Question

A simple random sample of size nequals40 is obtained from a population with muequals66 and sigmaequals14. ​(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample​ mean? Assuming that this condition is​ true, describe the sampling distribution of x overbar. ​(b) Assuming the normal model can be​ used, determine ​P(x overbarless than69.6​). ​(c) Assuming the normal model can be​ used, determine ​P(x overbargreater than or equals67.8​).

Answer 1

Solution:

We are given

n = 40

µ = 66

σ = 14

Part a

It must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample​ mean, that the sampling distribution of the sample mean follows an approximate normal distribution, although the original population have or don’t have normal distribution. If population doesn’t have normal distribution, sampling distribution will follow normal distribution. There is no effect on sampling distribution whether population is normal or not.

The mean and standard deviation for the sampling distribution of Xbar are given as below:

Mean of sampling distribution = µ = 66

Standard deviation of sampling distribution = σ/sqrt(n) = 14/sqrt(40) = 2.213594

Part b

Here, we have to find P(Xbar<69.6)

Z = (Xbar - µ) / [σ / sqrt(n) ]

Z = (69.6 – 66) / [14/sqrt(40)]

Z = 3.6 / 2.213594

Z = 1.626314

P(Z< 1.626314) = P(Xbar<69.6) = 0.948059

(By using z-table)

P(Xbar<69.6) = 0.948059

Required probability = 0.948059

Part c

Here, we have to find P(Xbar ≥ 67.8)

P(Xbar ≥ 67.8) = 1 – P(Xbar<67.8)

Find P(Xbar<67.8)

Z = (Xbar - µ) / [σ / sqrt(n) ]

Z = (67.8 – 66) / [14/sqrt(40)]

Z = 1.8 / 2.213594

Z = 0.813157

P(Z< 0.813157) = 0.791936

(by using z-table)

P(Xbar<67.8) = 0.791936

P(Xbar ≥ 67.8) = 1 – P(Xbar<67.8)

P(Xbar ≥ 67.8) = 1 – 0.791936

P(Xbar ≥ 67.8) = 0.208064

Required probability = 0.208064