Statistics

An airline reports that if has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows.

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  1. Mean = np = 150 * .15 = ?

    Standard deviation = √npq = √(150 * .15 * .85) = ?

    Note: q = 1 - p

    I'll let you finish the calculations.

    Once you have the mean and standard deviation, use z-scores:
    z = (x - mean)/sd
    Note: x = 20

    After you calculate z, check a z-table to find your probability. (Remember that the problem is asking for "fewer than 20" when you check the table.)

    I hope this will help get you started.

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