Econ_221-2_F2007_Assignment_2_Answers

Concordia University Department of Economics ECON 221/2 – Sections A, B, BB, C Instructors: N. Islam, I. Jamali, G. Tsoublekas Fall 2007 - ASSIGNMENT 2 - ANSWERS Due at noon on Friday, October 12, 2007 1. In a recent survey, 90 top executives were asked how many hours they spend each year in community service. The data are presented below. [The calculations and the graph for this exercise must be done on Excel. As soon as you get the data on class intervals and frequencies that are shown below, you will create a table on Excel with the necessary detailed column calculations that will be used as a basis to answer sub-questions (a), (d) and (e). You will paste your table here.] (20 points) a) Construct a cumulative frequency distribution, and a cumulative relative frequency distribution. (3 points) See table above. b) Using Excel, draw a histogram of the frequencies and paste it in the space provided below. (4 points)

c) On the basis of the shape of the histogram, make a rough assessment about symmetry in the data and the implication for the mean time spent in community services. (3 points) The histogram indicates a fairly symmetric distribution. Therefore, the mean is not distorted by skews (or outliers) in the data. d) Using the appropriate formula for grouped data, calculate the mean time spent in community services. Use the table provided above for your computations. (3 points) Using the data calculated above, Mean =  m i * f i / n = 6300 / 90 = 70 e) Using the appropriate formulas for grouped data, calculate the variance and the standard deviation. Use the table provided above for your computations. (4 points) Using the data calculated above, Variance = s 2 =  (m i –mean) 2 f i / ( n-1) = 76000 / 89 = 853.93 Standard Deviation =  853.93 = 29.22 f) Calculate the Coefficient of Variation. (3 points) The coefficient of variation expresses in a percentage form the standard deviation in terms of the mean. CV = (s / mean) * 100 = ( 29.22 / 70)*100 = 41.7% 2. A general contractor has submitted two bids for two projects; A and B . The probability of getting project A is 0.60. The probability of getting project B is 0.75. The probability of getting at least one of the projects is 0.85. (15 points) a. Construct the table or tree of joint probabilities. (3 points) We know that P (YA  YB) = P (YA) + P (YB) – P (YA  YB) = 0.60 + 0.75 – P (YA  YB) = 0.85 Therefore, P (YA  YB) = 0.50 Project B Project A Total Yes (YA) No (NA) Yes (YB) No (NB) 0.50 0.10 0.25 0.15 0.75 0.25 Total 0.60 0.40 1.00 2

b. Find the probability that the price of stock A will be $40 while the price of stock B is $55. (2 points) P (A=40  B=55) = P (A=40  B=55) / P (B=55) = 0.05 / 0.20 = 0.25 c. Find the probability that either of the two stock prices will be at their maximum level. (3 points) P (B=65  D=70) = P (B=65) + P (D=70) – P (B=65  D=70) = 0.35 + 0.35 – 0.00 = 0.70 d. Find the probability that both stock prices will be up to $60. (3 points) P (B≤60  A≤60) = P (A=40  B=50) + P (A=50  B=50) + P (A=60  B=50) + P (A=40  B=55)+ + P (A=50  B=55) + P (A=60  B=55) + P (A=40  B=60) + P (A=50  B=60) + P (A=60  B=60)= = 0.00 + 0.00 + 0.05 + 0.05 +0.00 +0.05 + 0.10 + 0.05 +0.00 = 0.30 e. Find the average price and variance of stock B. (3 points) E(Y) = μ y =  Y i * P(Y i ) = 58.25See calculations in table above. 35.69 See calculations in table above. Standard Deviation =  35.69 = 5.974 f. Find the average price and variance of stock A. (3 points) E(X) = μ x =  X i * P(X i ) = 55.00 See calculations in table above. 165.00 See calculations in table above. Standard Deviation =  165.00 = 12.845 g. Can you tell which of the two stocks offers the greatest potential for capital gains? (3 points) Stock A has a larger variance. Therefore, it offers a greater opportunity for capital gains as well as capital losses. 4

h. Find the covariance and correlation coefficient of the prices of the two stocks. (4 points) = = 3150 – (58.25*55.00) = – 53.75 This implies that there is a strong negative relationship between the prices of the two stocks. i. Find the average value of the entire portfolio. (3 points) This requires that we create a new variable (a combination of the two) as follows: W = 10 X + 10 Y = 10*55.00 + 10*58.25 = 1132.5 j. Find the standard deviation of the value of the entire portfolio and calculate a 68% range of its value. (4 points) (100 * 165) + (100 * 35.69) + [200 * (– 53.75)] = 9319 Standard Deviation =  9319 = 96.535 A 68% range is given by μ ± 1σ, or 1132.5 ± 96.535 Which means that there is 68% chance that the value of the entire portfolio will be between $1035.47 and $1229.03 4. There are 20 professors in the Psychology Department. 15 of them have received good evaluations from students last year, while 5 have received poor evaluations. You are planning to take four courses in the Psychology Department next semester. (15 points) a. What is the probability that all of your professors next semester have received good evaluations? (4 points) 5

b. Assuming that a student has already finished the coursework of the first two years, what is the probability that he will graduate in the next two academic years? (4 points) The studies of the first two years do not need to be considered, as the focus is on the last two years. So we apply the formula for the binomial distribution with n=2 and x=2, or we get the answer directly from the table. c. What is the probability that a new entrant will face some difficulties in finishing his coursework at any time throughout his studies? (4 points) To find the probability of a student facing some difficulty throughout his coursework we need to consider that this may happen at any or all of his regular four years of studies. d. Assuming that a student has already finished two years of coursework, what is the probability that he will not be able to graduate within the next two years? (4 points) The studies of the first two years do not need to be considered, as the focus is on the last two years. So we apply the formula for the binomial distribution with n=2 and x=1 or x=2, or we get the answer directly from the table. e. Consider two randomly chosen students that have already finished the coursework of the first two years. What is the probability that at least one of them will have some difficulty with his graduation in the next two years? (4 points) The probability of a student’s success within the remaining two years is 0.64(b). The probability that out of two randomly chosen students at least one will have difficulties is equivalent to 100% less the probability of success of both of them 7