Question

In: Statistics and Probability

A commonly used practice of airline companies is to sell more tickets than actual seats to...

A commonly used practice of airline companies is to sell more tickets than actual seats to a particular flight because customers who buy tickets do not always show up for the flight. Suppose that the percentage of no shows at flight time is 2%. For a particular flight with 380 seats, a total of 384 tickets were sold. Use normal approximation to find the probability that

(a) at most 375 passengers will show up.

(b) the airline overbooked this flight.

(c) between 4 and 8 passengers (both inclusive) will not show up

I just want to check my Answers: for (a) my ans: 0.3821 (b) my ans: 0.0643

If my answers are wrong then please give the correct solution

Expert Solution

Answer:

Given that:

Suppose that the percentage of no shows at flight time is 2%. For a particular flight with 380 seats, a total of 384 tickets were sold.

n= 384 , p= 0.02
here mean of distribution=

and standard deviation

for normal distribution z score =(X-μ)/σx
therefore from normal approximation of binomial distribution and continuity correction:

a) at most 375 passengers will show up.

P( at most 375 passengers will show up ) =P(At least 5 did not show up)

probability =

b) The airline overbooked this flight.

the airline overbooked this ﬂight =P(at most 3 did not show up)

probability =

c) between 4 and 8 passengers (both inclusive) will not show up

probability

orchestra answered 1 year ago

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