This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 88 s Y = 0.445 (5.55 2 ) + 0.555 (4.45 2 ) = 4.970 Thus, the correlation is r XY = 2.73 0.917 (4.970) = 0.60 7.11. a. The basic setup for this problem is the same as it is for the previous problem. We already have the joint probabilities, so we can start by calculating the expected values of X and Y : E( X ) = (2) + (1) + 0 + 1 + 2 5 = 0 E( Y ) = 0.2 (0) + 0.4 (1) + 0.4 (2) = 1.2 Thus, we have the following table: X Y X E( X ) Y E( Y ) ( XE( X ))( YE( Y )) 2 2 2 0.8 1.6 1 1 1 0.2 0.2 0 0 0 1.2 0 1 1 1 0.2 0.2 2 2 2 0.8 1.6 The covariance is the expected value of the numbers in the last column, each of which can occur with probability 1/5. Calculating this expected value gives a covariance of zero. Likewise, the correlation equals zero. b. P( Y = 2) = 0.4, but P( Y = 2  X = 2) = P( Y = 2  X = 2) = 1.0 and P( Y = 2  X = 0) = 0 c. P( X = 2) = 0.2, but P( X = 2  Y = 2) = 0.5 and P( X = 2  Y = 0) = 0 d. Clearly, X and Y are dependent. In fact, Y =  X . But it is not a linear relationship, and the covariance relationship does not capture this nonlinear relationship. 7.12. The influence diagram would show conditional independence between hemlines and stock prices, given adventuresomeness: Adventuresomeness Hemlines Stock Prices Thus (blatantly ignoring the clarity test), the probability statements would be P(Adventuresomeness), P(Hemlines  Adventuresomeness), and P(Stock prices  Adventuresomeness). 7.13. In many cases, it is not feasible to use a discrete model because of the large number of possible outcomes. The continuous model is a convenient fiction that allows us to construct a model and analyze it. 89 7.14. B B A A 0.204 0.006 0.476 0.314 0.68 0.32 0.79 0.21 1 P(B and A) = P(B  A) P(A) = 0.30 (0.68) = 0.204 P(B and A ) = P(B  A ) P(A ) = 0.02 (0.32) = 0.006 P(B  A) = P( B and A) P(A) = 0.476 0.68 = 0.70 P(B  A ) = P( B and A ) P(A ) = 0.314 0.32 = 0.98 P(B) = P(B and A) + P(B and A ) = 0.204 + 0.006 = 0.21 P(B ) = 1  P(B) = 1  0.21 = 0.79 P(A  B) = P(A and B) P(B) = 0.204 0.21 = 0.970 P(A  B) = 1  P(A  B) = 1  0.970 = 0.030 P(A  B ) = P(A and B ) P( B ) = 0.476 0.79 = 0.603 P(A  B ) = 1  P(A  B ) = 1  0.603 = 0.397 7.15. P(offer) = 0.50 P(good interview  offer) = 0.95 P(good interview  no offer) = 0.75 P(offer  good interview) = P(offer  good) = P(good  offer) P(offer) P(good  offer) P(offer) + P(good  no offer) P(no offer) = 0.95 (0.50) 0.95 (0.50) + 0.75 (0.50) = 0.5588 90 7.16. a. E( X ) = 0.05 (1) + 0.45 (2) + 0.30(3) + 0.20(4) = 0.05 + 0.90 + 0.90 + 0.80 = 2.65 Var( X ) = 0.05 (12.65) 2 + 0.45 (22.65) 2 + 0.30(32.65) 2 + 0.20(42.65) 2 = 0.05 (2.72) + 0.45 (0.42) + 0.30(0.12) + 0.20(1.82) = 0.728 X = 0.728 = 0.853 b. E( X ) = 0.13 (20) + 0.58 (0) + 0.29(100) = 2.60 + 0 + 29 = 26.40 Var( X ) = 0.13 (20  26.40) 2 + 0.58 (0  26.40) 2 + 0.29(100  26.40) 2 = 0.13 (2152.96) + 0.58 (696.96) + 0.29(5416.96) = 0....
View
Full
Document
 Spring '11
 John

Click to edit the document details