Question

A particular fruit's weights are normally distributed, with a mean of 298 grams and a standard deviation of 28 grams. If you pick 5 fruit at random, what is the probability that their mean weight will be between 279 grams and 305 grams

I am using excel for this problem.

Answer 1

Solution:

Given

µ = 298

σ = 28

n = 5

We have to find P(279<Xbar<305)

P(279<Xbar<305) = P(Xbar<305) – P(Xbar<279)

Find P(Xbar<305)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (305 - 298)/(28/sqrt(5))

Z = 0.559017

P(Z<0.559017) = P(Xbar<305) = 0.711925

(by using z-table or excel)

Find P(Xbar<279)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (279 - 298)/(28/sqrt(5))

Z = -1.51733

P(Z<-1.51733) = P(Xbar<279) = 0.064591

(by using z-table or excel)

P(279<Xbar<305) = P(Xbar<305) – P(Xbar<279)

P(279<Xbar<305) = 0.711925 - 0.064591

P(279<Xbar<305) = 0.647334

Required probability = 0.647334