

Answer 1

We have a normal distribution, $\mu=25.6, \sigma=3.3.$

Let's convert it to the standard normal distribution,

$z=\cfrac{x-\mu}{\sigma}=\cfrac{30-25.6}{3.3}=1.33;$

$P(X>30)=P(Z>1.33)=1-P(Z<1.33)=\\ =1-0.9082=0.0918\ \text{​(from z-table).}$