Use the normal distribution of fish lengths for which the mean is 8 inches and t

Question

Use the normal distribution of fish lengths for which the mean is 8 inches and the standard deviation is 2 inches Assume the variable x is normally distributed (a) What percent of the fish are longer than 12 inches? (b) If 300 fish are randomly selected, about how many would you expect to be shorter than 6 inches? (a) Approximately L1% of fish are longer than 12 inches. (Round to two decimal places as needed.) (b) You would expect approximately fish to be shorter than 6 inches. (Round to the nearest fish.)

Explanation / Answer

Solution :
Given that mean = 8, standard deviation = 2
(a) P(x > 12) = P(z > (x- ?)/?)
= P(z > (12-8/2)
= P(z > 2)
= 0.02

(b) P(x < 6) = P(z > (x- ?)/?)
= P(z > (6-8/2)
= P(z > -1)
= 0.1587
=> 0.1587*300 = 47.61 = 48(rounded)

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