Question

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?

Answer 1

It can be solved using a simple formula:

`z=(measured - mean)/(standard deviation)`

`z= (barx - mu)/sigma`

In this case, that would be `z= (142 - 100)/(15)` which is equal to `z=2.80`