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2 August, 00:37

A company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 20 components and accept the whole batch if there are fewer than 3 defectives. If a particular shipment of thousands of components actually has a 8% rate of defects, what is the probability that this whole shipment will be accepted? Round to three decimal places.

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  1. 2 August, 00:53
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    Answer: Probability that this whole shipment will be accepted is 0.629

    Step-by-step explanation:

    The company randomly select and test 20 components and accept the whole batch if there are fewer than 3 defectives.

    Applying normal distribution,

    z = (x-u) / s

    Where

    n = number of components that were tested.

    s = standard deviation

    u = mean = np

    p = probability that the component has defect

    q = probability that the component does not have defect

    x = r = the number of components

    From the information given,

    p = 8/100 = 0.08

    q = 1-q = 1-0.08 = 0.92

    n = 20

    u = 20*0.08 = 1.6

    s = √20 * 0.08 * 0.92 = √1.472

    s = 1.21

    Probability that this whole shipment will be accepted will be probability that fewer than 3 components have defects. This means that it must be lesser than or equal to 2 components.

    P (x lesser than 3) = P (x lesser than or equal to 2). Therefore,

    x = 2

    z = (2-1.6) / 1.21 = 0.33

    Looking at the normal distribution table, the corresponding z score is 0.629
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