Question

An airline, believing that 4% of passengers fail to show up for flights, overbooks.  Suppose a plane will hold 300 passengers, and the airline sells 310 tickets. What is the probability the airline will not have enough seats, so someone gets bumped?  Round your answer to 3 decimal places.  

Answer 1

here for binomial distribution parameter n=310 and p=1-0.04 =0.96

1st method:

from exact distribution:

P(X>=301)=1-P(X<=300)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.204

from 2nd method:

(Note from Normal approximation:

here mean of distribution=μ=np=		297.60
and standard deviation σ=sqrt(np(1-p))=		3.4502
for normal distribution z score =(X-μ)/σx
since np and n(1-p) both are greater than 5, we can use normal approximation of binomial distribution
therefore from normal approximation of binomial distribution and continuity correction:

P( airline will not have enough seats )=P(301 or more people book arrive):

probability =P(X>300.5)=P(Z>(300.5-297.6)/3.45)=P(Z>0.84)=1-P(Z<0.84)=1-0.7995=0.201