Question

The mean annual cost to attend a college is $44,324. Assume the distribution of  annual costs follows the normal probability distribution and the standard deviation is $4,590.

Ninety percent of all students pay less than what amount?

Answer 1

Let $X=$ the annual cost to attend a college: $X\sim N(\mu, \sigma^2)$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $\mu=\$44324, \sigma=\$4590$

$$P(X<x)=P(Z<\dfrac{X-44324}{4590})=0.9$$

$$\dfrac{X-44324}{4590}=1.28155157$$

$$X=50206.32$$

Ninety percent of all students pay less than $50206.32