Question

An airline 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.

Answer 1

Solution :

Given that,

p = 0.15

q = 1 - p = 1 - 0.15 = 0.85

n = 150

Using binomial distribution,

= n * p = 150 * 0.15 = 22.5

= n * p * q = 150 * 0.15 * 0.85 = 4.37

P(x < 20 )

= P[(x - ) / < (20 - 22.5) / 4.37]

= P(z < -0.57)

Using z table,

= 0.2843

answered by: anonymous