Use the normal distribution of fish lengths for which the mean is 9 inches and the standard deviation is 5 inches. Assume the variable x is normally distributed. (a) What percent of the fish are longer than 14 inches? (b) If 300 fish are randomly selected, about how many would you expect to be shorter than 6 inches?
(a) right parenthesis Approximately nothing% of fish are longer than 14 inches. (Round to two decimal places as needed.)
(b) If 300 fish are randomly selected, about how many would you expect to be shorter than 6 inches?
Part a
Solution:
We are given that the random variable X = length of fish follows normal distribution.
We are given mean = 9, SD = 5
We have to find P(X>14)
P(X>14) = 1 – P(X<14)
Z = (X – mean) / SD
Z = (14 – 9) / 5
Z = 5/5
Z = 1
P(Z<1) = P(X<14) = 0.841344746
(by using z-table/excel)
P(X>14) = 1 – P(X<14)
P(X>14) = 1 – 0.841344746
P(X>14) = 0.158655254
Required percentage = 15.87%
Part b
We are given that the random variable X = length of fish follows normal distribution.
We are given mean = 9, SD = 5, n = 300
First, we have to find P(X<6)
Z = (X – mean) / SD
Z = (6 – 9)/5
Z = -0.6
P(Z< -0.6) = P(X<6) = 0.274253
(by using z-table/excel)
Required number of fish = n* P(X<6) = 300*0.274253
Required number of fish = 82.2759
Required number of fish ≈ 82
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