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11 November, 17:50

Suppose that 11% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked.

probability x is at most 30? less than 30? between 15 and 25?

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  1. 11 November, 20:00
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    P = 0.11

    1-p =.89

    n = 200

    mean = n*p = 22

    std dev = n*sqrt (p (1-p) / n) = 4.425

    find z values for standard normal distribution

    Z = (30-22) / 4.425 = 1/808

    look up table to find P (Z less than 1.808) which is same P (x less than 30)

    for 2nd part, you need 2 Z values

    Z1 = (25-22) / 4.425 = 0.678

    Z2 = (15-22) / 4.425 = - 1.58

    P (-1.58 less than Z less than 0.678) = P (Z less than. 678) - P (Z less than - 1.58)
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