chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions76.1.9+9 / 2a. E(X) = = 10. 02423.8+(.81) .81e ; Var(X) = ⋅ ( e −1) = 125. 395e , σ x = 11.20b. P(X ≤ 10) = P(ln(X) ≤ 2.3026) = P(Z ≤ .45) = .6736P(5 ≤ X ≤ 10) = P(1.6094 ≤ ln(X) ≤2.3026)= P(-.32 ≤ Z ≤ .45) = .6736 - .3745 = .299177. The point of symmetry must be 1 1 12, so we require that f ( − µ ) = f ( + µ )221α−11β −11α −11β −1( − ) ( + µ ) = ( + µ ) ( − µ )2222, i.e.,µ , which in turn implies that α = β.78.55 + 257a. E(X) = = = . 714( )( )( ) ( )Γ 7b. f(x) =Γ 5 Γ 2⋅ x4⋅.210, V(X) = = . 0255(49)(8)4 5( 1−x) = 30( x − x )4 5so P(X ≤ .2) =∫ 30( x − x ) dx = . 00160for 0 ≤ X ≤ 1,.44 5c. P(.2 ≤ X ≤ .4) =∫ 30( x − x ) dx = . 03936.25 2d. E(1 – X) = 1 – E(X) = 1 - = = . 2867 779.a. E(X) =ΓΓ1 Γ( α + β ) α−1∫ x ⋅ x ( − x)0Γ( α) Γ( β )( α + β ) Γ( α + 1) Γ( β )⋅=( α ) Γ( β ) Γ( α + β + 1)b. E[(1 – X) m 1] =∫ ( − )0Γ( α + β ) 1=Γ( α ) Γ( β ) ∫ x0β −1 Γ( α + β ) 1α β −1dx =( )( ) ( ) ∫ x 1 − xΓ α Γ β0αΓ( α ) Γ( α + β ) α⋅=( α ) Γ( β ) ( α + β ) Γ( α + β ) α + β1 dxΓ( α + β )( α) Γ( β )Γ( − x)−−11 x m⋅ xα 1 1βdxΓα−1 1( − x)βFor m = 1, E(1 – X) = .α + βm+β−1Γdx =Γ( α + β ) ⋅ Γ( m + β )( α + β + m) Γ( β )155