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7 March, 21:34

An airline finds that 5% of the persons who make reservations on a certain flight do not show up for the flight. if the airline sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

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  1. 7 March, 21:59
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    Probability of success (showing up) = 1-0.05=0.95 is constant and known.

    Trials are Bernoulli (show or no show).

    Trials are independent and random (assumed from context)

    Number of trials is known, n=160.

    All being satisfied, we can then model with binomial distribution, where

    P (x) = C (n, x) p^x * (1-p) ^ (n-x)

    where C (n, x) = n! / (x! (n-x) !)

    Here we look for

    P (X<=155) = P (X=0) + P (X=1) + P (X=2) + ... + P (X=155)

    =0.9061461 (using technology, or add up 156 values calculated, or read from binomial distribution table).

    Alternatively, the normal approximation can be used, when n is large.

    mean=np=160*0.95=152

    standard deviation=sqrt (np (1-p)) = 2.75681

    Apply continuity correction, x=155.5

    Z = (155.5-152) / 2.75681=1.26958

    P (z<=Z) = 0.89788 (read from normal distribution tables)

    Error = (0.89788-0.9061461) * 100%=-0.83%

    The approximation is considered good considering p=0.95 is quite skewed, but compensated by n>>50.
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