MBA 503
TEXT: LEVINE, BERENSON & STEPHAN (LBS)
INSTRUCTOR: SCLOVE

CORRECTIONS AND IMPROVEMENTS TO SOLUTIONS TO SAMPLE FINAL EXAM QUESTIONS

PART 10. STOCK PORTFOLIO

                                    INVESTMENT
                                    A        B
                                ----------------
 variable representing % incr.      X        Y
              mean of  % incr.     10%       6%
            std dev of % incr.     10%       3%
                                ----------------
                                 Corr(X,Y) = -.6

22. Assuming X is distributed according to a normal distribution, P(X > 0) = ?

The purpose of the problem is to compare several portfolios, A alone, B alone, a fifty/fifty mix and a forty/sixty mix. The criterion in this case is the probability of showing an increase. The stock returns are taken as having a normal distribution. The z-scores are needed. For Problem 22:
z = (x - µ)/sigma = (0-10)/10 = -1; P(X>0) = P(Z >-1) = .8413

23. Assuming Y is distributed according to a normal distribution, P(Y > 0) = ?

z = (x - µ)/sigma = (0-6)/3 = -2; P(Y>0) = P(Z>-2) = .9772

24. Cov(X,Y) = ?

The z-score involves both the mean and the standard deviation. For finding the standard deviation, first the variance is found. The variance of a linear combination such as a portfolio involves the covariance:
It is given that Corr(X,Y) = -.6 So, Cov(X,Y) = Corr(X,Y)*SD(X)*SD(Y) = -0.6*10*3 = -18

25. Var(.5X + .5Y) = ?

Var(aX+bY) = (a2)Var(X) + (b2)Var(Y) + 2abCov(X,Y)
Var(.5X+.5Y) = (.52)(100) + (.52)(9) + 2(.5)(.5)(-18)
    = 25 + 2.25 - 9 =18.25

26. Var(.4X + .6Y) = ?

Var(aX+bY) = (a2)Var(X) + (b2)Var(Y) + 2abCov(X,Y)
Var(.4X+.6Y) = (.42)(100) + (.62)(9) + 2(.4)(.6)(-18)
= 16 + 3.24 - 8.64 = 10.6

27. What is P(.4X+.6Y > 0) ?

We'll need the z-score; this is z = (x - µ)/sigma. The mean is .4(10) + .6(6) = 7.6% .
The std.dev. is sqrt(10.6), or about 3.256.
z = (x-mean)/s.d. = (0-7.6)/3.256 = -2.33; the prob. of exceeding this is about .99 .
This portfolio has less chance of losing money than does either stock alone, because of its fairly high mean, combined with a low standard deviation due to the negative correlation of returns on the two stocks.

(Note. Mathematically speaking, in treating the linear combination as distributed according to a normal distribution, we are assuming that X and Y have a bivariate normal distribution.

In practical terms, however, if X and Y have distributions that are not too skewed, linear combinations will have distributions that are reasonably well approximated by normal distributions.)

PART 11. ANALYSIS OF VARIANCE

TABLE.  Production (no. of widgets), by Day of Week and Shift

                         SHIFT
               |   1st    2nd    3rd |MEAN
     ----------|---------------------|------
           M   |   100    150    200 |  150
 DAY       T   |   190    200    210 |  200
 OF        W   |   210    200    190 |  200
 WEEK      R   |   160    240    200 |  200
           F   |   200    220    180 |  200
     ----------|---------------------|------
        MEAN   |   172    202    196 |  190

28. The value expected for Monday's 2nd shift would be

Monday mean + 2nd shift mean - Overall mean .

What is its value?

150 + 202 - 190 = 162

29. What is the residual for Monday's 2nd shift?

residual = observed - expected = 150 - 162   =   -62 .

latest revision 10-Oct-97