Question

If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what is the probability of a random person scoring over 110?

Answer 1

`P(X>110) = P(z >(110-100)/15)`

Explanation:

`P(X>110) = P(z >(110-100)/15)`

Now, using a STANDARD NORMAL table:

`P(z>0.6667) =0.2525`

hope that helped