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8 August, 08:23

The life expectancy (in hours) of an electric bulb is normally distributed with a mean of 5000 and a standard deviation of 1000. Find the probability that a bulb lasts for more than 6300 hours. Round answer to four decimal places. (TIP: calculate z-value and use z-probability distribution table.)

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  1. 8 August, 08:49
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    P (z>1.3) = 0.9032

    Step-by-step explanation:

    We are given:

    Mean = 5000

    Standard deviation = 1000

    x = 6300

    P (x>6300) = ?

    z-score = ?

    z-score = x - mean/standard deviation

    z-score = 6300 - 5000/1000

    z - score = 1300/1000

    z-score = 1.3

    So, P (x>6300) = P (z>1.3)

    Looking at the z-probability distribution table and finding value:

    P (z>1.3) = 0.9032

    So, P (z>1.3) = 0.9032
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