The quality control manager at a computer manufacturing company believes that the mean life of a computer is 88 months, with a variance of 81. If he is correct, what is the probability that the mean of a sample of 84 computers would differ from the population mean by less than 1.39 months? Round your answer to four decimal places.

Probability = 0.8444

Step-by-step explanation:

To solve this problem, we need to obtain the z-value which corresponds to the probability that the mean of a sample of 84 computers would differ from the population mean by less than 1.39 months

Let x' denote the mean life of the computer.

Mean = 88 months

Variance = 81

Thus, standard deviation = √81 = 9

Thus, we obtain;

P(|x - μ| < 1.39) = P(-1.39 > x < 1.39)

For x' = - 1.39,

z = -1.39/(9/√84)

z = -1.39/0.982

z = -1.42

For x' = 1.39,

z = 1.39/(9/√84)

z = 1.39/0.982

z = 1.42

Thus, P(|x - μ| < 1.39) =

P(−1.42 < z < 1.42)

From the Z-distribution tables i attached, we obtain ;

0.92220 - 0.07780 = 0.8444

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