Key to Midterm II

1. The first question tested whether you could find expectations and variances of discrete random variables and linear combinations of discrete rvs.
   1. E(X) = 1*(1/21) + 2*(2/21) + ... + 6*(6/21) = 19/21
   2. E(X^2) = 1*(1/21) + 4*(2/21) + ... + 36*(6/21) = 431/21
      So VAR(X) = 431/21 - (19/21)^2 = 8690/441 approx = 19.70
   3. E(Z) = 3*E(X) - 2 = 19/7 - 14/7 = 5/7
   4. VAR(Z) = 9*VAR(X) = 78210/441 approx = 177.35
2. The second question tested whether you could find expectations and variances of jointly distributed continuous random variables.
   1. E(XY) = int int (1/18) * x^2 * y^2 dy dx
      The outer integral is evaluated from 0 to 3 and the inner from 1 to 3.
      The answer is 26/6.
   2. One approach is to first find the marginal of X, f(x) = int (1/18)*xy dy where the limits of the integral (int) are 1 to 3. The answer is f(x) = (2x/9).
      Now just find E(X) = int x * f(x) dx where the limits are 0 to 3.
      The answer is 2.
   3. f(y) = int (1/18)*xy dx, limits 0 to 3.
      f(y) = y/4, E(Y) = int y*f(y) dy = 26/12.
   4. X and Y are independent, since f(x)*f(y) = f(x,y). (You had to state and show this to get full credit), hence VAR(X+Y) = VAR(X) + VAR(Y).
      VAR(X) = E(X^2) - 4 = 81/2 - 4 = 73/4
      VAR(Y) = 80/16 - 169/36 = 11/36
      You could also find E(X+Y)^2 by doing the double integral.
3. Question 3 tested facets of the Poisson distribution.
   1. The time interval can be broken into smaller intervals such that the probability that two events occur in the same sub-interval is zero (or acceptably small), and the events of one interval are independent of the events of another (a necessary assumption, but possibly a little shaky here since if a person coughs in one interval, he or she may cough soon in the next), and two events can't happen simultaneously.
   2. Plug into the poisson density with lambda = 2.5:
      P(X=0) = e^(-2.5) = approx 0.082
      The P(X >2) = 1 - P(x <= 2) = 1 - (P(X=0) + P(X=1) + P(X=2)) = approx .456.
   3. The mean and the variance for a Poisson random variable are both equal to lambda: 2.5.
4. The fourth question concerned the normal distribution and your ability to deal with linear combinations of random variables.
   1. This is difficult to show in this format, but we did it in class. The answer is that E(Xbar) = mu. Ask to see this if you can't get it.
   2. Var(Xbar) = sigma^2/n. Again, we did this in class.
   3. Part a) P(7 < X < 13) = P(-.3 < Z < .3) = .6179-.3821
      Part b) P(7 < Xbar < 13) = P(-1.5 < Z < 1.5) = .9332 - .0668
      (We now divide everything by sigma/sqrt(n) = 2.)
   4. You should trust the person who takes five observations and reports the average. Since the variance of the average is less than the variance of any single observations, and since both the average and the observations are centered about the mean mu, the average is more likely to be closer to the true value. Also, intuitively, more information should always be more useful than less information.