Question

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

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:

Answer 2

z = -0.4.

Explanation:

Here's the formula for finding the z-score (a.k.a. standardized normal variable) of one measurement
`x` of a normal random variable
`X \sim N(\mu,\; \sigma^(2))`:

`\displaystyle z = (x-\mu)/(\sigma)`,

where

• 
  `z` is the z-score of this measurement,
• 
  `x` is the value of this measurement,
• 
  `\mu` is the mean of the random normal variable, and
• 
  `\sigma` is the standard deviation (the square root of variance) of the random normal variable.

Let the length of a trout in this lake be
`X` inches.

`X \sim N(30, \; 4.5^(2))`.

• 
  `\mu = 30`, and
• 
  `\sigma = 4.5`.

For this measurement,
`x = 28.2`. Apply the formula to get the z-score:

`\displaystyle z = (x - \mu)/(\sigma) = (28.2 - 30)/(4.5) = -0.4`.