# 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?

Nov 5, 2015

$P \left(X > 110\right) = P \left(z > \frac{110 - 100}{15}\right)$

#### Explanation:

$P \left(X > 110\right) = P \left(z > \frac{110 - 100}{15}\right)$

Now, using a STANDARD NORMAL table:

$P \left(z > 0.6667\right) = 0.2525$

hope that helped