The quality control manager at a computer manufacturing company believes that the mean life of a computer is 80 months, with a standard deviation of 10 months. If he is correct, what is the probability that the mean of a sample of 73 computers would be less than 83.28 months? Round your answer to four decimal places.

0.4998

Explanation:

Given that:

Mean life of computer μ = 80

standard deviation σ  = 8

Mean of sample n = 73

x = 83.28

We have to find P(Mean of sample is less than 83.38 months) =

P(X<x=83.38), to do this we have to first determine the z score.

Since we are dealing with multiple samples, the z score will :

z = (x – μ) / (σ / √n)

z = (83.28 - 80)/ (8/√73)

z = 3.28/0.936

z = 3.5

P(X<x=83.38) = P ( Z < z = 3.5)

Use the standard normal table to find P ( Z < 3.5 )

We will have P (Z < 3.5 ) = 0.4998

The probability that the mean of a sample of 73 computers would be less than 83.28 months is 0.4998

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