1. Suppose that the unemployment rate in the general population is 5%. Suppose also that in families with two working adults the probability that one adult is unemployed is independent of whether or not the other adult is unemployed. What percentage of families with two working adult members have at least one adult unemployed? Show your reasoning.
Solution: For a family with two working adults there are four possible outcomes. In the table below U stands for 'unemployed' and E stands for 'employed'.
| Outcome | UU | UE | EU | EE |
|---|---|---|---|---|
| Probability | (0.05)(0.05) | (0.05)(0.95) | (0.95)(0.05) | (0.95)(0.95) |
There are two ways to compute
Note: Many people tried to use this argument:
That argument does not work because the two events on the right are not disjoint. Two events are disjoint if they can not possibly happen at the same time. Since it is possible for both adults to be unemployed at the same time, these two events are not disjoint and the addition rule is not valid.
The events are independent, which means that knowing whether or not the first adult is unemployed tells you nothing about whether or not the second adult is unemployed. Independent is not the same thing as disjoint.
2. The table below shows a distribution for people living in households of various sizes.
| Household Size | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Proportion | 0.25 | 0.32 | 0.17 | 0.15 | 0.07 | 0.03 | 0.01 |
How many people live in the average household?
Solution: The problem is essentially asking us to compute the mean of a probability distribution. If X is the number of members in a randomly selected household,
3. A random variable Z has mean υZ = 1.2 and variance ¤Z2 = 0.4. We form a new random variable W from Z by repeating the process Z two times and averaging the results. (W = (Z + Z)/2). Using the formulas
compute the mean and the variance for the random variable W.
Solution: If we introduce the random variable U = Z/2, the formulas tell us that
In terms of U, W = U + U. The formulas tell us that
This result tells us that when we average together two trials of a random variable the mean stays the same, but the variance decreases by a factor of 1/2.
Note: A number of people tried to argue that since
the mean and the variance would stay the same. The flaw in that argument is that Z is a random variable, not an ordinary mathematical variable. The statement above is certainly true for ordinary variables. However, a random variable represents a process that generates a stream of random numbers. In the case of a random variable, (Z+Z)/2 means "use the random process Z to generate a pair of random numbers and then average them together". This produces a stream of random numbers that looks slightly different than the stream of random numbers generated by Z alone.