the lengths of trout in a lake are normally distributed with a mean of 30 inches and a standard deviation of…

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.

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

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Identify values

We are given that $\mu = 30$, $\sigma=4.5$, and $x = 28.2$.

Step3: Substitute values into formula

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

Step4: Calculate z - score

$z=-0.4$

Answer:

$-0.4$