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

-0.4

Explanation:

z score is:

z = (x - μ) / σ

For x = 28.2, μ = 30, and σ = 4.5:

z = (28.2 - 30) / 4.5

z = -0.4

Answer 2

Answer: -0.4

Explanation:

Given: Mean :
`\mu=30\text{ inches}`

Standard deviation :
`\sigma=4.5\text{ inches}`

The formula to calculate z-score is given by :-

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

For X = 28.2 inches, we have

`z=(28.2-30)/(4.5)\\\\\Rigahtarrow\ z=-0.4`

Hence, the z-score of a trout with a length of 28.2 inches.= -0.4