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The average score on exam 1 is 73 with a standard deviation of 14. The average score on exam 2 is 70 with a standard deviation of 14. Let Z= The difference of the scores on exam 1 and exam 2. What is the mean of Z?

A.0.

B.1.5

C.3.0

D.9.0

E.Do not have enough information to tell

I thought it was (70 + 73)/2 but then got confused how to do this because it's the mean of a difference. I was going to use aE(X)+bE(Y)+c but I don't know the probabilities of the scores. Where do I start? Thanks so much!

Relatedly,

If the correlation between the two exams were higher than 0.5 what effect would this have on the mean and standard deviation of Z?

A.This would not change the mean and reduce the standard deviation.

B.This would reduce the mean and not change the standard deviation.

C.This would reduce both the mean and standard deviation.

D.This would increase both the mean and standard deviation.

E.This would not change the mean and increase the standard deviation.

I think the answer is A but I am not sure.

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  • $\begingroup$ Basketball team 1 averages 73 points per game. Basketball team 2 averages 70 points per game. What is the average difference in their scores? $\endgroup$
    – whuber ♦
    Commented Dec 13, 2017 at 15:06

1 Answer 1

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First question

Expected value is a linear function. Let X be the random variable representing the score for test 1 and Y for exam 2. E[z] = E[x - y] = E[x] - E[y]. In your case, this gives 73-70 = 3

Second question

A. is correct. Since the averages from exam 1 and 2 will not change, the average of Z will not change. As for the standard deviation we can look at the Variance:

Var[X,Y] = Var[X] + Var[Y] - 2Cov[X,Y].

Since Cov[X,Y] > 0.5, it reduces the variance and thus reduces the standard deviation.

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  • $\begingroup$ Welcome to our site! Because you're new, please consult the guidance available by linking to self-study. $\endgroup$
    – whuber ♦
    Commented Dec 13, 2017 at 16:45
  • $\begingroup$ Sorry, I was not familiar with the tag. $\endgroup$
    – Brandon LS
    Commented Dec 14, 2017 at 15:16

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