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=(X-\mu)/(\sigma)`

Using our information, we have:

`z=(28.2-30)/(4.5)= (-1.8)/(4.5)=-0.4`