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

x would be the value that we are "interested" in, so the 28.2 inch trout

μ is the mean, σ is the standard deviation (these are already given in the problem)

plug in these values to find z

Answer 2

\[\frac{ 28.2-30 }{ 4.5 }\]

Answer 3

would that be 21.53?

Answer 4

check your calculations again, remember your order of operations

subtraction comes before division, so subtract 28.2 - 30 and ~then~ divide the result by 4.5

Answer 5

oh okay i'll do that rq

Answer 6

-0.4?

Answer 7

yeah -0.4 is right, my comp bugged out for a second

Answer 8

cool thanks!