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24 February, 10:10

A local concert center reports that it 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. 24 February, 10:15
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    Given Information:

    Number of advanced reservations = n = 150

    Probability of success = p = 15% = 0.15

    Required Information:

    Probability that there will be fewer than 20 no-shows = ?

    Answer:

    P (X < 19.5) = 24.83 %

    Explanation:

    The mean is given by

    μ = np

    Where n is the number of advanced reservations and p is the probability of no shows

    μ = 150*0.15

    μ = 22.5

    since np ≥ 5 and n (1-p) = 127.5 ≥ 5 we can proceed with normal approximation to binomial distribution

    The standard deviation is given by

    σ = √npq

    Where q is given by

    q = 1 - p

    q = 1 - 0.15

    q = 0.85

    σ = √ (150*0.15*0.85)

    σ = 4.37

    We want to find out the probability of fewer than 20 no shows among the 150 advanced reservations given the success rate of no shows is 15% and correction factor of 0.5

    X = 20 - 0.5 = 19.5

    P (X < 19.5) = P (z < (x - μ) / σ)

    P (X < 19.5) = P (Z < (19.5 - 22.5) / 4.37)

    P (X < 19.5) = P (Z < - 0.68)

    The z-score corresponding to - 0.68 is found from the z-table that is 0.2483

    P (X < 19.5) = 0.2483

    P (X < 19.5) = 24.83%

    Therefore, there is 24.83% probability that there will be fewer than 20 no-shows among the 150 advanced reservations having a 15% rate of no-shows.
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