Week 9

# 251 ln4 597 b p1 x 3 f3 f1 966 597

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Unformatted text preview: 2 3 20 0 ( 3 8 ∫x 1 3 80 4 ) x 2 dx = dx = ( 31 85 ∫ 1 3 80 x 3 dx = x 5)]0 = 2 12 5 ( 31 84 x4 )] 2 0 = 3 = 1.5 2 = 2.4 = .15 σx = .3873 µ ± σ = (1.1127, 1.8873). Thus, P(µ - σ ≤ X ≤ µ + σ) = F(1.8873) – F(1.1127) = .8403 .1722 = .6681 136 Chapter 4: Continuous Random Variables and Probability Distributions 1 9. a. P(X ≤ 1) = F(1) = .25[1 + ln(4)] ≈ .597 b. P(1 ≤ X ≤ 3) = F(3) – F(1) ≈ .966 - .597 ≈ .369 c. f(x) = F′ (x) = .25 ln(4) - .25 ln(x) for o &lt; x &lt; 4 a. For 0 ≤ y ≤ 5, F(y) = 2 0. ∫ 1 y2 udu = 25 50 y 0 For 5 ≤ y ≤ 10, F(y) = ∫ y 0 = f (u ) du = ∫ f ( u) du + ∫ f ( u) du 5 y 0 5 y 2 1 u 2 y2 + ∫ − du = y − −1 2 0 5 25 5 50 F (x1) 1.0 0.5 0.0 0 5 10 x1 y2 p ⇒ y p = (50 p )1/ 2 b. For 0 &lt; p ≤ .5, p = F(y p ) = c. E(Y) = 5 by straightforward integration (or by symmetry of f(y)), and similarly V(Y)= 50 y2 2 p For .5 &lt; p ≤ 1, p = yp − − 1 ⇒ y p = 10 − 5 2(1 − p ) 5 50 50 = 4.1667 . For the waiting time X for a single bus, 12 25 E(X) = 2.5 and V(X) = 12 2 1. ( ) 2 3 2 πr 1 − (10 − r ) dr 4 3 11 3 11 2 2 2 3 4 = π ∫ r 1 − (100 − 20r + r ) dr = π ∫ − 99r + 20r − r dr = 100 ⋅ 2π 4 9 49 E(area) = E( πR2 ) = ( ∫ ∞ −∞ πr f ( r ) dr = 2 ∫ 11 9 ) 138 Chapter 4: Continuous Random Variables and Probability Distributions 2 2. x 1 1 1 a. For 1 ≤ x ≤ 2, F(x) = ∫ 21 − 2 dy = 2 y + = 2 x + − 4, s o 1 y y 1 x 0 x &lt;1 F(x) = 2( x + 1 ) − 4 1≤ x ≤ 2 x 1 x&gt;2 x b. 1 − 4 = p ⇒ 2xp 2 – (4 – p)xp + 2 = 0 ⇒ xp = 1 [ 4 + p + 2 x p + 4 xp ~ ~ find µ , set p = .5 ⇒ µ = 1.64 p 2 + 8 p ] To 2 c. E(X...
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