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?
+128475 (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
