Question

Iq scores are normally distributed with a mean of 100 and a standard deviation of 15.

Answer 1

Mean is $\mu=100$ and a standard deviation is $\sigma =15$, then the variable $X\sim N(100,15^2)$.

Use substitution

$Z=\dfrac{X-\mu}{\sigma},\\ \\ Z=\dfrac{X-100}{15}$.

This substitution gives you that $Z\sim N(0,1)$.

a. For X=130, $Z=\dfrac{130-100}{15}=2$ and $Pr(X>130)=Pr(Z>2) =0.9772$ (the decimal value is taken from the Standard Normal Distribution Table).

b. For X=90, $Z=\dfrac{90-100}{15}=-\dfrac{2}{3}$ and for X=110, $Z=\dfrac{110-100}{15}=\dfrac{2}{3}$. Then $Pr(90$ (the decimal value is taken from the Standard Normal Distribution Table).