# Solution note that p x 0 2 4 1 3 1 6 p x 1 2 2 4 2 3

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Solution: Note that P ( X = 0) = 2 4 × 1 3 = 1 6 , P ( X = 1) = 2 2 4 × 2 3 = 2 3 , and P ( X = 2) = 2 4 × 1 3 = 1 6 . Therefore, E ( X ) = 0 × 1 6 + 1 × 2 3 + 2 × 1 6 = 1 , E ( X 2 ) = 0 × 1 6 + 1 × 2 3 + 4 × 1 6 = 4 3 , V ar ( X ) = E ( X 2 ) - ( E ( X )) 2 = 4 3 - 1 2 = 1 3 , and finally SD ( X ) = p V ar ( X ) = 1 3 . Page 3

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Math 431: Exam 1 5. [10 points] Ten people sit randomly around a circular table. Six are men and four are women. How many men do you expect are not sitting next to a woman? Solution: Let I n be the indicator that the person in the i th chair is a man and he has men sitting on both sides. Then, for each n ∈ { 1 , . . . , 10 } E ( I n ) = P ( I n = 1) = 6 10 × 5 9 × 4 8 = 1 6 . The number of men not sitting next to a woman is X = I 1 · · · I 10 and E ( X ) = 10 E ( I 1 ) = 10 6 = 5 3 . 6. Let X and Y be independent random variables with E ( X ) = 4, SD ( X ) = 3, E ( Y ) = 6, and SD ( Y ) = 4. (a) [4 points] Find E ( X + Y ) and SD ( X + Y ). Solution: E ( X + Y ) = E ( X ) + E ( Y ) = 10. Because X, Y are independent V ar ( X + Y ) = V ar ( X ) + V ar ( Y ) = 3 2 + 4 2 = 25, so SD ( X + Y ) = 25 = 5. (b) [8 points] Find an upper bound for P ( X + Y 25). Solution: Let μ = E ( X + Y ) and σ = SD ( X + Y ). Using Chebychev’s in- equality, P ( X + Y 25) = P ( X + Y - μ 15) P ( | X + Y - μ | ≥ 15) = P ( | X + Y - μ | ≥ 3 σ ) 1 3 2 = 1 9 . (c) [2 points] Assuming that X and Y are symmetric (hence also X + Y ), improve the answer to part (b). Solution: By symmetry (meaning symmetry around μ ), P ( X + Y - μ 3 σ ) = P ( X + Y - μ ≤ - 3 σ ). Since P ( X + Y - μ 3 σ ) + P ( X + Y - μ ≤ - 3 σ ) = P ( | X + Y - μ | ≥ 3 σ ) 1 9 , each of these probabilities is at most 1 / 18. 7. [10 points] Assume that each job you have lasts an average of 5 years with a standard deviation of 2. Approximate the probability that you will have more than 10 jobs in 40 years. ( Hint : find the probability that your first 10 jobs last less than 40 years.) Page 4
Math 431: Exam 1 Solution: Let J i be the length of the your i
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