practice-exam2-sol

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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) = ± 0 . 75 f ( x ) dx = ± 1 0 . 75 xdx + ± 2 1 1 2 dx = 23 32 (or write P ( 0 . 75) = 1 - ² 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 = ± 1 0 x 4 · xdx + ± 2 1 x 4 · 1 2 dx - ³ ± 1 0 x 2 · xdx + ± 2 1 x 2 · 1 2 dx ´ 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 0i f y 5 0; i.e., if y 0 ² y/ 5 0 xdx = y 2 50 if 0 y 5 1; i.e., if 0 y 5 ² 1 0 xdx + ² 5 1 1 2 dx = y 10 if 1 y 5 2; i.e., if 5 y 10 1i f 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 ( 15) = ² 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 µ i =4 1000 i · 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) = µ 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
(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 )= ± -∞ f ( x,y ) dy = ² ³ 3 - x x 4 9 dy = 4 3 - 8 x 9 if 0 x 3 2 0 otherwise (b). Draw a Fgure! In the Fgure (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 ( 2 X ± 1 0 ± 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 ± 3 / 2 0 ± 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 ± = f X (1) · f Y (0 . 5) > 0.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/06/2010 for the course EC 11 taught by Professor All during the Spring '10 term at UCLA.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online