Matthew S. answered 03/14/20
PhD in Ops Research with industry experience in Data Science
a) Each instance is a success (on time) or a failure (not on time). The number of successes goes according to a binomial distribution.
b) Let X = # successes in 24 trials. P(X=14) = (24 choose 14) * .7514 * .2510 ≈ .0333
In words, this says
(1) That the probability of any sequence with 14 successes and 10 failures is Prob(success)14 * Prob(failure)10, and
(2) According to the binomial theorem, there are (24 choose 14) such sequences
c) P(X < 14) is the sum of P(X=0) + P(X=1) + ... + P(X=13) ≈ .0213
where each of the terms in the sum is computed in similar fashion to what we did for part b)
d) P(X ≥ 14) = 1 - P(X < 14) ≈ 0.979
e) P(12 ≤ X ≤ 14) = P(X = 12) + P(X = 13) + P(X = 14) ≈ .0526
The probabilities P(X=14) and P(X<14) are small because the probability of an individual flight being on time is much greater than 1/2. This exercise quantifies how much the distribution is "skewed to the right".