Question

The weight W of fish in a given pond is normally distributed with mean 8.5 pounds and standard deviation 1.2 pounds. If you randomly select 5 fish from the pond, what is the probability that the mean weight of the fish is between 8 and 9 pounds?

Answer 1

Solution:

We are given

µ = 8.5

σ = 1.2

n = 5

We have to find P(8<Xbar<9) = P(Xbar<9) – P(Xbar<8)

Z = (Xbar - µ)/[σ /sqrt(n)]

Z = (9 – 8.5)/[1.2/sqrt(5)]

Z = 0.5/ 0.536656

Z = 0.931695

P(Z< 0.931695) = P(Xbar<9) = 0.824253

(by using z-table)

Now find P(Xbar<8)

Z = (8 – 8.5)/[1.2/sqrt(5)]

Z =-0.931695

P(Z< -0.931695) = P(Xbar<8) = 0.175747

(by using z-table)

P(8<Xbar<9) = P(Xbar<9) – P(Xbar<8)

P(8<Xbar<9) = 0.824253 - 0.175747

P(8<Xbar<9) = 0.648506

Required probability = 0.648506