# How do you find the z-score for an IQ test score of 142 when the mean is 100 and the standard deviation is 15?

$z = \frac{m e a s u red - m e a n}{s \tan \mathrm{da} r d \mathrm{de} v i a t i o n}$
$z = \frac{\overline{x} - \mu}{\sigma}$
In this case, that would be $z = \frac{142 - 100}{15}$ which is equal to $z = 2.80$