The no-show rate for passengers with reservations on a flight

[rudyx61]

Was hoping someone could help me out with a question. It was on a online test and wasnt sure how to do it.

The question was as follows:

The no-show rate for passengers with reservations on a flight run by wizair is 16%. The next flight has 42 reservations.

Find the probability of there being 4 or more no-shows on this flight

 

[SEngstrom]

You need to consider a binomial distribution with n=42 and p=0.16. The cumulative distribution up to 3 is the probability of there being less than four no-shows: Reaching for Mathematica to put a number to it:
dist = BinomialDistribution[42, 0.16]
1 - CDF[dist, 3]
(0.921048, quite likely)

 

[rudyx61]

ok but how would i do that on paper, say its a written test with no access to mathematica

 

[SEngstrom]

add up the four first terms of the sum for the CDF: (n choose k)*p^k*q^(n-k) for k=0,1,2,3.

 

[rudyx61]

yeah i get it now, thanks

 

[HallsofIvy]

By the way, SEngstrom is assuming, probably correctly, that saying "The no-show rate for passengers with reservations on a flight run by wizair is 16%" means that anyone passenger has a .16 probability of not showing up- so the probability distribution is a binomial distribution with mean .16.

I would have been inclined to interpret it as meaning that the mean value of the number of passenger who do show up is .16 but that leaves the distribution itself unknown. I might then have been inclined to use a normal distribution but, of course, for reasonably large number of people (such as 42) that well approximates the binomial distribution.

In fact, my reason for choosing the normal distribution (and the reason it is so common) is that, by the "Central Limit Theorem", in a case like this, the binomial distribution is a good approximation to just about every distribution!