Question

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

a. calculate the probability of finding a genius (iq > 130).
b. calculate the probability of average intelligence (between 90 and 110).

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=(X-\mu)/(\sigma),\\ \\ Z=(X-100)/(15)`.

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

a. For X=130,
`Z=(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=(90-100)/(15)=-(2)/(3)` and for X=110,
`Z=(110-100)/(15)=(2)/(3)`. Then
`Pr(90<X<110)=Pr\left(-(2)/(3)<Z<(2)/(3)\right)=0.2454-(-0.2454)=0.4908` (the decimal value is taken from the Standard Normal Distribution Table).