Use the normal distribution of fish lengths for which the mean is 10 inches and

Question

Use the normal distribution of fish lengths for which the mean is 10 inches and the standard deviation is 5 inches. Assume the variable x is normally distributed (a) What percent of the fish are longer than 12 inches? (b) If 500 fish are randomly selected, about how many would you expect to be shorter than 7 inches? (a) Approximately [ of fish are longer than 12 inches. Round to two decimal places as needed.) (b) You would expect approximatelyfish to be shorter than 7 inches. Round to the nearest fish.) Enter your answer in each of the answer boxes

Explanation / Answer

Solution:

We are given that the fish lengths follow a normal distribution.
We are given
Mean µ = 10 and SD ? = 5

(a) We have to find P(X>12)

P(X > 12) = 1-P(X < 12)

The formula for z-score is given as below:
Z = (X – µ) / ?
Z = (12 – 10)/5 = 2/5 = 0.4

P(Z < 0.4) = 0.6554

P(X < 12) = P(Z < 0.4) = 0.6554
P(X > 12) = 1 – P(X < 12)
= 1 – 0.6554
= 0.3446

(b) Here, we are given n = 500

We have to find P(X < 7)

The z-score formula is given as below:

Z = (X – µ) / ?

Z = (7 – 10)/5

Z = -3/5 = -0.6

P(Z < -0.6) = 0.2743

Expected number = 500* 0.2743
= 137.15

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