Exam One Answers
Econ 180: Quantitative Methods
Professor Robert J. Lemke
Spring 2006
Add up all of the elements in the set to get 66. As there are 11 elements in the set, the sample mean is then 66 / 11 = 6.
Square all of the elements in the set, and then add them up: 16 + 36 + ... + 100 = 476. The sample variance is then [ 476 - 11*(6*6) ] / 10 = 8.
The coefficient of variation = the sample standard deviation divided by the sample mean = sqrt(8) / 6 = 47.14%.
The z-score for x = 8 is [ 8 - 6 ] / sqrt(8) = sqrt(2) = 0.7071.
The second element of the ordered sample is the second smallest element in the sample, which is 3.
| x | 1 | 2 | 3 | 4 | 5 | ||||||||||
| p(x) | 0.40 | 0.30 | 0.15 | 0.10 | 0.05 |
P ( X > 3 ) = p(4) + p(5) = 0.10 + 0.05 = 0.15.
P ( X <= 2 ) = p(1) + p(2) = 0.40 + 0.30 = 0.70.
P ( X = 5 ) = p(5) = 0.05.
To begin, add two rows to the table.
x 1 2 3 4 5 Total p(x) 0.40 0.30 0.15 0.10 0.05 1.00 xp(x) 0.40 0.60 0.45 0.40 0.25 2.10 x2p(x) 0.40 1.20 1.35 1.60 1.25 5.80 So now we are in a position to answer this and the next question. The mean of the distibution is the sum of the third row, which equals 2.1.
The variance of the distribution is the sum of the fourth row less the mean squared, which equals 5.8 - 2.1*2.1 = 1.39.
This is a standard Binomial question where the number of students who receive financial aid is distributed Binomial with a probability of success of 0.70 and 8 draws. Let X be the number of students in the sample who receive financial aid. Then X ~ Bin(0.7,8). Finally, P(X=7) = (8 choose 7) * (0.7)7 * (0.3)1 = 0.197 = 19.7%.
With the same set-up as the previous part, the question is P(X>=6) = p(6) + P(7) + P(8) = (8 choose 6) * (0.7)6 * (0.3)2 + (8 choose 7) * (0.7)7 * (0.3)1 + (8 choose 0) * (0.7)8 * (0.3)0 = 0.296 + 0.197 + 0.058 = 0.551 = 55.1%.
Note that education and charitable giving are independent. Thus, P(NE and C) = P(NE)xP(C|NE) = (0.6)(0.75) = 0.45, where C = gives to charity and NE = not educated with a college diploma. Thus, 200(0.45) = 90 households give to charity and are not college educated.
Because of independence,P(C|E) = P(C) = 0.75, where E means college educated. Thus, 80(0.75) = 60 of the 80 college-educated households give to charity on a regular basis.
| Number of Job Offers | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| 0 | 1 | 2 | 3 | 4
| GPA | 3.50 - 4.00 | 2 | 8 | 26 | 34 | 3
| 3.00 - 3.49 | 9 | 25 | 78 | 62 | 8
| 2.00 - 2.99 | 10 | 29 | 13 | 4 | 1
| 0.00 - 1.99 | 13 | 14 | 8 | 0 | 0
| | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Summing up the counts in the top three rows, we find that 2 + 8 + ... + 4 + 1 = 312 students graduated with at least a 2.00 grade point average. Alternatively, we know that 347 students were surveyed. As 13 + 14 + 8 = 35 students had a GPA under 2.0, we immediately know that 347 - 35 = 312 students graduated with at least a 2.00 grade point average.
The CDC's analysis is correct, while the student's analysis is wrong. The student is making two errors. First, he is looking at raw counts, which is always a bad idea. One should look at probabilities (or percentages). The second mistake being made is that the goal is not necessarily to get one job offer, but to get many job offers. Looking at the middle two rows, for example, reveals that there are 9 + 25 + 78 + 62 + 8 = 182 students in the 3.0 to 3.49 range and there are 10 + 29 + 13 + 4 + 1 = 57 students in the 2.00 to 2.99 range. Thus, of the first group, 25 / 182 = 13.7% received one job offer, 78 / 182 = 42.9% received two job offers, 62 / 182 = 34.1% received three job offers, and 8 / 182 = 4.4% received four job offers. In comparison, in the 2.00 to 2.99 GPA range, 29 / 57 = 50.1% received one job offer, 13 / 57 = 22.8% received two job offers, 4 / 57 = 7.0% received three job offers, and only 1 / 57 = 1.8% received four job offers. Clearly one is more likely to be offered more jobs if one is in the higher GPA range.
The probability of receiving 2 or more job offers if one's GPA is at least 3.5 is (26+34+3) / (2+8+26+34+3) = 63 / 73 = 86.3%.
The probability of receiving at least one job offer in the range of 3.50 to 4.00 GPAs is (8+26+34+3) / (2+8+26+34+3) = 71 / 73 = 97.3%. The probability of receiving at least one job offer in the range of 3.00 to 3.49 GPAs is (25+78+62+8) / (9+25+78+62+8) = 173 / 182 = 95.1%. The probability of receiving at least one job offer in the range of 2.00 to 2.99 GPAs is (29+13+4+1) / (10+29+13+4+1) = 47 / 57 = 82.4%. The probability of receiving at least one job offer in the range of 0.00 to 1.99 GPAs is (14+8+0+0) / (13+14+8+0+0) = 22 / 35 = 62.9%.
One way to calculate these probabilities is to create pdf tables. Consider the following:Thus, students in the 3.00 to 3.49 range receive 2.194 job offers on average, while students in the 2.00 to 2.99 range receive 1.247 job offers on average.
GPA: 3.00 to 3.49 x 0 1 2 3 4 Total p(x) 9/182 = 0.049 25/182 = 0.137 78/182 = 0.429 62/182 = 0.341 8/182 = 0.044 1.00 xp(x) 0.0 0.137 0.858 1.023 0.176 2.194 GPA: 2.00 to 2.99 x 0 1 2 3 4 Total p(x) 10/57 = 0.175 29/57 = 0.509 13/57 = 0.228 4/57 = 0.070 1/57 = 0.018 1.00 xp(x) 0.0 0.509 0.456 0.210 0.0.72 1.247