t4f07solns[1]

# t4f07solns[1] - 1 NAME → ISyE 3770 — Test 4 Solutions...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 NAME → ISyE 3770 — Test 4 Solutions — Fall 2007 This test is 90 minutes long. You are allowed one cheat sheet. Do not look at or start the test until you are told to do so. When we ask you to return the test, stop immediately, hand the test in, and do not utter a word to anyone. Do not show any work other than your answers on this sheet. Check your answers — we won’t be giving any credit for any answers that you transfer incorrectly from your worksheets. Good luck! 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 2 1. Suppose U 1 ,U 2 ,U 3 are i.i.d. U(0,1). Find P ( U 1 < U 2 < U 3 ). Solution: There are six possible random orderings of the U i ’s. Therefore, the probability of this particular ordering is 1/6. ♦ 2. Let X be the outcome of a die toss. Find E [1 /X ] to 3 significant digits. Solution: E [1 /X ] = ∑ 6 i =1 (1 /i ) P ( X = i ) = 1 6 ∑ 6 i =1 1 i = 0 . 408 ♦ 3. Suppose that the lifetime of a light bulb is exponential with a mean of 10000 hours. Further suppose that the bulb has already survived 20000 hours. Find the probability that it survives at least another 10000 hours. Solution: By the memoryless property, P ( X > 30000 | X > 20000) = P ( X > 10000) = e- λt = e- 1 = 0 . 368 . ♦ 4. Suppose that X and Y are both Nor(1,4) random variables with Cov ( X,Y ) = 1. Find E [ XY ]. Solution: E [ XY ] = Cov ( X,Y ) + E [ X ] E [ Y ] = 1 + 1 = 2. ♦ 5. If X and Y are both Nor(0 , 1) with Cov ( X,Y ) = 0 . 5. What’s the distribution of X + Y + 1? Solution: E [ X + Y + 1] = E [ X ] + E [ Y ] + 1 = 1 and Var ( X + Y + 1) = Var ( X ) + Var ( Y ) + 2 Cov ( X,Y ) = 3. Thus, X + Y + 1 ∼ Nor(1 , 3). ♦ 6. Suppose that the number of accidents at a factory is Poisson with rate 2/month....
View Full Document

## This note was uploaded on 08/20/2008 for the course ME 3770 taught by Professor Goldsman during the Spring '08 term at Georgia Tech.

### Page1 / 7

t4f07solns[1] - 1 NAME → ISyE 3770 — Test 4 Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online