1181 Words5 Pages

100 Points Total
Answer the following questions as well as you can. LATE HOMEWORKS ARE NEVER ACCEPTED. You may meet/consult with colleagues in the class. But the assignment you turn in needs to be your own work. You should show some (though not necessarily every bit) of work for any substantial calculations.
1. (Each part 5 points) Suppose . That is, X has a normal distribution with μ=30 and σ2=144.
1a. Find a transformation of that will give it a mean of zero and a variance of one (ie., standardize ).
Let the transformed variable be named Z. We desire μz=0, σ2z=1. This means 0=a+b μX and b2 σ2X=1. One solution to this system of equations is b=1/12 and a=-5/2.
Of course, if you recognized the fact that our standard*…show more content…*

4a. If X and W are uncorrelated, find the mean and variance of . The value of the mean is 30+2*40=110 Since they are uncorellated, the Correlation term is zero. Thus, the variance is 144+4*225=1044 4b. Find the probability that . With the mean and variance calculated above, the z-value matching 120 is (120-110)/32.31=.31 The probability that z exceeds .31 is .3783. Henceforth, suppose that X and W have a correlation coefficient ρ=-.25. 4c. What is the covariance of X and W? 4d. Find the probability that . Using the formula from part a: Var[X+2W]=864, making the z-value (120-110)/29.39=.34. The probability of exceeding .34 is .3669. 4e. Find the probability that . At 50, the z-value is -60/29.39=-2.04 The probability that z is less than this number is .0207. Since the probability of being less than 120 was implied in part d of being .6331, the probability of landing between the two is the difference: .6124. 5. (Each part 5 points) Our bank from Question 2 has decided to look more deeply into the matter of customer wait times. In addition to information on the waiting times, the bank has compiled information about the credit scores of the applicants. That is, the bank has 20 observation of the following 2 variables: Observation 1 2 3 4 5 6 7 8 9 10 Wait Time 5 7 22 4 12 9 9 14 3 6 Credit Score 740 730 550 700 650 660 630 600 760 730 Observation 11 12 13 14 15 16

4a. If X and W are uncorrelated, find the mean and variance of . The value of the mean is 30+2*40=110 Since they are uncorellated, the Correlation term is zero. Thus, the variance is 144+4*225=1044 4b. Find the probability that . With the mean and variance calculated above, the z-value matching 120 is (120-110)/32.31=.31 The probability that z exceeds .31 is .3783. Henceforth, suppose that X and W have a correlation coefficient ρ=-.25. 4c. What is the covariance of X and W? 4d. Find the probability that . Using the formula from part a: Var[X+2W]=864, making the z-value (120-110)/29.39=.34. The probability of exceeding .34 is .3669. 4e. Find the probability that . At 50, the z-value is -60/29.39=-2.04 The probability that z is less than this number is .0207. Since the probability of being less than 120 was implied in part d of being .6331, the probability of landing between the two is the difference: .6124. 5. (Each part 5 points) Our bank from Question 2 has decided to look more deeply into the matter of customer wait times. In addition to information on the waiting times, the bank has compiled information about the credit scores of the applicants. That is, the bank has 20 observation of the following 2 variables: Observation 1 2 3 4 5 6 7 8 9 10 Wait Time 5 7 22 4 12 9 9 14 3 6 Credit Score 740 730 550 700 650 660 630 600 760 730 Observation 11 12 13 14 15 16

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