Suppose that 15% of people dont show up for a flight, and suppose that their decisions are independent. how many tickets can you sell for a plane with 144 seats and be 99% sure that not too many people will show up.

The book says to do this by using the normal distribution function and that the answer is selling 157 tickets.

The binomial distribution B(n, p) is approximately close to the normal i.e. N(np, np(1 − p)) for large 'n' and for 'p' and neither too close to 0 nor 1 .

Now, Let us assume 'n' = n

and we are given

p=0.15

So now B(n,0.15n) follows Normal distribution

u=n

[tex]\sigma^{2}[/tex] = 0.15n

We have to calculate P(X<144) with 99% accuracy

P(X<144) = P(Z<z)

where;

z= [tex](144-\bar{X})\div \sigma[/tex]

z score for 99% is 2.33

i.e.

[tex](144-\bar{X})\div \sigma = (144-n)/\sqrt{np} = 2.33\\\ (144-n)^{2} = \ np*2.33^{2}\\\ 20736 +n^{2} - 288n = 0.15n*5.43\\\ n^{2} - 288.81n + 20736=0[/tex]

solving this we will get one root nearly equal to 157 and other root as 133

Hence the answer is 157.