chapter 1

Chapter 4: Continuous Random Variables and Probability Distributionsd.3232E(x)= ∞∞x ⋅ f ( x)dx = x ⋅ dx = ∫∞∫ ∫( x + 4 − 4) ⋅3−∞∞ 32∞ 32= ∫ dx − 4= 8 − 4 = 40 2 ∫ dx0 3−∞( x + 4) ( x + 4)3 0( x + 4) ( x + 4)dx∞ 100 32e. E(salvage value) = = ∫ ⋅ dx = 3200x + 4∫y + 4( ) ( y + 4)13200dx =(3)(64)∞=0 3 0 416.6795.a. By differentiation,⎧⎪7f(x) = ⎨⎪4⎩2x3−0 4x0 ≤ x < 171≤y ≤3otherwise1 ⎛ 7 ⎞⎛7b. P(.5 ≤ X ≤ 2) = F(2) – F(.5) = 1−⎜ − 2⎟⎜−2 ⎝ 3 ⎠⎝434⎞⋅ 2⎟−⎠(.5)3311= = .91712172 3 ⎛ 7 3 ⎞ 131c. E(X) =∫ x ⋅ x dx + ⋅ ⎜ − ⎟ = = 1. 2130 ∫ x x dx1⎝ 4 4 ⎠ 10896. µ = 40 V; σ = 1.5 V⎛ 42 − 40 ⎞ ⎛ 39 − 40 ⎞a. P(39 < X < 42) = Φ⎜⎟ − Φ⎜⎟⎝ 1.5 ⎠ ⎝ 1.5 ⎠= Φ(1.33) - Φ(-.67) = .9082 - .2514 = .6568b. We desire the 85 th percentile: 40 + (1.04)(1.5) = 41.56⎛ 42 − 40 ⎞c. P(X > 42) = 1 – P(X ≤ 42) = 1 − Φ⎜⎟ = 1 - Φ(1.33) = .0918⎝ 1.5 ⎠Let D represent the number of diodes out of 4 with voltage exceeding 42.⎛ 4⎞P(D ≥ 1 ) = 1 – P(D = 0) = 1− ⎜ ⎟( .0918 ) 0 (.9082 ) 4 =1 - .6803 = .3197⎝ 0 ⎠164