6 139 chapter 4 continuous random variables and

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Unformatted text preview: ) = 2 E(X ) = d. ∫ 2 1 2 x2 1 1 x ⋅ 21 − 2 dx = 2∫ x − dx =2 − ln( x) = 1.614 2 1 x x 1 2∫ 2 1 ( ) Amount left = max(1.5 – X, 0), so E(amount left) = 2 3. 2 x3 8 x − 1 dx = 2 − x = ⇒ Var(X) = .0626 3 1 3 2 ∫ 2 1 1 .5 1 max( 1.5 − x,0) f ( x ) dx =2 ∫ (1.5 − x )1 − 2 dx = .061 1 x 9 X + 32, s o 5 2 9 9 Var X + 32 = ⋅ ( 2) 2 = 12.96 , 5 5 With X = temperature in °C, temperature in °F = 9 9 E X + 32 = (120) + 32 = 248, 5 5 so σ = 3.6 139 Chapter 4: Continuous Random Variables and Probability Distributions 3 5. a. b. 30 + 5( -1.555) = 22.225 c. 3 6. µ + σ⋅(91st percentile from std normal) = 30 + 5(1.34) = 36.7 µ = 3.000 µm; σ = 0.140. We desire the 90th percentile: 30 + 1.28(0.14) = 3.179 µ = 43; σ = 4.5 40 − 43 P z ≤ = P(Z < -0.667) = .2514 4 .5 60 − 43 P(X > 60) = P z > = P(Z > 3.778) ≈ 0 4 .5 a. b. 3 7. P(X < 40) = 43 + (-0.67)(4.5) = 39.985 P(damage) = P(X < 100) = 100 − 200 P z < = P(Z < -3.33) = .0004 300 P(at least one among five is damaged) 3 8. = 1 – P(none damaged) = 1 – (.9996)5 = 1 - .998 = .002 From Table A.3, P(-1.96 ≤ Z ≤ 1.96) = .95. Then P(µ - .1 ≤ X ≤ µ + .1) = .1 .1 .1 − .1 = .0510 P < z < implies that = 1.96, and thus that σ = σ σ 1.96 σ 3 9. Since 1.28 is the 90th z percentile (z.1 = 1.28) and –1.645 is the 5th z percentile (z.05 = 1.645), the given information implies that µ + σ(1.28) = 10.256 and µ + σ(-1.645) = 9.671, from which σ(-2.925) = -.585, σ = .2000, and µ = 10. 4 0. a. P(µ - 1.5σ ≤ X ≤ µ + 1.5σ) = P(-1.5 ≤ Z ≤ 1.5) = Φ(1.50) - Φ(-1.50) = .8664 b. P( X < µ - 2.5...
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This note was uploaded on 03/02/2014 for the course MATH 311 taught by Professor Yu during the Spring '08 term at Drexel.

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