Econ 57
Gary Smith
Fall 2009
Midterm Answers
1.
a. Scatter diagram.
b. Scatter diagram.
c. Boxplot.
d. Time series.
e. Scatter diagram.
2.
Higher Mean 
Higher Median 
Higher Standard Deviation 

Case 1  same 
same 
same 
Case 2  B 
same 
B 
Case 3  same 
same 
B 
Case 4  B 
B 
same 
3.
a. Women may be less likely to marry and more likely to divorce men who are in poor health.
b. Maybe men who don’t marry are in bad health to begin with. Maybe men who marry and have their health deteriorate the most get divorced.
c. For a controlled experiment, we could allow a random sample of men to marry and a random sample of married men to divorce.
4. Collecting data:
Taxable Income Bracket  Observations  Frequency  Frequency/Interval Width 
$0 to $16,000  1 
0.1 
0.1/16,000 
$16,000 to $64,000  6 
0.6 
0.6/(64,000  16,000) = 0.2/16,000 
$64,000 to $128,000  3 
0.3 
0.3/(128,000  64,000) = 0.075/16,000 
$128,000 to $208,000  0 

$208,000 to $350,000  0 

$350,000 to no limit  0 

Total  10 
1.0 
5. The minimum is $15,000; the maximum is $120,000; the median is $48,000; the first quartile is $32,000; and the third quartile is $80,000. Here is a box plot:
6.
a. This is a binomial problem with π = 0.1, n = 8, and x ≥ 3:
b. The probability of at least one lucky dollar among 4 dollar bills is equal to 1 minus the probability of no lucky dollars:1  (1  0.813)^4 = 0.999
7. Using Bayes Rule,
Using a contingency table with 100,000 women:
++ 
not ++ 
Total 

malignant  1,000*0.8*0.8 = 640 
1,000  640 = 360 
1,000 
benign  99,000*0.1*0.1 = 990 
99,000  990 = 98,010 
99,000 
Total  1,630 
98,370 
100,000 
The probability that a woman who gets two positive readings has a malignant lump is 640/1,630 = 0.3926.
8. The probabilities are
Number of Dice 
Probability 
0 
(5/6)(5/6)(5/6) = 0.5787 
1 
3(1/6)(5/6)(5/6) = 0.3472 
2 
3(1/6)(1/6)(5/6) = 0.0694 
3 
(1/6)(1/6)(1/6) = 0.0046 
Total 
1.0000 
The expected value of the payoff is about an 8 percent loss per dollar wagered: 0(0.5787) + 2(0.3472) + 3(0.0694) + 4(0.0046) = 0.921.
9.
a. The probability that the 6 numbers chosen on Sept 10 will be the same 6 numbers chosen on Sept 6 (not necessarily in the same order) is equal to the probability that the 6 numbers chosen on Sept 10 will be any prespecified 6 numbers:
b. For an event with probability P, the expected wait until it happens is 1/P. Here, after the first lottery there is a probability P = 1/(5,245,787) probability that any lottery will match the previous one. So, the expected wait is 1 + 1/P = 5,245,787.
10.
a. (ii) because her Z value is Z = (70  66)/2 = 2 and the absolute value of his Zvalue is less than 2: Z = (66  70)/2.5.
b. (i) because A is more likely than A and B, unless B is certain.
c. (ii) because A or B is more likely than A, unless B is impossible.
c. (ii) because is (i) is only one of the many ways that (ii) could be true. Think of it this way: (ii) is (i) and her height is more than 66 inches, and we know that A is more likely than A and B, unless B is certain.