This Question: 10 pts 170131 (13 complete) | Use the normal distribution of fish

Question

This Question: 10 pts 170131 (13 complete) | Use the normal distribution of fish lengths for which the mean is 11 inches and the standard deviat (a) What percent of the fish are longer than 14 inches? (b) If 200 fish are randomly selected, about how many would you expect to be shorter than 9 inches? ion is 2 inches. Assume the variable x is normally distributed (a) Approximately % of fish are longer than 14 inches Round to two decimal places as needed) (b) You would expect approximately fish to be shorter than 9 inches. Round to the nearest fish.)

Explanation / Answer

Mean is 11 and s is 2

z can be found as (x-mean)/s

a) P(X>14)= P(z>(14-11)/2)=P(z>1.5) or 1-P(z<1.5) =1-0.9332 =0.07

b) P(x<9)=P(z<(9-11)/2)=P(z<-1) or 1-P(z<1) or 1-0.8413 or 0.1587

thus fish number is 0.1587*200=31.74 or 31

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