5. A certain rare species of tuna has mean weight-20 lbs, with standard deviatio

Question

5. A certain rare species of tuna has mean weight-20 lbs, with standard deviation =2.25 lbs. a. A group of aquatic biologists at sea will catch 150 of these tuna. What is the probability that the total weight of the catch exceeds 2900 lbs.? What assumption are you making about the 150 tuna that will be caught in order to answer this question? b. Once a year, a different group of aquatic biologists catches 200 of these tuna. Exactly ten percent(10%) of the time, the 200 tuna will weigh a total of at least how many pounds? What is the probability that (a random sample of) five of these tuna will have a total weight exceeding 110 pounds?

Explanation / Answer

a)

we are assuming that weights are indepdnent and sample of 150 is taken randomely

for 150 tuna; mean weight =150*20 =3000

std deviation =2.25*(150)1/2 =27.56

probability that weight exceeds 2900 lbs

b)

or 150 tuna; mean weight =200*20 =4000

std deviation =2.25*(200)1/2 =31.82

for highest 10% ; at 90th percentile z =1.2816

hence correpsonding weight =mean +z*std deviation =4000+1.2816*31.82 =4040.7787

c)

for normal distribution z score =(X-)/ here mean=       = 3000.000 std deviation   == 27.56

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