Question

A certain population has a mean of 300 and a (population) standard deviation of 40. If a sample of size 200 is taken, what is the probability that the sample average is within plus-or-minus 2 of the population average?

Answer 1

Solution:

We are given

µ = 300

σ = 40

n = 200

We have to find P(300 - 2 < x̄ < 300+2) = P(298 < x̄ < 302)

P(298 < x̄ < 302) = P(x̄<302) - P(x̄<298)

Find P(x̄<302)

Z = (x̄ - µ)/[σ/sqrt(n)]

Z = (302 - 300)/(40/sqrt(200))

Z = 0.707107

P(Z<0.707107) = P(x̄<302) = 0.76025

(by using z-table)

Now find P(x̄< 298)

Z = (x̄ - µ)/[σ/sqrt(n)]

Z = (298 - 300)/(40/sqrt(200))

Z = -0.707107

P(Z<-0.707107) = P(x̄<298) = 0.23975

(by using z-table)

P(298 < x̄ < 302) = P(x̄<302) - P(x̄<298)

P(298 < x̄ < 302) = 0.76025 - 0.23975

P(298 < x̄ < 302) = 0.5205

Required probability = 0.5205