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 answer is -0.4....

Answer 2

We have been given that the lengths of trout in a lake are normally distributed with a mean of 30 inches and a standard deviation of 4.5 inches.

We are asked to find the z-score of a trout with a length of 28.2 inches.

To find z score of the given length  we will use z score formula.

$z=\frac{x- \mu }{\sigma}$, where x is the given value of which we want to find out z score, $\mu$ is mean of the given data and $\sigma$ is standard deviation of data.

After substituting our given values in this formula we will get,

$z=\frac{28.2-30}{4.5}\\ \\ z=\frac{-1.8}{ 4.5}\\ \\ z=\frac{-2}{5}=-0.4$

Therefore, z  score of a trout with a length of 28.2 inches is -0.4.