chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions93.a. For 0 ≤ Y ≤ 25, F(y) =124⎧ 0⎪31 ⎛2 y ⎞F(y) = ⎨⎜ y −⎟⎪48⎝ 18 ⎠⎪⎩1∫0y2⎛ u ⎞⎜u− ⎟ =⎝ 12 ⎠y < 0y > 121240 ≤ y ≤ 12b. P(Y ≤ 4) = F(4) = .259, P(Y > 6) = 1 – F(6) = .5P(4 ≤ X ≤ 6) = F(6) – F(4) = .5 - .259 = .2412⎛ u⎜⎝ 23u ⎞⎤− ⎟⎥36 ⎠⎦y0. Thus3 41 122 ⎛ y ⎞ 1 ⎡ y y ⎤c. E(Y) = 1= 624 ∫ y ⎜ − ⎟dy=0 12 24⎢ −3 48⎥⎝ ⎠ ⎣ ⎦E(Y 2 ) = 3 ⎛ y ⎞1 ⎟ = 43. 224 ∫ y ⎜ − dy , so V(Y) = 43.2 – 36 = 7.20⎝ 12 ⎠d. P(Y < 4 or Y > 8) = 1 - P(4 ≤ X ≤ 8) = .518e. the shorter segment has length min(Y, 12 – Y) soE[min(Y, 12 – Y)] =+∫126∫12012min( y ,12 − y)⋅ f ( y)dy =0∫06min( y,12− y)⋅ f ( y)dy61290min( y ,12 − y)⋅ f ( y)dy = ∫ y ⋅ f ( y)dy + ∫ (12 − y)⋅ f ( y)dy = = .3. 75062494.a. Clearly f(x) ≥ 0. The c.d.f. is , for x > 0,x∫ −∞F(x)= f ( y)dy = ∫( F(x) = 0 for x ≤ 0.)∫ ∞ −∞x322( y + 4) ( y + 4) ( x + 4) 20 3dy = −12⋅32Since F(∞) = f ( y)dy = 1,f(x) is a legitimate pdf.⎤⎥⎦x0= 1−16b. See abovec.16 ⎛ 16 ⎞P(2 ≤ X ≤ 5) = F(5) – F(2) = 1 − − ⎜1− ⎟ = . 24781 ⎝ 36 ⎠(continued)163