practice-exam2-sol

practice-exam2-sol - AMS 311 Joe Mitchell Practice Exam 2...

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AMS 311 Joe Mitchell Practice Exam 2 – Solution Notes Statistics: n = 35, μ = 72 . 8, median 86, σ = 28 . 0; score range: 8–100 1. (a). P ( X > 0 . 75) = integraldisplay 0 . 75 f ( x ) dx = integraldisplay 1 0 . 75 xdx + integraldisplay 2 1 1 2 dx = 23 32 (or write P ( X > 0 . 75) = 1 integraltext 0 . 75 0 xdx ) (b). The area is given by A = X 2 , since it is an X -by- X square. Thus, we compute: var ( X 2 ) = E ( X 4 ) [ E ( X 2 )] 2 = integraldisplay 1 0 x 4 · xdx + integraldisplay 2 1 x 4 · 1 2 dx bracketleftbiggintegraldisplay 1 0 x 2 · xdx + integraldisplay 2 1 x 2 · 1 2 dx bracketrightbigg 2 (c). (First, note that since X is always in (0,1) or (1,2), then Y = 5 X is always in (0,5) or (5,10).) We want to compute F Y ( y ) = P ( Y y ) = P (5 X y ) = P ( X y 5 ) = F X ( y 5 ), which requires looking at cases: F Y ( y ) = 0 if y 5 0; i.e., if y 0 integraltext y/ 5 0 xdx = y 2 50 if 0 y 5 1; i.e., if 0 y 5 integraltext 1 0 xdx + integraltext y/ 5 1 1 2 dx = y 10 if 1 y 5 2; i.e., if 5 y 10 1 if y 5 2; i.e., if y 10 2. X is exponential with parameter 1 5 years - 1 (since E ( X ) = 5 years). (a). The probability that a radio lasts more than 15 years is P ( X > 15) = integraltext 15 1 5 e - x/ 5 dx = e - 15 / 5 = e - 3 (b). Let Y be the number (among the 1000) that last more than 15 years. Then, Y is Binomial(1000, p ), where p = e - 3 was found in part (a). We want P ( Y 4) = 1000 summationdisplay i =4 parenleftBigg 1000 i parenrightBigg p i (1 p ) 1000 - i (c). Since 1000 is “large” and e - 3 is “small”, we know that Y has a distribution that is approximated by that of Z , a Poisson(1000 p ) random variable. Thus, P ( Y 4) P ( Z 4) = summationdisplay i =4 e - 1000 p (1000 p ) i i ! (Note: you could also stop the summation at i = 1000; it does not really matter for the approximation.) 3. X is Normal(75,2 2 ). We want P (74 . 2 X 75 . 6) = P ( 74 . 2 75 2 X 75 2 75 . 6 75 2 ) = P ( 0 . 4 Z 0 . 3) = Φ(0 . 3) Φ( 0 . 4) = Φ(0 . 3) (1 Φ(0 . 4)) = . 6179 (1 . 6554) = . 2733 4. (20 points) We begin by plotting the support set (below, left, in red). y=x y=3-x y=2x y=3-x y=x 1 3/2 3/2 3 3 3 3 1

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(a). There are various cases, depending on the value of x : think of sweeping a vertical line across the plane, and see how it interacts with the support set: f X ( x ) = integraldisplay -∞ f ( x, y ) dy = braceleftbigg integraltext 3 - x x 4 9 dy = 4 3 8 x 9 if 0 x 3 2 0 otherwise (b). Draw a figure! In the figure (above, right) we plot and highlight in green that portion of the support set where y > 2 x . To compute the probability P ( Y > 2 X ), we integrate the joint density over the green region: P ( Y > 2 X ) = integraldisplay 1 0 integraldisplay 3 - x 2 x 4 9 dydx = 2 3 (c). In order to compute the expected value, we integrate the product of the function, xy , and the joint density function, over the entire support set: E ( XY ) = integraldisplay 3 / 2 0 integraldisplay 3 - x x xy · 4 9 dydx (d). X and Y are NOT independent, since the support set is not a rectangle; in particular, f (1 , 0 . 5) = 0 negationslash = f X (1) · f Y (0 . 5) > 0.
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