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Homework 2 Solution

# Homework 2 Solution - to detect an attack would become...

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Homework 2 Solution Spring 2011 Dr. Zhang 1. #7 on page 89 Let x be the monthly demand. a. Monthly order quantity = E ( x ) = Σ xf ( x ) = 300(0.20) + 400(0.30) + 500(0.35) + 600(0.15) = 445 The monthly order quantity should be 445 units. b. Cost: 445 @ \$50 = \$22,250 Revenue: 300 @ \$70 = 21,000 \$ 1,250 Loss c. μ = 445 x f ( x ) ( x - μ ) ( x - μ )2 ( x - μ )2 f ( x ) 300 0.20 -145 21025 4205.00 400 0.30 - 45 2025 607.50 500 0.35 + 55 3025 1058.75 600 0.15 +155 24025 3603.75 Variance: σ 2 = 9475.00 Standard deviation σ = 9475 = 97.34 2. #13 on page 90 a. 0.90 b. P (at least 1) = f (1) + f (2) f (1) = 2! (0.9) 1 (0.1) 1 1!1! = 2 (0.9) (0.1) = 0.18 f (2) = 2! (0.9) 2 (0.1) 0 2!0! = 1 (0.81) (1) = 0.81 So P (at least 1) = 0.18 + 0.81 = 0.99 Alternatively P (at least 1) = 1 - f (0) = 1- (0.9) 0 (0.1) 2 0!2! 2! =1-0.01 = 0.99 c. P (at least 1) = 1 - f (0) = 1- (0.9) 0 (0.1) 3 0!3! 3! =1-0.001 = 0.999 d. Yes; P (at least 1) becomes very close to 1 with multiple systems and the inability

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Unformatted text preview: to detect an attack would become catastrophic. 3. #25 on page 92 Let r. v. X be the stock price X ∼ N (30, 8.2 2 ) a. At X = 40, 40 30 1.22 8.2 z − = = P (X ≤ 40) = P ( z ≤ 1.22) =.50 + .3888 = .8888 P (X ≥ 40) = P ( z ≥ 1.22) = 1- P (X ≤ 40) = 1 - .8888 = .1112 b. At X = 20, 20 30 1.22 8.2 z − = = − P ( x ≤ 20) = P ( z ≤-1.22) = P ( z ≥ 1.22) = 0.1112 c. Suppose let x be the stock price. P (Z ≥ z) = 0.1 ⇒ P (0 ≤ Z ≤ z) = 0.4 ⇒ z = 1.28 A z-value of 1.28 cuts off an area of approximately 10% in the upper tail. x = 30 + 8.2(1.28) = 40.50 A stock price of \$40.50 or higher will put a company in the top 10%. 4. #33 on page 93 a. 2 minutes (1/ μ =0.5 ? μ =2) b. let X be the time between telephone calls. P (X ≤ 30 seconds) = P (X ≤ 0.5 minutes) = 1 - e-0.5/2 = 0.221 c. P (X ≤ 1) = 1 - e-1/2 = 0.393 d. P ( x ≥ 5) = 1- P (X ≤ 5) = 1 -(1 - e-5/2) = 1-0.918 = 0.082...
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Homework 2 Solution - to detect an attack would become...

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