



 
First note that any solution to this problem either needs more data or some (not necessarily realistic) assumptions. In particular, you are presumably meant to assume that noshows are completely independent [a dangerous assumption, as many people fly in groups] and that the airline does not use this knowledge to overbook planes [in which case the number of passengers would exceed 175.] With these assumptions, the number of noshows is a binomial RV. What is N? What is p? This gives the easiest answer for (c). For the other questions you probably want to use a normal approximation, using one or two table lookups instead of a long sum. Approximate the distribution by a normal RV with the same mean and standard deviation (your text will have formulae for the mean and SD of a binomial RV.) And break halfway between the extreme legal values and the first illegal ones. So if your mean was 20 and your SD was 5, you would find P(13 ≤ X ≤ 23) by breaking at 12.5 (=20  1.5 × 5) and 23.5 (= 20 + 0.7 × 5) and finding as the probability that a standard normal RV is between 1.5 and 0.7. Good Hunting!  


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