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Unformatted text preview: Homework 8 • Homework must be answered in the order shown here (else please make a note telling the reader where it is). • Write your answers neatly • Work must be shown for full credit • No late homework accepted under any circumstances • Homework must be stapled. 1. A firm that sells word processing systems keeps track of the number of customers who call on any one day and the number of orders placed on any one day. Let X denote the number of calls, let Y denote the number of orders placed, and let P(X, Y) denote the joint probability function for (X, Y). Records indicate that P(0,0) = 0.02, P(0,1)=0.01, P(1,0) = 0.1, P(1,1) = 0.19, P(1,2)=0.25, P(2,0) = 0.2, P(2,1)=0.03, P(2,2) = 0.2. Thus, for any given day, the probability of, say, two calls and one order is 0.30. (a) Find the Expected Value of the Product of the two random variables, E(XY). (b) Find the Expected number of calls and the variability of the number of calls. (c) Find the Expected number of orders. (d) Compute the covariance between the number of calls and the number of orders. (e) Is the number of calls independent of the number of orders? (f) Find the conditional expected number of orders when there is 2 call. . X0 1 2 Y 0 1 2 .02 .01 0 .1 .19 .25 .2 .03 .2 (a) E [ XY ] = (1) (.19 ) + ( 2 ) (.25 ) + ( 2 ) (.03) + ( 4 ) (.2 ) = 1.55
(b) E [ X ] = ( 0 ) (.03) + (1) (.54 ) + ( 2 ) (.43) = 1.4 Var ( X ) = 2.26 − (1.4 )
(c) E ⎡ X 2 ⎤ = 0 2 (.03) + 12 (.54 ) + 2 2 (.43) = 2.26 ⎣⎦
2 () () () E [Y ] = ( 0 ) (.32 ) + (1) (.23) + ( 2 ) (.45 ) = 1.13
1 (d) Cov ( X , Y ) = E [ XY ] − E [ X ] E [Y ] = 1.55 − (1.4 ) (1.13) = −0.032 (e) P ( X = 0, Y = 0 ) = 0.02 ≠ .0096 = P ( X = 0 ) P (Y = 0 ) ⇒ not independent
(f)
E ⎡Y X = 2 ⎤ = ( 0 ) (.465 ) + (1) (.0698 ) + ( 2 ) (.465 ) = .9998 ⎣ ⎦ 2.
Problem not in the reader (but similar). A certain market has both an express checkout register and a superexpress register. Let X be the number of customers queueing at the express register at a particular weekday time and let Y be the number of customers queueing at the superexpress register at that same time. Suppose that the joint probability distribution P(X,Y) is given in the accompanying table Y X 0 1 2 3 4 0.08 0.06 0.05 0.00 0.00 0.07 0.15 0.04 0.03 0.01 0.04 0.05 0.1 0.04 0.05 0.00 0.04 0.06 0.07 0.06 0 1 2 3 Calculate the probability that the number of customers at the express register is larger than the number of customers at the other register. P(customers at express larger than other register) = P ( X = 1, Y = 0 ) + P ( X = 2, Y = 0 ) + P ( X = 2, Y = 1) + P ( X = 3, Y = 0 ) + P ( X = 3, Y = 1)
= .06 + .05 + .04 + .00 + .03 + .04 + .00 + .01 + .05 + .06 = 0.34 + P ( X = 3, Y = 2 ) + P ( X = 4, Y = 0 ) + P ( X = 4, Y = 1) + P ( X = 4, Y = 2 ) + P ( X = 4, Y = 3) 2 3.
Suppose that 3 balls are chosen without replacement from an urn consisting of 6 white and 7 red balls. Let X i equal 1 if the ith ball selected is white, and let it equal 0 otherwise. Give the joint probability mass function of (a) X 1 , X 2 ; (b) X 1 , X 2 , X 3 ; (a) 65 5 P (1,1) = = 13 12 26 67 7 P (1, 0 ) = = 13 12 26 76 7 P ( 0,1) = = 13 12 26 76 7 P ( 0, 0 ) = = 13 12 26 (b) 654 10 P (1,1,1) = = 13 12 11 143
657 35 = 13 12 11 286 676 21 P (1, 0, 0 ) = P ( 0,1, 0 ) = P ( 0, 0,1) = = 13 12 11 143 765 35 P ( 0, 0, 0 ) = = 13 12 11 286 P (1,1, 0 ) = P (1, 0,1) = P ( 0,1,1) = 4.
