Question

A random variable X is normally distributed with a mean of 45 and standard deviation of 3.If the area to the right of Z is 0.9292, what will be the value of its random variable X

Answer 1

$\mu=45,\sigma=3,P(Z>z)=0.9292,\\ P(Z<z)=1-0.9292=0.0708.$

From z-table we find $z=-1.47.$

So, $P(Z<z)=P(X<x), \\ z=\cfrac{x-\mu} {\sigma},\\ x=z\sigma+\mu=-1.47\cdot3+45=40.59.$