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.
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:
$-0.4$