Question

A population of values has a normal distribution with μ=114.1μ=114.1 and σ=17σ=17. You intend to draw a random sample of size n=123n=123.

Find the probability that a sample of size n=123n=123 is randomly selected with a mean between 110.6 and 112.6.
P(110.6 < M < 112.6) =

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution:

P(110.6 < M < 112.6) = P(M<112.6) – P(M<110.6)

We are given µ = 114, σ = 17, n = 123

Z = (Xbar - µ)/[σ/sqrt(n)]

First find P(M<112.6)

Z = (112.6 – 114)/[17/sqrt(123)]

Z = -1.4/ 1.532838

Z = -0.91334

P(Z< -0.91334) = P(M<112.6) = 0.180532

(by using z-table)

Now fond P(M<110.6)

Z = (110.6 – 114)/[17/sqrt(123)]

Z = -3.4/1.532838

Z = -2.21811

P(Z<-2.21811) = P(M<110.6) = 0.013274

(by using z-table)

P(110.6 < M < 112.6) = P(M<112.6) – P(M<110.6)

P(110.6 < M < 112.6) = 0.180532 - 0.013274

P(110.6 < M < 112.6) = 0.16726

Required probability = 0.1673