# QZ 6-D - Var X – Y = Var X Var Y = 24 2 7 2 = 625...

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STAT 400 Fall 2011 Version D Name ANSWERS . Quiz 6 (10 points) Be sure to show all your work; your partial credit might depend on it. No credit will be given without supporting work. 1. The price of Abitrisky, Inc. stock ( X ) varies from day to day according to normal distribution with mean \$70 and standard deviation \$24. The price of Surething Corp. stock ( Y ) also varies from day to day according to normal distribution with mean \$65 and standard deviation \$7. Assume the prices of the two stocks are independent. a) (5) Find the probability that on a given day, the price of Surething Corp. stock is higher than the the price of Abitrisky, Inc. stock. That is, find P ( Y > X ). P ( Y > X ) = P ( X – Y < 0 ). X – Y has Normal distribution with mean E ( X – Y ) = 70 – 65 = \$5 and variance
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Unformatted text preview: Var ( X – Y ) = Var ( X ) + Var ( Y ) = 24 2 + 7 2 = 625 ( standard deviation = \$25 ). P ( X – Y < 0 ) = -< 25 5 Z P = P ( Z < – 0.20 ) = Φ ( – 0.20 ) = 0.4207 . b) (5) Alex buys 7 shares of Abitrisky, Inc. stock and 10 shares of Surething Corp. stock. What is the probability that the value of this portfolio will exceed \$1000? That is, find P ( 7 X + 10 Y > 1000 ). 7 X + 10 Y has Normal distribution with mean E ( 7 X + 10 Y ) = 7 ⋅ 70 + 10 ⋅ 65 = \$1140 and variance Var ( 7 X + 10 Y ) = 49 ⋅ Var ( X ) + 100 ⋅ Var ( Y ) = 49 ⋅ 12 2 + 100 ⋅ 5 2 = 33124 ( standard deviation = \$182 ). P ( 7 X + 10 Y > 1000 ) = -> 182 1140 1000 Z P = P ( Z > – 0.77 ) = 1 – Φ ( – 0.77 ) = 0.7794 ....
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