Question

A population has a mean of 74 with a standard deviation of 9.8.

a) What is the probability that one element of the population selected at random is between 70 and 91?

b) What is the probability that a random sample of 36 from this population has a sample mean between 73 and 79?

Answer 1

Solution:

Part a

We are given

µ = 74

σ = 9.8

We have to find P(70<X<91) = P(X<91) – P(X<70)

Find P(X<91)

Z = (X - µ)/σ

Z = (91 - 74)/9.8

Z = 1.734693878

P(Z<1.734693878) = 0.958602479

(by using z-table)

Now find P(X<70)

Z = (70 - 74)/9.8

Z = -0.408163265

P(Z<-0.408163265) = 0.341576908

(by using z-table)

P(70<X<91) = 0.958602479 - 0.341576908 = 0.617025571

Required probability = 0.617025571

Part b

We have to find P(73<Xbar<79) = P(Xbar<79) – P(Xbar<73)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = (79 - 74)/(9.8/sqrt(36)) = 3.061224

Z = (73 - 74)/(9.8/sqrt(36)) =-0.61224

P(73<Xbar<79) = P(Z<3.061224) – P(Z<-0.61224)

P(73<Xbar<79) = 0.998898 - 0.270188 = 0.72871

(by using z-table)

Required probability = 0.72871