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.

Accepted Solution

Answer:z = -0.4.Step-by-step explanation:Here's the formula for finding the z-score (a.k.a. standardized normal variable) of one measurement [tex]x[/tex] of a normal random variable [tex]X \sim N(\mu,\; \sigma^{2})[/tex]:[tex]\displaystyle z = \frac{x-\mu}{\sigma}[/tex],where[tex]z[/tex] is the z-score of this measurement, [tex]x[/tex] is the value of this measurement, [tex]\mu[/tex] is the mean of the random normal variable, and[tex]\sigma[/tex] is the standard deviation (the square root of variance) of the random normal variable.Let the length of a trout in this lake be [tex]X[/tex] inches.[tex]X \sim N(30, \; 4.5^{2})[/tex].[tex]\mu = 30[/tex], and[tex]\sigma = 4.5[/tex].For this measurement, [tex]x = 28.2[/tex]. Apply the formula to get the z-score:[tex]\displaystyle z = \frac{x - \mu}{\sigma} = \frac{28.2 - 30}{4.5} = -0.4[/tex].