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?

Nov 15, 2017

$z = 2.8$

Explanation:

we use the standardising formula

$z = \frac{X - \mu}{\sigma}$

$X =$the score

$\mu =$the mean

$\sigma =$the standard deviation

$\therefore z = \frac{142 - 100}{15}$

$z = 2.8$