Calculate the expected proportion of denied boardings 
An airline company knows that 20% of its passengers will not show up
  for their scheduled flights. A plane has 80 seats. If the airline
  company accepts 100 bookings, what is the expected proportion of
  passengers to which the company will deny boarding?

This is what I thought:
The number of people who will show up is given by a binomial distribution: $S\sim Bin(100,0.8)$ which can be approximated by a normal distribution $S\sim N(80,16)$. The number of people who will not be able to fly is: $max(0,S-80)$. So the answer should be:
$\frac{E[max(0,S-80)]}{E[S]}$
Of course, $E[S]=80$.
How can I calculate $E[max(0,S-80)]$?         
 A: If we take the normal approximation to the binomial as is, half the time no passengers will be denied boarding and half the time we'll be looking at a value from the right half of a normal distribution. 
[The exact calculation using normal approximation has been left out deliberately -- there should be something for you to still do, but if you do it correctly it comes out a little below 1.6]
Here's a simulation:
 mean(pmax(rbinom(100000, 100, 0.8)-80,0))
[1] 1.59157

Exact binomial calculation gives 
  sum(dbinom(80:100,100,.8)*0:20)
 [1] 1.588803

whuber's comment about the continuity correction (which I mentioned earlier) is right -- I was misled by the visual appearance of the binomial near the peak to think it would help given we are calculating with a cut-off near the peak of the binomial, but it doesn't. It does tend to help - even in the tail - when the parameter is close to 1/2 but we're not in that situation, so more caution on my part (like double checking the correction would actually improve the approximation) was called for.
