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Ryerson Polytechnic University QMS400: Introduction to Management Statistics Assignment # 2 Instructor: Moez Hababou Exercise 1:Finance and Accounting (35points) Problem II: Finance and Accounting (35 points) (a) the joint frequencies table of grades for both students is: (5pts) grade in Finance grades in Accounting 0-50 50-70 70-100 Totals (b) (3pts) Grades in Accounting frequency 0-30 30-60 60-80 80-100 Totals 0 3 1 4 3 0 2 5 2 7 6 15 3 10 3 16 8 20 12 40 0-50 8 Grades in Finance frequency (c) (5pts) Grades in Accounting probability 0-30 4 Grades in Finance frequency 0-30 10% 0-50 20% 50-70 20 30-60 5 70-100 12 60-80 15 50-70 50% 30-60 12.5% 80-100 16 70-100 30% 60-80 37.5% 80-100 40% total 40 Total 40 total 100% Total 100% In absolute terms, it is easier to have higher grades in Finance. Yes, as there are more students with high grades in Finance (in the interval 80 to 100) whereas students obtain grades that are lower (mostly in the interval 50-70) in Accounting. (d) (5pts) Let GF be the grade received in Finance and GA refer to the grade received in Accounting. P(GF>85)=40%*(100-85)/(100-80)=30% P(GA>75)=30%*(100-75)/(100-70)=25% There is thus a higher probability of getting an A in Finance. Most students would thus enroll in Finance if both courses were elective. (e) (7.5pts) grade in Finance grades in Accounting 0-50 50-70 70-100 Totals 0-30 30-60 60-80 0.00% 7.50% 2.50% 10.00% 7.50% 0.00% 5.00% 12.50% 5.00% 17.50% 15.00% 37.50% 80-100 7.50% 25.00% 7.50% 40.00% Totals 20.00% 50.00% 30.00% 100.00% From the joint probability table, we see that there is a probability of 7.5% for someone to have a grade in accounting higher than 70 and a grade in Finance higher than 80. If we assume that these students are evenly distributed in the interval 70-100, with respect to their grade in Accounting, we can deduce that the proportion if people receiving both grades higher than 80 is equal to: P(GF>80 and GA>80)=7.5%*(100-80)/(100-70)=5%. Now, build the joint probability distribution table. What is the probability of someone P(GF>80 given that GA>80)= P(GF>80 and GA>80)/P(GA>80) By linear extrapolation, P(GA>80)=30%*(100-80)/(100-70)=20% We obtain thus P(GF>80 given that GA>80)= 5%/20%=25% We know from (c) that P(GF>80)=40%. P(GF>80 given that GA>80)= 25/40=5/8 Therefore, P(GF>80 given that GA>80)P(GF>80). These two events are not independent. (f) (5pts) There is no real way to compute a coefficient of correlation. However, we observe that most observations are located in entries with high grades in both courses (call it southeastern cells of the joint probability table). For instance, 60% of the population has grades higher than 50 in accounting and higher than 60 in Finance. We can conclude then that grades are more plausibly positively correlated. (g) (5pts) It is somehow unrealistic, as grades in absolute terms are irrelevant. For instance, a final grade of 80 is equivalent to an A in accounting. However, this grade is worth only a B in Finance!!! Consequently, correlation can not really be tested and can be tested only based upon final letter grades. Exercise 2: (35 pts) Returns 1990 1991 1992 1993 1994 1995 1996 1997 1.13% 1998 1999 A 28.57% -13.33% 2.56% 26.00% 5.16% B C -35.83% 14.29% 27.27% 8.93% 11.48% -7.35% 23.81% 12.82% 14.20% 11.94% 4.17% 42.40% -75.00% -50.00% 233.33% -48.00% 150.00% -60.00% 200.00% 8.97% Statistics average A 8.86% B 6.08% C 50.05% geo mean N/A N/A N/A variance 2.94% 3.62% 145.34% return period 10.67% 8.64% 11.76% Covariance A -0.00846 -0.03502 5.88% Sharp ratio 0.353828 0.091228 0.28681 Based on the Sharpe coefficient, which stock would you invest in? Stock A Explain your approach in computing the Sharpe ratio=> compute first standard-deviation and average return for the whole period 1990-1999 Exercise 3 (30 pts) The probability distribution of the number of Computer systems sales for a salesman at Future Shop is given by the following: X=number of daily sales Probability 1 .20 2 .35 3 .30 4 .15 (a) Plot the above probability distribution and briefly describe it. Almost symmetrical Probability 0.4 0.35 0.3 0.25 0.2 Probability 0.15 0.1 0.05 0 1 2 3 4 (b)-(c) X=number of daily sales Probability 1 2 3 4 0.2 0.35 0.3 0.15 EV(X) x.P(x) 0.2 0.7 0.9 0.6 2.4 x-EV(x) -2.2 -1.7 -1.5 -1.8 Variance st-dev (x-EV(x))^2 4.84 2.89 2.25 3.24 13.22 3.64 CV=1.52 => very high variation (b) Let p refer to the probability that at least 3 daily sales are achieved. Out of 5 salesmen, what is the probability that: At least 4 of them succeed in getting 3 daily sales? Exactly 2 employees achieve 3 daily sales? Strictly less then one employee achieves 3 daily sales p=.45 Binomial process with p=.45 and n=5 At least 4 of them succeed in getting 3 daily sales? => Prob(X4)=1-P(X3)=1-.869=.131 Exactly 2 employees achieve 3 daily sales? => Prob(X=2)= P(X3)- P(X2)=.593-.256=.337 Strictly less then one employee achieves 3 daily sales => Prob(X<1)=Prob(X=0)=.050 (c) Assuming that the previous distribution is asymptotically normal (that is if the number of salesmen increase, it becomes more and more symmetrical), if you have 100 salesmen, using the normal table, what is the probability that: X follows normal distribution with mean (nEV(Xi)=100*2.4=240) and standard-deviation (nStdev(Xi)=100*3.64=364) Prob(between 250 and 350 computers are sold) = P(250X350)=P(.022Z.30)=.6179-.5080=.1099 Prob(less than 100 computers are sold)=P(X100)= P(Z-.3846)=0 Prob(more than 400 computers are sold)=P(X400)=P(Z.4395)=1-.6664=.3336