017 p z 225 0122 c p20 x 002 0026 029 0329 617 5

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Unformatted text preview: or X > µ + 2.5σ) = 1 - P(µ - 2.5σ ≤ X ≤ µ + 2.5σ) = 1 - P(-2.5 ≤ Z ≤ 2.5) = 1 - .9876 = .0124 c. P(µ - 2σ ≤ X ≤ µ - σ or µ + σ ≤ X ≤ µ + 2σ) = P(within 2 sd’s) – P(within 1 sd) = P(µ 2σ ≤ X ≤ µ + 2σ) - P(µ - σ ≤ X ≤ µ + σ) = .9544 - .6826 = .2718 144 Chapter 4: Continuous Random Variables and Probability Distributions 4 9. P: µ: σ: a. .5 12.5 2.50 .6 15 2.45 .8 20 2.00 P(15≤ X ≤20) P(14.5 ≤ normal ≤ 20.5) .5 .212 P(.80 ≤ Z ≤ 3.20) = .2112 .6 .577 P(-.20 ≤ Z ≤ 2.24) = .5668 .8 .573 P(-2.75 ≤ Z ≤ .25) = .5957 b. P(X ≤15) P(normal ≤ 15.5) .885 P(Z ≤ 1.20) = .8849 .575 P(Z ≤ .20) = .5793 .017 P( Z ≤ -2.25) = .0122 c. P(20 ≤X) .002 .0026 .029 .0329 .617 5 0. P(19.5 ≤ normal) .5987 P = .10; n = 200; np = 20, npq = 18 30 + .5 − 20 Φ = Φ(2.47) = .9932 18 a. P(X ≤ 30) = b. P(X < 30) =P(X ≤ 29) = c. P(15 ≤ X ≤ 25) = P(X ≤ 25) – P(X ≤ 14) = 29 + .5 − 20 Φ = Φ(2.24) = .9875 18 Φ(1.30) - Φ(-1.30) = .9032 - .0968 = .8064 5 1. 25 + .5 − 20 14 + .5 − 20 Φ − Φ 18 18 N = 500, p = .4, µ = 200, σ = 10.9545 a. P(180 ≤ X ≤ 230) = P(179.5 ≤ normal ≤ 230.5) = P(-1.87 ≤ Z ≤ 2.78) = .9666 b. P(X < 175) = P(X ≤ 174) = P(normal ≤ 174.5) = P(Z ≤ -2.33) = .0099 147 Chapter 4: Continuous Random Variables and Probability Distributions c. d. We want a value of X for which F(X;4)=.99. In table A.4, we see F(10;4)=.990. So with β = 6, the 99th percentile = 6(10)=60. We want a value of X for which F(X;4)=.995. In the table, F(11;4)=.995, so t = 6(11)=66. At 66 weeks, only .5% of all transistors would still...
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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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