Question

An airline believing that 5% of passengers fail to show up for flights, overbook. Suppose a plane holds 255 passengers and the airlines sells 265 tickets, what's the probability the airline will not have enough seats so someone gets bumped?

b)find the exact probability using the binomial model

c)find an approx. probability using the normal approximation to the binomial

Answer 1

b)

Using exact Binomial distribution,

X~Binomial (n,p)

Where n = 265 , p = 0.95 (probability of show up) (Since 5% fail to show up)

P(X) = ⁿC_x p^x ( 1 - p)^n-x

P( X > 255) = 1 - P( X <= 255)

= 1 - 0.8567

= 0.1433

b)

Using normal approximation,

Mean = np = 265 * 0.95 = 251.75

Standard deviation = sqrt ( np( 1 - p))

= sqrt( 265 * 0.95 * 0.05)

= 3.5479

Using normal approximation,

P( X > x) = P( Z > x+0.5 - mean / SD)

So,

P( X > 255) = P( Z > 255.5 - 251.75 / 3.5479)

= P( Z > 1.0570)

= 0.1453