Question

A flight company is interested in evaluating how many flights that are overbooked. Assume that whether one flight is overbooked or not has no influence on whether other flights are overbooked. The probability of an overbooked flight is 20%. If the company evaluates 100 flights, what is the probability that over 32 flights are overbooked?

Answer 1

Solution:

Given that,

P = 0.20

1 - P = 0.80

n = 100

Here, BIN ( n , P ) that is , BIN (100 , 0.20)

then,

n*p = 100*0.20 = 20 > 5

n(1- P) = 100*0.80 = 80 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 80

Standard deviation = =n*p*(1-p) = 100*0.20*0.80 = 16= 4

We using countinuity correction factor

P(x > a ) = P( X > a + 0.5)

P(x > 20.5) = 1 - P(x < 20.5)

= 1 - P((x - ) / < (20.5 - 20 ) / 4)

= 1 - P(z < 0.125)

= 1 - 0.5497   

= 0.4503

The probability that over 32 flights are overbooked is 0.4503