# Penn STAT 101 - STAT 101 Solutions Midterm Exam (2 pages)

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## STAT 101 Solutions Midterm Exam

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Lecture Notes

- Pages:
- 2
- School:
- University of Pennsylvania
- Course:
- Stat 101 - Introductory Business Statistics.

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Solutions Midterm Exam II Fall 2004 Statistics 101 1 A First compute E XY 1 0 1 2 0 1 3 0 4 0 3 6 0 3 9 0 2 5 1 Hence Cov X Y 5 1 2 2 2 2 0 26 and so Cor X Y 0 26 0 46 0 565 Thus ratings tend to go in the same direction there is a moderate linear relationship between the two ratings B Plainly the expectation will be the same as the original expectation viz 2 2 To compute the standard deviation we use the relationship Var X 2 Y 2 1 4 Var X 1 4 Var Y 1 2 Cov X Y Thus from the first part we have Var X 2 Y 2 1 2 46 1 2 0 26 0 36 Thus SD 0 6 Since there is a positive correlation independence would reduce the overall variance but have no e ect on the mean C First we compute the conditional probabilities P rating k same and these are 0 1 0 6 1 6 0 3 0 6 1 2 and 0 2 0 6 1 3 for k 1 2 3 respectively Hence E rating same 1 1 6 2 1 2 3 1 3 2 16 2 A P X 9 R 10 9 3x2 1000dx 0 271 B i We assume a binomial model Thus X B 100 0 271 and P 20 P100 k n k X k 20 100 k 0 271 0 729 ii We use the central limit theorem p X N 27 1 27 1 0 729 and compute P 20 X P 20 27 1 4 4476 Z P 1 597 Z 0 9452 C i OK integrate xk to get xk 1 k 1 Evaluating the definite integral we obtain 10k 1 k 1 and so the normalizing constant is k 1 10k 1 R 10 ii E X k 1 10k 1 0 xk 1 dx 10 k 1 k 2 We set this equal to 8 and solve to get k 3 3 A Of course it depends on one s stomach for risk If we consider average risk then since expected profit is 14 0 2 1 0 8 2 0 we would go for it However if we consider downside risk then since the probability of losing million smackers is 0 8 which is really quite large so we may decide against this venture B Hopefully we have automatic response to this problem CLT Use a normal approximation the E profit on one venture is 2 and the standard deviation is easily worked out to be 6 lose 1 unit with probability 8 and gain 14 units with probability 2 Thus since n 25 using the central limit theorem with X N 50 302 we compute P X 0 P Z 50 30 0 0475 C i Our friend Bayes theorem once again we

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