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Math 5010-1, Spring 2005 Assignment 6 Problems #2, p. 171. The possible values of X are 1, 2, 3, . . . , 36. The probabilities are routine computations. For instance, P {X = 1} = P (1, 1) = 1/36, P {X = 2} = p(1, 2) + p(2, 1) = 2/36, P {X = 3} = p(1, 3) + p(3, 1) = 2/36, P {X = 4} = p(2, 2) + p(1, 4) + p(4, 1) = 3/36, P {X = 5} = p(1, 5) + p(5, 1) = 2/36, P {X = 6} = p(2, 3) + p(3, 2) + p(1, 6) + p(6, 1) = 4/36, etc. #3, p. 171. The possible values of X are 3, 4, 5, . . . , 18. The probabilities are routine calculations. For instance, P {X = 3} = p(1, 1, 1) = 3/63 , P {X = 4} = p(1, 1, 2) + p(1, 2, 1) + p(2, 1, 1) = 3/63 , etc. #4, p. 171. The problem can be interpreted in the following way: Lay down 10 seats in a row. Number them from left to right (1 to 10), each number denoting a rank. Then seat, completely at random, the five men and the five women. In this way you will find that 5 , P {X = 1} = P (W1 ) = 10 5 5 P {X = 2} = P (M1 ∩ W2 ) = P (M1 )P (W2 |M1 ) = × , 10 9 5 4 5 P {X = 3} = P (M1 ∩ M2 ∩ W3 ) = × × , 10 9 8 5 4 3 5 P {X = 4} = P (M1 ∩ M2 ∩ M3 ∩ W4 ) = × × × , 10 9 8 7 5 4 3 2 5 P {X = 5} = · · · = × × × × , 10 9 8 7 6 5 4 3 2 1 P {X = 6} = · · · = × × × × . 10 9 8 7 6 where Wj = {woman in seat j} and Mj = {man in seat j}. The probability that X = j, for any other value of j, is zero. R1 #7, p. 229. We know two things: First 0 (a + bx2 ) dx = 1. Therefore, a+ Also, we know that R1 0 b = 1. 3 x(a + bx) dx = (3/5). But a b 3 + = . 2 3 5 (1) R1 0 x(a + bx) dx = a 2 + 3b . Therefore, (2) Subtract (2) from (1) to find that (a/2) = (2/5), or a = (4/5). This shows that b = (3/5) too. 1 #9, p. 229. (Example 4b of Chapter 4—Please note the typo) As in Chapter 4 (p. 134), bX − (s − X)`, if X ≤ s, P (s) = sb, otherwise. Therefore, Z s Z ∞ E [P (s)] = {bx − (s − x)`} f (x) dx + sbf (x) dx 0 s Z s Z s Z ∞ = (b + `) xf (x) dx − s` f (x) dx + s` f (x) dx 0 0 s Z s = (b + `) xf (x) dx − s`F (s) + sb [1 − F (s)] . 0 Therefore, by the fundamental theorem of calculus, d E [P (s)] = (b + `)sf (s) − ` {sf (s) + F (s)} + b {1 − F (s) − sf (s)} ds = −`F (s) + b − bF (s) = b − (b + `)F (s). Differentiate again to see that d2 E [P (s)] = −(` + b)f (s) ≤ 0. ds2 So if we set (d/ds)E[P (s)] to zero we have a max. The rest follows immediately from these computations. #12, p. 229. Let X denote the position of the breakdown-point. Let g(x) denote the distance to the nearest service station, if the breakdown occurs at x. [X is random; x is a number.] If the service stations are in A, B, and at the 50-mile mark then Z ∞ Z 100 1 E[g(X)] = g(x)f (x) dx = g(x) dx 100 0 −∞ Z 25 Z 50 Z 75 Z 100 1 = x dx + (50 − x) dx + (x − 50) dx + (100 − x) dx 100 0 25 50 75 Z 25 1 = 4× x dx 100 0 = 12.5. If we put the service stations at 25, 50, and 75-mile positions (as suggested), then Z 25 Z 37.5 1 (25 − x) dx + (x − 25) dx E[g(X)] = 100 0 25 Z 50 Z 67.5 Z 75 + (50 − x) dx + (x − 50) dx + (75 − x) dx 37.5 50 67.5 Z 100 + (x − 75) dx 75 Z 25 Z 37.5 1 = (25 − x) dx + 4 (x − 25) dx = 9.375. 2 100 0 25 2 The second method is better. #14, p. 229. Evidently, n 1 Z xn dx = E [X ] = 0 1 . n+1 To do this directly, we need first the pdf of X n . But g(x) = xn is increasing and differentiable, so the pdf of Y = g(X) = X n is d −1 g (y) (y = g(x) some x) fY (y) = fX g −1 (y) dy 1 = y (1/n)−1 , if 0 ≤ y ≤ 1. n I have used the fact that g −1 (y) = y 1/n , so that (d/dy)g −1 (y) = (1/n)y (1/n)−1 . Therefore, Z E[Y ] = 0 1 1 yfY (y) dy = n 1 Z y 1/n dy = 0 1 1 = , n(1 + (1/n)) n+1 as before. #15, p. 229. (b) Written in terms of N (0, 1), we have 4−µ 16 − µ P {4 < X < 16} = P < N (0, 1) < σ σ 16 − 10 4 − 10 =Φ −Φ 6 6 = Φ(1) − Φ(−1) = 2Φ(1) − 1 ≈ (2 × 0.8413) − 1 ≈ 0.6826. 3