Convoy Airlines has recently instituted a policy of booking 400 passengers for a flight on an airplane that only has a capacity of 350. Recent studies show that only 75% of passengers booked for a particular flight actually arrive for the flight. What is the probability that if Convoy Airlines books 400 passengers for a particular flight that not enough seats will be available?

The answer i came up with is 1.06 which i think is wrong.

I used the binomial formula nCx (p)^x (q)^n-x to solve this problem, can some one help me with this.

The number who turn up has a binomial distribution B(400, 0.75) the

probability that the flight has insufficient seats is the probability that

351 or more passengers turn up:

p(insuff) = sum_{n=351 to 400} b(n; 400, 0.75)

and b(n; 400, 0.75) = 400!/[n!(400-n)!] 0.75^n 0.25^(400-n)

Now we can use brute force here to compute the sum, or we can use the

normal approximation to the binomial distribution. For this we need the mean

and standard deviation for the binaomial distribution, which are:

mu = 400*0.75 = 300

sigma = sqrt(400*0.75*0.25) ~= 8.66

So now we compute the z score for n=350.5:

z = (350.5 - 300)/8.66 ~= 5.83

Which there is no need to look up in a normal table, the probability of

getting a z score this size or larger is negligable (ie close to 0).

Hence the probability that their are more passengers than seats is ~=0.0

RonL