chapter 1

Chapter 4: Continuous Random Variables and Probability DistributionsSection 4.566.⎛ 1 ⎞ 1 ⎛ 1 ⎞⎜ + ⎟ ⎜ ⎟⎝ 2 ⎠ 2 ⎝ 2 ⎠⎡2⎛ 1 ⎞⎤9⎢Γ 1 + 1 − Γ ⎜1+ ⎟ = 1.2⎥⎣ ⎝ ⎠⎦a. E(X) = 3 Γ 1 = 3 ⋅ ⋅ Γ = 2. 66 ,Var(X) = ( ) 926α2−(6/β )−(6/3)−4b. P(X ≤ 6) = 1−e = 1−e = 1−e= . 98222−(6/3)−(1.5/3) −.25−4c. P(1.5 ≤ X ≤ 6) = 1−e − [ 1−e] = e − e = . 76067.a.2.5−(250/200)−1.75P(X ≤ 250) = F(250;2.5, 200) = 1−e= 1−e ≈ . 8257P(X < 250) = P(X ≤ 250) ≈ .82572.5−(1.5)P(X > 300) = 1 – F(300; 2.5, 200) = e = . 0636b. P(100 ≤ X ≤ 250) = F(250;2.5, 200) - F(100;2.5, 200) ≈ .8257 - .162 = .6637c. The median µ ~ is requested. The equation F( µ ~ ) = .5 reduces to~ 2.52.5−(µ~/ 200)⎛ µ ⎞.5 = e , i.e., ln(.5) ≈ − ⎜ ⎟ , so µ ~ = (.6931) .4 (200) = 172.727.⎝ 200 ⎠68.a. For x > 3.5, F(x) = P( X ≤ x) = P(X – 3.5 ≤ x – 3.5) = 1 -e( −3 .5 )[ ] 2− x1.5⎛ 3 ⎞b. E(X – 3.5) = .5Γ⎜⎟⎝ 2 ⎠1 = 1.329 so E(X) = 4.829⎛ 3 ⎞⎤⎜ ⎟2⎥⎝ ⎠⎦22Var(X) = Var(X – 3.5) = ( 1.5) Γ ( 2) − Γ = . 483⎡⎢⎣−1−1c. P(X > 5) = 1 – P(X ≤ 5) = 1−[ 1−e ] = e = . 368−9−1−1−9d. P(5 ≤ X ≤ 8) = 1−e − [ 1−e ] = e − e = .3679 −.0001=. 3678152