Question

The lengths of trout in a lake are normally distributed with a mean of 30 inches and a standard deviation of 4.5 inches. Enter the z-score of a trout with a length of 28.2 inches.

Answer 1

The z-score would be -0.4.

The formula for z-scores is

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

Using our information, we have:
$z=\frac{28.2-30}{4.5}= \frac{-1.8}{4.5}=-0.4$