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18 January, 08:26

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.

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  1. 18 January, 08:48
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    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
    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
    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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