A warehouse contains TV sets, of which 5% are defective, 60% are in working condition but used, and the rest are brand new. What is the probability that in a random sample of five TV sets from this warehouse, there are exactly one defective and exactly two brand new sets? Show work. P ( x1 = 1, x2 = 2, x3 = 2 n = 5, p1 = .05, p2 = .60, p3 = .35 ) ⎛5⎞ 5! =⎜ .051 .60 2 .35 2 = .051 .60 2 .35 2 = 0.06615 ⎝ 1, 2, 2 ⎟ ⎠ 1!2!2! ( )( )( ) ( )( )( ) 5.
The joint density function of X and Y is given by 3 Are X and Y independent? Yes, yes. f ( x) = f ( y) =
∞ ∫ xe
0 − x− y dy = xe− x , dx = e− y , x >0 y >0
∞ ∫ xe
0 − x− y f ( x, y ) = f ( x ) f ( y )
What if f(x, y) were given by ⎧2 f ( x, y ) = ⎨ ⎩0 0 < x < y ≤1 elsewhere no. f ( x) = f ( y) = ∫ 2dy = 2(1 − x ),
x y 1 0 < x <1 0 < y <1
∫ 2dx = 2 y,
0 f ( x, y ) ≠ f ( x ) f ( y )
6. The joint probability density function of X and Y is given by f ( x, y ) = c( y 2 − x 2 )e − y − y ≤ x ≤ y, 0 < y < ∞ (a) Find c. (Hint: a Gamma function appears somewhere… try to find it). (b) Find the marginal density of Y. (c ) Find the expected value of X. 4 (a) 1= X ,Y ∫∫ c ( y 2 − x 2 )e− y dxdy = c ∫ 0 ∞ [∫ y −y ( y 2 − x 2 )e− y dx dy ⎤ ∞⎡ 1 y y = c ∫ ⎢ y 2e− y x − y + e− y x 3 − y ⎥dy 0 3 ⎣ ⎦ 4∞ = c ∫ y 3e− y dy notice that this integral is the Gamma(4) 30 See page 237 of the textbook, (new edition) 4 = c × 3!= 8c 3 1 c= 8 (b) fY ( y ) = 1 8 ∫ ( y 2 − x 2 )e− y dx = −y y 1 2 −y y 1 1 −y 3 y 1 3 −y y e x −y + e x − y = y e , y ∈ (0, ∞) 8 83 6 7.
The joint probability density function of X and Y is given by f ( x, y ) = e − ( x + y ) 0 ≤ x < ∞, 0 ≤ y < ∞ Find (a) P{X < Y} (b) P{X < a}. (a) P( X < Y ) = = ∫ [∫
∞ 0 y − x− y o e dx dy = e dy − ∫ [e
∞ 0 −y (−e− x ) y dy 0 ∫ [e
∞ 0 −y − e−2 y ] dy = ∫ ∞ −y 0 1 2 ∫ ∞ −2 y 0 e d2 y = 1− 11 = 22 (b) a ∞ a P( X < a ) = ∫ ⎡ ∫ e − x − y dy ⎤ dx = ∫ e − x (−e − y ) ∞ dx 0 ⎥ 0⎢o 0 ⎣ ⎦ [ = ∫ e dx = −e
−x 0 a −x a 0 = 1− e −a 8. The joint density of X and Y is given by 5 f ( x, y ) = e−x / ye− y , 0 < x < ∞, 0 < y < ∞ y Compute E[X2  Y=y]. 1 −y − y ee x 1 −y y f X Y ( x  y ) = =e e− y y E ( X 2  Y = y) =
x 2 x − y ∫ ∞ 0 ∞⎛ x ⎞ 1 x x 2 e dx = y 2 ∫ ⎜ ⎟ e d = 2 y 2 0y y y ⎝⎠ x − y 9.
Gasoline is to be stocked in a bulk tank once each week and then sold to customers. Let X denote the proportion of the tank that is stocked in a particular week, and let Y denote the proportion of the tank that is sold in the same week. Due to limited supplies, X is not fixed in advance but varies from week to week. Suppose that a study of many weeks shows the joint relative frequency behavior of X and Y to be such that the joint density function provides an adequate model: f(x, y) = 3 x 0 ≤ y ≤ x ≤ 1 (a) Find the conditional density function of Y given that X=0.5, f(Y X=0.5) and the conditional expectation of Y given that X=0.5. (b) Among all weeks in which the tank was half full, immediately after stocking, how often did sales amount to less than 20% of the tank ( a) f ( x, y ) f (X) 3x 1 = = 0<y<x 2 [3 x ] x f (y  X) = f ( y  X = 0.5) = 2 E [Y  X = 0.5] =
0.5 0 < y < 0.5 ∫ (2 y )dy = y
0 2 0.5 0  = 0.25 (b) P (Y < .20  X = 0.5) =
.20 ∫ 2 dy = 2 y
0 .20 0 = .40 6 ...
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 Winter '09

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