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

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