Mathematics 126 -- Calculus for Social Science 2

Discussion 3: Working with Normal Distributions

Or, ``What do you mean I don't have a seat on this flight -- I have a ticket!''

November 7, 2001

Background

We have discussed expected values and variances for both discrete and continuous random variables. We have also discussed the normal probability density functions which give us good models for continuous random variables with a given expected value and variance in many situations. Today we will use ideas about probabilities and the normal distribution to look at a kind of question that many businesses face in planning their activities.

Suppose you are the management of Super-Saver Airline. On average, you have determined that 20% of the people who buy tickets on your flights are ``no-shows'' -- that is, they do not end up using their tickets for whatever reason (usually random events like illness, last-minute changes in plans, ``cold feet'' about flying, etc.) Of course, you do not want your planes flying 80% full on a regular basis because that will result in losses to the company -- your fares are set so low that to break even you need to have flights 90% full at least.

So you quietly institute a policy of overbooking flights -- that is, selling more tickets than there are seats on the flights. That way, even if there are no-shows, your flights will still be closer to the break-even point. But on the other hand, you don't want to overbook flights by too much, since your customers will get disgusted with you if they are turned away from flights on which they have tickets too often(!)

In the discussion today, you will see how to determine answers to questions like:

How many tickets can be sold on a flight while keeping the probability that no one gets ``bumped'' 95% or larger?
How many tickets should be sold to make the probability that a flight is at least 90% full be at least 85%?

Discussion Questions

A) To analyze questions like those above, we need to think about some discrete probability issues first. To each person holding a ticket on a flight, we can associate a discrete random variable X_i with two possible values: 1 with probability .8 (the ticket holder shows up for the flight) and 0 with probability .2 (the ticket holder is a no-show). In probability lingo, X_i is often called a ``Bernoulli trial.'' What are the expected value and variance of each X_i?
We have one discrete random variable as in part A for each ticket holder on a flight. If some number n of tickets are sold, then we will have another random variable X (depending on n) representing the number of people who actually show up. Strictly speaking, this is not a continuous random variable (why not?). But if n is large enough, then we can think of X as approximately a normally distributed continuous random variable with expected value (mean) .8 n, and variance v n, where v is the variance of each of the discrete X_i you found in part A. (This is justified if we think of the individual ticket holders as independent -- their decisions to show or not are not connected, so the variance of the sum of the X_i is the sum of the variances.)
B) On a plane with capacity 300, what is the largest number of tickets that can be sold while keeping the probability that no one gets ``bumped'' .95 or larger? (Note: this means that we want the probability that more than 300 people show up to take the flight to be .05 or less.) Hints: You will need to use the probability condition to set up an equation to be solved for n. Recall that the variance is the square of the standard deviation. Also recall our formulas for converting a normally distributed random variable with given mean and standard deviation into the ``standard'' normal with mean 0 and sigma = 1 in the table in the text.
C) On a plane with capacity 200, what is the smallest number of tickets that can be sold to make the probability that a flight is at least 90% full be at least .85?

Assignment

Solutions will be due in class, Monday, November 12. One set of solutions per group.