Solutions HWB07

4.4:

F(x) = 0 if x < 0

F(x) = 1/27 if 0 <= x < 1

F(x) = 7/27 if 1 <= x < 2

F(x) = 19/27 if 2 <= x < 3

F(x) = 27/27 if 3 <= x

4.10:

mu = SUM{xf(x)} = {(0)(1/27) + (1)(6/27) + (2)(12/27) + (3)(8/27)} = {0/27 + 6/27 + 24/27 + 24/27} = 54/27 = 2

sigma^2 = SUM{(x-mu}^2f(x)} = {(-2)^2(1/27) + (-1)^2(6/27) + (0)^2(12/27) + (1)^2(8/27)} = {4/27 + 6/27 + 0 + 8/27} = 18/27 = 2/3

Or you can use the simpler formula

sigma^2 = E(X^2) – mu^2 = (14/3) – (4) = 2/3

sigma = sqrt[2/3] = sqrt[6/9] = sqrt[6]/3 = 0.8165

4.18: Draw three cards without replacement from the box [A, A, N, N, N, N, N, N]

4.18a:

x f(k) = Pr(K = k)

0 Pr(NNN) = (6/8)(5/7)(4/6) = 120/336 = 10/28

1 Pr(ANN or NAN or NNA) = (2/8)(6/7)(5/6) + (6/8)(2/7)(5/6) + (6/8)(5/7)(2/6) = 180/336 = 15/28

2 Pr(AAN or ANA or NAA) = (2/8)(1/7)(6/6) + (2/8)(6/7)(1/6) + (6/8)(2/7)(1/6) = 36/336 = 3/28

3 Pr(K = AAA) = (2/8)(1/7)(0/6) = 0/28

4.18b:

F(k) = 0 if k < 0

F(k) = 10/28 if 0 <= k < 1

F(k) = 25/28 if 1 <= k < 2

F(k) = 28/28 if 2 <= k

 The graph is an increasing step function that has jumps of 10/28, 15/28, and 3/28 at 0, 1, and 2, respectively.

4.18c:

 mu = SUM{kf(k)} = {(0)(10/28) + (1)(15/28) + (2)(3/28)} = {0/28 + 15/28 + 6/28} = 21/28 = 3/4

sigma^2 = SUM{(k-mu}^2f(k)} = {(-3/4)^2(10/28) + (1/4)^2(15/28) + (5/4)^2(3/28)} = {90 + 15 + 75}/{(16)(28)} = 180/{(16)(28)} = 45/112

Or you can use the simpler formula

sigma^2 = E(K^2) – mu^2 = 108/112 – (9/16) = 45/112

 sigma = sqrt[45/112] = 0.6339