ST 558 Homework 4

ST 558 Homework 4 - Kelly Harrison 2/14/10 ST 558 D...

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Unformatted text preview: Kelly Harrison 2/14/10 ST 558 D Homework 4 PROBLEM 1 Determine the value c so each of the following functions can serve as a probability distribution of the discrete random variable x. a) f(x) = c(x2+4), x = 0, 1, 2, 3 f(x) = 1 c(02+4) + c(12+4) + c(22+4) + c(32+4) = 1 c[4 + 5 + 8 + 13] = 1 30c = 1 c = 1/30 b) f(x) = c(2,x)(3,3-x) for x = 0, 1, 2 f(x) = 1 c(2,0)(3,3-0) + c(2,1)(3,3-1) + c(2,2)(3,3-2) = 1 c(1)(1) + c(2)(3) + c(1)(3) = 1 c(1 + 6 + 3) = 1 10c = 1 c = 1/10 PROBLEM 2 The shelf life, in days, for bott les of a certain prescribed medicine is a random variable having the densit y funct ion: f(x) = 20,000/(x+100)3, 0, P( x 200) = x>0 elsewhere dx = 20, 000 lim p 200 x + 100 ) ( 20, 000 3 200 ( x p + 100 ) -3 dx = -10, 000 lim p 1 (x + 100 ) 2 | p 200 = -10, 000 lim p 1 1 1 1 - = -10, 000 0 - = 2 2 300 2 9 ( p + 100 ) ( 200 + 100 ) b) P(80 < x < 120) = 120 80 x + 100 ) ( 20, 000 dx = 20, 000 3 120 80 ( x + 100 ) -3 dx = -10, 000 120 1 | 2 80 ( x + 100 ) = -10, 000 1 1 - = .10203 2 2 ( 120 + 100 ) ( 80 + 100 ) PROBLEM 3 The total number of hours, measured in units of 100 hours, that a family runs a vacuum cleaner over a period of one year is a contin uous random variable x that has the density function: x, f(x) = 2-x, 0, a) P( x < 1.2) = = 1 2 0<x<1 1x<2 elsewhere. 0 1 xdx + 1.2 (2 - x ) dx 1 = x 2 1 2 |+ - 2x 0 x 2 2 1.2 1.2 1 12 0 2 | = - + 2(1.2) - - 2(1) - 1 2 2 2 2 2 2 + 1.68 - 1.5 = .68 b) P( .5 < x < 1) = xdx = .5 1 x 2 2 | = .5 1 1 2 2 - .5 2 2 = .375 PROBLEM 4 The proport ion of people who respond to a certain mail- order solicitat ion is a continuous random variable x that has the density function: f(x) = 2(x+2)/5, 0, a) P(0 < x < 1) = = 1 5 1 0<x<1 elsewhere. 2 5 (x + 2 ) dx = 2 ( x + 2) 5 2 2 | = 0 1 1 5 (x + 2) | = 2 0 1 0 + 2)2 - ( 0 + 2)2 (1 5 1 [ 9 - 4] = 1 5 ( 5) .5 =1 b) P( .25 < x < .5) = PROBLEM 5 2 5 ( x + 2 ) dx = 2 ( x + 2) 5 2 2 .5 .25 .25 | = 1 5 ( x + 2) 2 .25 | = .5 1 .5 + 2 ) 2 - ( .25 + 2 ) 2 = .2375 ( 5 A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of 3 of the sets. If x is the number of defective sets purchased by the hotel, find the probabilit y distribu tion of x. Express the results graphically as a probabilit y histogram. 2 5 x -x 3 P( x) = , x = 0, 1, 2 7 3 2 5 0 1(10) 3 2 P(0) = = = 7 35 7 3 2 5 1 2(1 0) 2 4 P(1) = = = 7 35 7 3 2 5 2 1 ( 5 ) 1 1 P(2) = = = 7 35 7 3 PROBLEM 6 The waiting time, in hours, between successive speeders spot ted by a radar unit is a continuous random variable with a cumulative distribution: F(x) = 0, x0 1-e , x 0 -8x 12 minutes = 1/5 hour a) P(x < 12) = 1 e-8(1/5)= .7981 b) This is an exponential distribution with f(x) = 8e-8x, x 0 0, x<0 P( x < 12) = 1/ 5 8e -8 x 0 - 1 -8 x 1 |/ 5 = -e -8 x 1 |/ 5 = - e -8 ( 1 / 5 ) - e 0 dx = 8 e 0 8 0 ( ) = .7981 PROBLEM 7 a) P(1 < x < 3) = 3 1 2 dx = 1 2 x| = 1 3 1 2 (3 - 1) = 1 2 ( 2) =1 1 b) P(2 < x < 2.5) = 2.5 1 2 dx = 1 2 x | = 2 2.5 1 2 (2.5 - 2) = 1 2 ( .5 ) = .25 2 c) P( x ( 1.6) = 1.6 1 2 dx = 1 2 x | = 1 1.6 1 2 (1.6 - 1) = 1 2 ( .6 ) = .3 1 ...
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This note was uploaded on 05/06/2010 for the course ST 558 taught by Professor Staff during the Spring '08 term at NMT.

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