Econ 57 Fall 2001 Midterm Answers

1. The probability of five heads is(1/2)5 = 1/32; the probability of five tails is (1/2)5 = 1/32. Therefore, the probability of five heads or five tails is 2/32 = 0.0625.

2. Using Bayes’ Rule,

3. All three cases can be answered by seeing which city has the SMALLER standardized Z value:

a. ZA = (100 - 80)/20 = 1, ZB = (100 - 80)/30 = 2/3. So, B.
b. ZA = (100 - 80)/20 = 1, ZB = (100 - 70)/20 = 3/2. So, A.
c. ZA = (100 - 80)/20 = 1, ZB = (100 - 70)/30 = 1. So, equal.

4. The total area of the bars is not 1.

5. Because Mark has a higher probability of snow, he should bet on snow and Mindy should bet against snow. So, we will make the bet that Mindy pays Mark \$X if it snows and Mark pays Mindy \$Y if it doesn’t.

From Mark’s viewpoint, his expected value is (X)(4/5) + (-Y)(1/5). For this to be positive, he needs X/Y > 1/4. From Mindy’s viewpoint, her expected value is (-X)(2/3) + (Y)(1/3). For this to be positive, she needs X/Y < 1/2. Thus, any bet with 1/4 < X/Y < 1/2 will give both persons positive expected values. For example, Mindy pays Mark \$1 if it snows and Mark pays Mindy \$3 if it doesn’t.

a. scatter diagram
b. scatter diagram
c. side-by-side boxplots
d. time series graph
e. scatter diagram

7. Marilyn’s correct answer: “I’d choose heads/tails (HT), because I’d win three out of four times! That’s because there are four different sequence combinations: HH, HT, TH, and TT. If tails/tails (TT) were to appear at the very start, you’d win, but that would happen only one-fourth of the time. For TT to appear any time afterward, it would have to be preceded by H. Which means that I’d win before you ever saw your sequence come up at all!”

8. Using the binomial distribution with p = 0.2 and n = 5:

a. P[X = 5] = 0.25.
b. Similarly,

9. The mean is 2.3, the median is 2, the standard deviation is 1.