Chapt 3b

# Chapt 3b - 3.4 Moments of Random Variables and the Variance...

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Unformatted text preview: 3.4 Moments of Random Variables and the Variance 95 Finally, the second moment of K is given by E bracketleftbig K 2 bracketrightbig = Î» e- Î» âˆž summationdisplay k = 1 k Î» k- 1 ( k- 1 ) ! = (Î» e- Î» )( 1 + Î») e Î» = Î» + Î» 2 And the variance of K is given by Ïƒ 2 K = E bracketleftbig K 2 bracketrightbig- ( E [ K ] ) 2 = Î» + Î» 2- Î» 2 = Î» trianglesolid Thus, the expected value and variance of K have identical values. Example 3.8 A test engineer discovered that the CDF of the lifetime of an equipment in years is given by F X ( x ) = braceleftbigg x < 1- e- x / 5 â‰¤ x < âˆž a. What is the expected lifetime of the equipment? b. What is the variance of the lifetime of the equipment? Solution From the definition of its CDF, we can see that X is a random variable that takes only nonnegative values. Thus, (a) The expected lifetime of the equipment is given by E [ X ] = integraldisplay âˆž P [ X > x ] dx = integraldisplay âˆž [ 1- F X ( x ) ] dx = integraldisplay âˆž e- x / 5 dx = 5 (b) To find the variance, we first evaluate the PDF: f X ( x ) = d dx F X ( x ) = braceleftBigg 1 5 e- x / 5 x â‰¥ otherwise 96 Chapter 3 Moments of Random Variables Thus, the second moment of X is given by E bracketleftbig X 2 bracketrightbig = integraldisplay âˆž-âˆž x 2 f X ( x ) dx = 1 5 integraldisplay âˆž x 2 e- x / 5 dx Let u = x 2 â‡’ du = 2 xdx , and let dv = e- x / 5 dx â‡’ v = - 5 e- x / 5 . Thus, E bracketleftbig X 2 bracketrightbig = braceleftbigg- 5 x 2 e- x / 5 5 bracerightbigg âˆž + 10 integraldisplay âˆž xe- x / 5 5 dx = + 2 integraldisplay âˆž xe- x / 5 dx = 2 integraldisplay âˆž xe- x / 5 dx Let u = x â‡’ du = dx , and let dv = e- x / 5 dx â‡’ v = - 5 e- x / 5 . Then we have that E bracketleftbig X 2 bracketrightbig = 2 braceleftbig- 5 xe- x / 5 bracerightbig âˆž + 10 integraldisplay âˆž e- x / 5 dx = + 10 bracketleftbig- 5 e- x / 5 bracketrightbig âˆž = 50 Finally, the variance of X is given by Ïƒ 2 X = E bracketleftbig X 2 bracketrightbig- { E [ X ]} 2 = 50- 25 = 25 trianglesolid Example 3.9 A shopping cart contains ten books whose weights are as follows: There are four books with a weight of 1.8 lbs each, one book with a weight of 2 lbs, two books with a weight of 2.5 lbs each, and three books with a weight of 3.2 lbs each. a. What is the mean weight of the books? b. What is the variance of the weights of the books? Solution The total number of books is 10. The fractions of books in each weight category are as follows: Fraction of books with weight 1.8 lbs is 4 / 10 = . 4 Fraction of books with weight 2.0 lbs is 1 / 10 = . 1 Fraction of books with weight 2.5 lbs is 2 / 10 = . 2 Fraction of books with weight 3.2 lbs is 3 / 10 = . 3 Let Y be a random variable that denotes the weights of the books. Since these fractions are essentially the probabilities of occurrence of these weights, we have that E [ Y ] = ( . 4 Ã— 1 . 8 ) + ( . 1 Ã— 2 . ) + ( . 2 Ã— 2 . 5 ) + ( . 3 Ã— 3 . 2 ) = 2 . 38 3.4 Moments of Random Variables and the Variance 97 Ïƒ 2 Y = 4...
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Chapt 3b - 3.4 Moments of Random Variables and the Variance...

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