# chapter2 - 1 Chapter 2 21 RX ={0 1 2 3 4 P(X = 0 = 4 0 48 5...

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1 Chapter 2 2–1. R X = { 0 , 1 , 2 , 3 , 4 } , P ( X = 0) = 4 0 48 5 52 5 P ( X = 1) = 4 1 48 4 52 5 , P ( X = 2) = 4 2 48 3 52 5 , P ( X = 3) = 4 3 48 2 52 5 , P ( X = 4) = 4 4 48 1 52 5 2–2. μ = 0 · 1 6 + 1 · 1 6 + 2 · 1 3 + 3 · 1 12 + 4 · 1 6 + 5 · 1 12 = 26 12 σ 2 = 0 2 · 1 6 + 1 2 · 1 6 + 2 2 · 1 3 + 3 2 · 1 12 + 4 2 · 1 6 + 5 2 · 1 12 - 26 12 2 = 83 36 2–3. Z 0 ce - x dx = 1 c = 1 , so f ( x ) = e - x if x 0 0 otherwise μ = Z 0 xe - x dx = - xe - x | 0 + Z 0 e - x dx = 1 σ 2 = Z 0 x 2 e - x dx - 1 2 = - x 2 e - x | 0 + Z 0 2 xe - x dx - 1 = - 2 xe - x | 0 + Z 0 2 e - x dx - 1 = 2 - 1 = 1 2–4. F T ( t ) = P T ( T t ) = 1 - e - ct ; t 0 f T ( t ) = F 0 T ( t ) = ce - ct if t 0 0 otherwise

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2 2–5. (a) Yes (b) No, since G X ( ) 6 = 1 and G X ( b ) 6≥ G X ( a ) if b a (c) Yes 2–6. (a) f X ( x ) = F 0 X ( x ) = e - x ; 0 < x < = 0; x 0 (c) h X ( x ) = H 0 X ( x ) = e x ; -∞ < x 0 = 0; x > 0 2–7. Both are since p X ( x ) 0, all x ; and Σ all x p X ( x ) = 1 2–8. The probability mass function is x p X ( x ) - 1 1 5 0 1 10 +1 2 5 +2 3 10 ow 0 E ( X ) = - 1 · 1 5 + 0 · 1 10 + 1 · 2 5 + 2 · 3 10 = 4 5 V ( X ) = ( - 1) 2 · 1 5
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