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 no-shows 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 no-shows 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!
RD