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Unformatted text preview: Math 361/Stat 351 X1 Homework 10 Solutions Spring 2003 Graded problems: 1(d), 3(c), 4(b), 7 Problem 1. [4.2.4] Suppose component lifetimes are exponentially distributed with mean 10 hours. Find (a) the probability that a component survives 20 hours; (b) the median component lifetime (see p. 165 for a definition of the median of a distribution); (c) the SD of component lifetime; (d) the probability that the average lifetime of 100 independent components exceeds 11 hours. Solution. (a) We need to compute P ( X 20), where X is the life time of the component. We are given that X has exponential distribution with mean E ( X ) = 10 (measured in hours). Since for an exponential( ) distribution E ( X ) = 1 / , we must have = 1 /E ( X ) = 1 / 10. Thus P ( X &gt; 20) = 1 P ( X &lt; 20) = 1 F (20) = 1 (1 e (1 / 10) 20 ) = e 2 , using the formula F ( x ) = 1 e x ( x &gt; 0) for the c.d.f. of an exponential distribution. (b) The median lifetime is the value of t for which P ( X t ) = 1 / 2 = P ( X t ). Now, by the same argument as in (a), we have for t P ( X t ) = e 5 / 10 , and setting this value equal to 1 / 2 we get e t/ 10 = 1 / 2 or t = 10ln2 = 6 . 93. (c) Since the lifetime X is exponentially distributed with parameter = 1 / 10, its standard deviation is, by the formula for the variance of an exponential distribution, SD ( X ) = p Var( X ) = p 1 / 2 = 1 / = 10 . (d) Let X 1 ,X 2 ,... ,X 100 be the lifetimes of the 100 independent components. The random variables X i are i.i.d., each having exponential distribution with = 1 / 10 and = E ( X 1 ) = 10, = SD ( X 1 ) = 10. We need to compute the probability P ( A 100 &gt; 11), where A 100 = 1 100 100 X i =1 X i = 1 100 S 100 is the average lifetime (in the naive sense of average) of the 100 components. By normal approx imation this is P ( A 100 &gt; 11) = 1 P ( A 100 11) = 1 P ( S 100 1100) 1 1100 100 10 100 10 = 1 (1) = 0 . 1587 . Problem 2. [4.R:4] Let X be a random variable with density f ( x ) = 0 . 5 e x  ( &lt; x &lt; ). Find (a) P ( X &lt; 1), (b) E ( X ) and SD ( X ), (c) the c.d.f. of X 2 . Solution. P ( X &lt; 1) = Z 1 f ( x ) dx = 1 2 Z 1 e x  dx = 1 2 Z e x dx + 1 2 Z 1 e x dx (a) = 1 2 e x + 1 2( 1) e x 1 = 1 2 + 1 2 ( e 1 + 1 ) = 1 1 2 e ....
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This note was uploaded on 04/03/2008 for the course STAT 134 taught by Professor Aldous during the Spring '03 term at University of California, Berkeley.
 Spring '03
 aldous
 Probability

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