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410Hw02ans - STAT 410 Summer 2009 Homework#2(due Wednesday...

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STAT 410 Summer 2009 Homework #2 (due Wednesday, June 24, by 4:00 p.m.) 1. The time, T , that a manufacturing system is out of operation has cumulative distribution function F ( t ) = - otherwise 0 2 if 2 1 2 t t The resulting cost to the company is Y = T 2 . Determine the density function of Y . t > 2 Y = T 2 y > 4 F Y ( y ) = P ( Y y ) = P ( T 2 y ) = P ( T y ) = y 4 1 - , y > 4. f Y ( y ) = F Y ' ( y ) = otherwise 0 4 if 4 2 y y OR f T ( t ) = F T ' ( t ) = otherwise 0 2 if 8 3 t t g ( t ) = t 2 g 1 ( y ) = y d t / d y = y 2 1 = 2 1 y 1 / 2 f Y ( y ) = f T ( g 1 ( y ) ) y t d d = ( 8 y 3 / 2 ) ( 2 1 y 1 / 2 ) = 4 y 2 , y > 4
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2. Grades on the last STAT 410 exam were not very good. Graphed, their distribution had a shape similar to the p.d.f. f X ( x ) = 000 , 5 1 ( 100 – x ) , 0 x 100, zero elsewhere. a) What was the class average, E ( X ) ? E ( X ) = ( 29 - 100 0 100 000 , 5 1 dx x x = 3 33.33 . As a way of “curving” the results, the professor announces that he will replace each person’s grade, X, with a new grade, Y = g ( X ) , where g ( X ) = 10 X . b) Find the p.d.f. that describes the new grades, Y. y = 10 x x = 100 2 y dy dx = 50 y f Y ( y ) = 50 100 100 000 , 5 1 2 y y - = ( 2 000 , 10 000 , 000 , 25 1 y y - , 0 y 100. c) Has the professor’s strategy been successful in raising the class average above 60? What is the new class average, E ( Y ) ? E ( Y ) = ( 29 - 100 0 100 000 , 5 1 10 dx x x = 3 53.33 . OR E ( Y ) = ( - 100 0 2 000 , 10 000 , 000 , 25 1 dy y y y = 3 53.33 .
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3. Suppose that X follows a uniform distribution on the interval [ π / 2 , π / 2 ] . Find the c.d.f. and the p.d.f. of Y = tan X. f X ( x ) = < < - o.w. 0 2 2 1 π π π x F X ( x ) = < - + - < 2 1 2 2 2 1 2 0 π π π π π x x x x F Y ( y ) = P ( Y y ) = P ( tan X y ) = P ( X arctan ( y ) ) = ( 29 2 1 1 arctan + y π , < y < . f Y ( y ) = ( 2 1 1 y + π , < y < . ( Standard ) Cauchy distribution. OR g ( x ) = tan x g 1 ( y ) = arctan ( y ) d x / d y = 2 1 1 y + f Y ( y ) = f X ( g 1 ( y ) ) y x d d = + 2 1 1 1 y π = ( 2 1 1 y + π , < y < . F Y ( y ) = ( + - y du u 2 1 1 π = ( 29 2 1 1 arctan + y π , < y < .
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4. 1.7.21 If the p.d.f. of X is f ( x ) = 2 x e x 2 , 0 < x < , zero elsewhere, determine the p.d.f. of Y = X 2 . f X ( x ) = < < - o.w. 0 0 2 2 x e x x F X ( x ) = - < - 0 1 0 0 2 x e x x Y = X 2 . y < 0 F Y ( y ) = P ( Y y ) = P ( X 2 y ) = 0. y 0 F Y ( y ) = P ( Y y ) = P ( X 2 y ) = P ( X y ) = F X ( y ) . = 1 – e y . F Y ( y ) = - < - 0 1 0 0 y e y y f Y ( y ) = < - 0 0 0 y e y y . OR g ( x ) = x 2 one-to-one, differentiable on ( 0, ) g 1 ( y ) = y d x / d y = y 2 1 f Y ( y ) = f X ( g 1 ( y ) ) y x d d = ( - y y y e 2 1 2 = e y , y > 0.
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5. 1.7.24 Let f ( x ) = 3 1 , 1 < x < 2, zero elsewhere, be the p.d.f.
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