Use the normal distribution of fish lengths for which the mean is

10

inches and the standard deviation is

4

inches. Assume the variable x is normally distributed.

(a)

What percent of the fish are longer than

11

inches?

(b)

If

400

fish are randomly selected, about how many would you expect to be shorter than

8

inches?

Guest Mar 23, 2020

#1**+1 **

(a) longer than 11 inches

[ 11 - 10]

________ = 1/4 = .25

4

The percentage associaed with this z score is .5987

So approx 1 - .5987 = .4013 ≈ 40.13% fish are longer than 11 inches

(b) shorter than 8 inches

[ 8 - 10]

_______ = -1/2 = -.50

4

The percentage associated with this z score is .3085

So.....the number of fish out of 400 that are shorter than 8 inches =

400 (.3085) ≈ 123

CPhill Mar 23, 2020