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STA 4321/5325 – Spring 2007 – Exam 4 – 6
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The following table gives the joint probability distribution for the numbers of washers (X 1 ) and dryers (X 2 ) sold by an appliance store salesperson in a day. Washers\Dryers x2=0 x2=1 x2= x1=0 0.25 0.08 0. x1=1 0.12 0.20 0. x1=2 0.03 0.07 0. Give the marginal distributions of numbers of washers and dryers sold per day. Give the Expected numbers of Washers and Dryers sold in a day. w\d 0 1 2 f(w) wf(w) E(W) 0 0.25 0.08 0.05 0.38 0 0. 1 0.12 0.2 0.1 0.42 0. 2 0.03 0.07 0.1 0.2 0. f(d) 0.4 0.35 0. df(d) 0 0.35 0. E(D) 0.** Give the covariance between the number of washers and dryers sold per day. wdf(w,d) 0 1 2 0 0 0 0 1 0 0.2 0. 2 0 0.14 0. E(WD) 0. COV(W,D) = 0.94-(0.820.85) = 0. If the salesperson makes a commission of $100 per washer and $75 per dryer, give the average daily commission. E[C] = E[100W+75D] = 100(0.82) +75*(0.85) = $147.
The number of stops (X 1 ) in a day for a delivery truck driver is Poisson with mean . Conditional on their being X 1 =x 1 stops, the expected distance driven by the driver (X 2 ) is Normal with a mean of x 1 miles, and a standard deviation of x 1 miles. Give the mean and variance of the numbers of miles she drives per day. E[X 1 ] = = V[X 1 ] E[X 2 |X 1 =x 1 ] = x 1 V[X 2 |X 1 =x 1 ] = ^2 x 12 E[X 2 ] = EX1{E[X 2 |X 1 ]} = E[X 1 ] = E[X 1 ] = V[X 2 ] = EX1{V[X 2 |X 1 ]} + VX1{E[X 2 |X 1 ]} = E[^2 X 12 ] + V[X 1 ] = ^2 E[X 12 ] + ^2 V[X 1 ] = ^2 ( The joint distribution of X 1 and X 2 is: 0 otherwise 0 , 1 ( 1 , 2 )^12 k x x f x x Give k k = 1 Give the marginal distributions of X 1 and X 2
1 (^20) 1 1 1 0 2
f x dx x x f x x
Are X 1 and X 2 independent? Yes… f(x 1 ,x 2 )=f 1 (x 1 )f 2 (x 2 ) The joint distribution of X 1 and X 2 is: 0 otherwise 2 0 1 ( 1 , 2 )^12 x x f x x and the marginal distribution of X 1 is: 0 otherwise 2 ( 1 ) 0 1 1 (^1 )^11 x x f x
Show that when X 1 and X 2 are independent then the moment generating function of Y=X 1 +X 2 is: MY(t) = MX1(t)MX2(t) MY(t) = E(etY) = E(et(X1+X2)) = E(etX1etX2) = (indep) = E(etX1)E(etX2) MX1(t)MX2(t) Use the result from the previous part to obtain the distribution of Y. MY(t) = (pet+(1-p))n1(pet+(1-p))n2^ = (pet+(1-p))n1+n Y~Binomial(n 1 +n 2 , p) The location of stores along a highway follows the density function: 1 10 ( 10 ) 20 10 10 0 10 ( ) 0 elsewhere
- 05 10 10 ( ) x x x x F x x f x
A courier (whose office is located at point 0) incurs a cost of U=4X^2 when making a delivery to a firm located at point X. Obtain the density function of costs by the courier by completing the following parts: What range of values can U take on? U [0,400] When U=u, what are the 2 locations where the delivery is located at (this is a function of u). 4x^2 = u x = ± u0.5/ When U ≤ u, what is the range of delivery locations? U [ - u0.5/2, + u0.5/2] Obtain FU(u)=P(U≤u) by obtaining the probability the store is in the range of locations in the previous part. 20 20 2 10 20 2 10 2 2 ( ) u u u u X u FU u P ^ ^ Obtain the density function fU(u) from the cumulative distribution function FU(u) 0 400 40 1 20 ( ) ( ) u u u du d du dF u f (^) U u U
