An airline has a policy of booking as many as 150 persons on a plane that seats 140. past studies indicate that only 85% of booked passengers show up for their flight. find the probability that if the airline books 150 persons for a 140-seat plane, not enough seats will be available.

Binomial distribution will be used to model situation because all following criteria are satisfied:
1. Bernoulli trials, i.e. exactly two possible outcomes (show up or not)2. Number of trials is known before and constant throughout the experiment, i.e. independent of outcomes (150 booked)3. All trials are independent of each other. (assumed from context)4. Probability of success is known, and remain constant throughout trials. (85%)Since all criteria are satisfied, we can model with binomial distribution, where the probability of x successes out of N trials each with probability of success p is given byP(x)=C(N,x)(p^x)(1-p)^(N-x)and,C(N,x) is number of combinations of selecting x objects out of N.
P(>140)=P(141)+P(142)+...+P(150)
=.00035620+0.00012783+0.00004056+0.00001117+0.00000262+0.00000051+0.00000008+0.00000001+0.00000000
=0.00053907

The above calculation is not very practical unless we have a statistical calculator or a software.

For practical hand calculations, we usually use the binomial approximation.
mean=np=150*0.85=127.5
variance = np(1-p)=150*.85*.15=19.125  (>10, possible to use binomial approximation, but accuracy is marginal).
We also need to use the continuity correction, which means that the portion above 140 is counted as 140.5 to 150 passengers.  Then we calculate
Z=(140.5-127.5)/sqrt(19.125)=2.973
P(x>=140.5)
=P(z>=Z)
=1-P(z<Z)
=1-P(z<2.973)
=1-0.99852
=0.00148
The answer is quite a bit off from the exact value of 0.000539, but an approximation is an approximation especially for skew distribution like in the present case (p=0.85 far from 0.5, and x=140 far from mean).

We need the pictures of the graphs