420Hw05ans - STAT 420 Homework #5 (due Friday, September...

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STAT 420 Fall 2007 Homework #5 (due Friday, September 28, by 3:00 p.m.) 1. Suppose that company A and company B are in the same industry sector, and the prices of their stocks, $X per share for company A and $Y per share for company B, vary from day to day randomly according to a bivariate normal distribution with parameters μ X = 45, σ X = 5.6, μ Y = 25, σ Y = 5, ρ = 0.8. a) What is the probability that on a given day the price of stock for company B ( Y ) exceeds $33? Y has Normal distribution with mean μ Y = 25 and standard deviation σ Y = 5. P ( Y > 33 ) = ± ² ³ ´ µ - > 5 25 33 Z P = P ( Z > 1.60 ) = 1 – Φ ( 1.60 ) = 1 – 0.9452 = 0.0548 . b) Suppose that on a given day the price of stock for company A ( X ) is $52. What is the probability that the price of stock for company B ( Y ) exceeds $33? Given X = 52, Y has Normal distribution with mean ( ) X X Y Y ± ± ² & - + x = ( ) 45 52 6 . 5 5 8 . 0 25 - + = 30 and variance ( ) 2 Y 2 ± ² 1 - = ( ) 2 2 5 8 . 0 1 - = 9 ( standard deviation = 3 ). P ( Y > 33 | X = 52 ) = ± ² ³ ´ µ - > 3 30 33 Z P = P ( Z > 1.00 ) = 1 – Φ ( 1.00 ) = 1 – 0.8413 = 0.1587 .
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c) Alex bought 5 shares of company A stock and 3 shares of company B stock. What is the probability that on a given day the value of his portfolio ( 5 X + 3 Y ) is below $250? Portfolio = 5 X + 3 Y. Portfolio has Normal distribution with mean 5 μ X + 3 μ Y = 5 45 + 3 25 = 300 and variance Var ( 5 X + 3 Y ) = Cov ( 5 X + 3 Y , 5 X + 3 Y ) = Cov ( 5 X , 5 X ) + Cov ( 5 X , 3 Y ) + Cov ( 3 Y , 5 X ) + Cov ( 3 Y , 3 Y ) = 25 2 X + 30 σ XY + 9 2 Y = 25 2 X + 30 ρ σ X σ Y + 9 2 Y = 25 5.6 2 + 30 0.8 5.6 5 + 9 5 2 = 1681 ( standard deviation = 1681 = 41 ). P ( Portfolio < 250 ) = ± ² ³ ´ µ - < 41 300 250 Z P = P ( Z < – 1.22 ) = Φ ( 1.22 ) = 0.1112 . d) What is the probability that 1 share of company A stock is worth more than 2 shares of company B stock? Want P ( X > 2 Y ) = P ( X – 2 Y > 0 ) = ? X – 2 Y has Normal distribution, E ( X – 2 Y ) = μ X – 2 μ Y = 45 – 2 25 = 5, Var ( X – 2 Y ) = 2 X – 4 σ XY + 4 2 Y = 2 X – 4 ρ σ X σ Y + 4 2 Y = 5.6 2 – 4 0.8 5.6 5 + 5 2 = 41.76 ( standard deviation 6.462 ). P ( X – 2 Y > 0 ) = P ( Z > 462 . 6 5 0 + ) = P ( Z > 0.77 ) = 1 – Φ ( 0.77 ) = 1 – 0.7794 = 0.2206 .
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From the textbook: 3.1 (a), (b), (c), (d) 3.2 3.1 For the prostate data, fit a model with lpsa as the response and the other variables as predictors. a)
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420Hw05ans - STAT 420 Homework #5 (due Friday, September...

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