The quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a standard deviation of 10 months. If he is correct, what is the probability that the mean of a sample of 90 computers would be less than 117.13 months? Round your answer to four decimal places.

P(x < 117.13) = 0.0033

Step-by-step explanation:

Let us assume that the mean life of the computers are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

x = lifetime of the computers in months.

µ = mean lifetime

σ = standard deviation

n represents the number of samples

From the information given,

µ = 120 months

σ = 10 months

n = 90

The probability that the mean life of a sample of 90 computers would be less than 117.13 months is expressed as

P(x < 117.13)

For x = 117.13

z = (117.3 - 120)/10/√90

z = - 2.87/1.0541 = - 2.72

Looking at the normal distribution table, the probability corresponding to the z score is 0.0033

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step-by-step explanation:

the answer is a

step-by-step explanation:

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