chapter 1

Chapter 3: Discrete Random Variables and Probability DistributionsM 30⋅N 50⎛ 35 ⎞ ⎛ 30 ⎞⎜ ⎟ 9 ⎜1− ⎟ = 2.⎝ 49 ⎠ ⎝ 50 ⎠d. E(X) = n = 15 ⋅ = 9V(X) = ( ) 5714σ x = 1.6036e. Let Y = 15 – X. Then E(Y) = 15 – E(X) = 15 – 9 = 6V(Y) = V(15 – X) – V(X) = 2.5714, so σ Y = 1.603667.a. Possible values of X are 5, 6, 7, 8, 9, 10. (In order to have less than 5 of the granite, therewould have to be more than 10 of the basaltic).⎛10⎞⎛10⎞⎜ ⎟⎜⎟⎝ 5 ⎠⎝10⎠⎛20⎞⎜ ⎟⎝15⎠P(X = 5) = h(5; 15,10,20) = = . 0163 .Following the same pattern for the other values, we arrive at the pmf, in table formbelow.x 5 6 7 8 9 10p(x) .0163 .1354 .3483 .3483 .1354 .0163b. P(all 10 of one kind or the other) = P(X = 5) + P(X = 10) = .0163 + .0163 = .032610c. E(X) = = 15 ⋅ = 7. 5σ x = .9934n ; V(X) = ( 7.5) 1 = . 9868⋅NM20⎛ 5 ⎞⎜ ⎟⎝19⎠µ ± σ = 7.5 ± .9934 = (6.5066, 8.4934), so we wantP(X = 7) + P(X = 8) = .3483 + .3483 = .6966⎛ 10 ⎞⎜ − ⎟⎝ 20 ⎠;68.a. h(x; 6,4,11)⎛ 4 ⎞⎜ ⎟⎝11⎠b. 6 ⋅ = 2. 18115