Question

the mean per capita income is 21053 dollars per annum with a standard deviation of 805 dollars per annum, what is the probability that the samele mean would differ from the true mean by greater than 167 is a sample of 71 persons is selected

Answer 1

Solution:

We are given

µ = 21053

σ = 805

n = 71

We have to find P((Xbar - µ)< -167) + P((Xbar - µ)> 167)

First find P((Xbar - µ)< -167)

Z = (Xbar - µ)/[σ/sqrt(n)]

Z = -167/(805/sqrt(71))

Z = -1.74803

P(Z<-1.74803) = 0.040229

P((Xbar - µ)< -167) = 0.040229

Now, find P((Xbar - µ)> 167)

P((Xbar - µ)> 167) = 1 - P((Xbar - µ) < 167)

Z = 167/(805/sqrt(71))

Z = 1.74803

P(Z<1.74803) = 0.959771

P((Xbar - µ) < 167) = 0.959771

P((Xbar - µ)> 167) = 1 - 0.959771

P((Xbar - µ)> 167) = 0.040229

P((Xbar - µ)< -167) + P((Xbar - µ)> 167) = 0.040229 + 0.040229

P((Xbar - µ)< -167) + P((Xbar - µ)> 167) = 0.080458

Required probability = 0.080458

answered by: anonymous