chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions60. Proof of E(X) = np:E(X) ====⎛n⎞nnx n−xx n−x∑ x ⋅⎜⎟ p (1 − p)= x ⋅ p − px=x∑(1 )0 ⎝ ⎠x=1 x!(n − x)!nnn! x n−x( n −1)!x−1∑ p (1 − p)= np∑p (1 − p)x=1 ( x −1)!(n − x)!x=1 ( x −1)!(n − x)!n( n −1)!y n−1−ynp∑p (1 − p)(y replaces x-1)( y)!(n −1−y)!y=0⎧np ⎨⎩n∑ − 1y=0⎛n−1⎞⎜ ⎟p⎝ y ⎠y(1 − p)n−1−y⎫⎬⎭The expression in braces is the sum over all possible values y = 0, 1, 2, … , n-1 of a binomialp.m.f. based on n-1 trials, so equals 1, leaving only np, as desired.n!n−x61.a. Although there are three payment methods, we are only concerned with S = uses a debitcard and F = does not use a debit card. Thus we can use the binomial distribution. So n= 100 and p = .5. E(X) = np = 100(.5) = 50, and V(X) = 25.b. With S = doesn’t pay with cash, n = 100 and p = .7, E(X) = np = 100(.7) = 70, and V(X)= 21.62.a. Let X = the number with reservations who show, a binomial r.v. with n = 6 and p = .8.The desired probability isP(X = 5 or 6) = b(5;6,.8) + b(6;6,.8) = .3932 + .2621 = .6553b. Let h(X) = the number of available spaces. ThenWhen x is: 0 1 2 3 4 5 66E[h(X)] = ∑x=0H(x) is: 4 3 2 1 0 0 0h ( x)⋅ b(x;6,.8)= 4(.000) + 3(.002) = 2(.015 + 3(.082) = .277c. Possible X values are 0, 1, 2, 3, and 4. X = 0 if there are 3 reservations and none show or…or 6 reservations and none show, soP(X = 0) = b(0;3,.8)(.1) + b(0;4,.8)(.2) + b(0;5,.8)(.3) + b(0;6,.8)(.4)= .0080(.1) + .0016(.2) + .0003(.3) + .0001(.4) = .0013P(X = 1) = b(1;3,.8)(.1) + … + b(1;6,.8)(.4) = .0172P(X = 2) = .0906, P(X = 3) = .2273,P(X = 4) = 1 – [ .0013 + … + .2273 ] = .6636112