math3C_Towsner_s11_midterm2.pdf

Xander and yolanda take calls for different products

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time 6 minutes. Xander and Yolanda take calls for different products, so their waiting times are independent. At precisely noon, they both hang-up from their last call and are waiting for the next call to come in. X is the amount of time Xander waits, Y is the amount of time Yolanda waits. ( a ) What is the probability that Yolanda waits at least 10 minutes. That is, what is P ( Y > 10)? This is an exponential distribution with parameter 1 / 6, so the cdf is 1 - e - x / 6 , and therefore P ( Y > 10 ) is 1 - ( 1 - e - 10 / 6 ) = e - 10 / 6 = e - 5 / 3 . (continued on next page) 2 of 7

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( b ) What is the chance that Xander gets a call after 5 minutes, but Yolanda is still waiting after 10 minutes. That is, what is P ( X < 5 and Y > 10)? These events are independent, and X is an exponential distribution with parameter 1 / 10, so P ( X < 5 and Y > 10 ) = P ( X < 5 ) P ( Y > 10 ) = ( 1 - e - 5 / 10 ) e - 5 / 3 ( c ) After 10 minutes, Yolanda has not gotten a call. What is the chance that she waits another 10 minutes before getting a call. That is, what is P ( Y > 20 | Y > 10)? Since the exponential is memoryless, P ( Y > 20 | Y > 10 ) = P ( Y > 10 ) = e - 5 / 3 ( d ) Assume that calls are very short, so the number of calls Xander gets is a Poisson process with mean 1 / 10 calls per minutes and the number of calls Yolanda gets is a Poisson process with mean 1 / 6 calls per minute. What is the chance that between noon and 1pm (that is, over one hour), Xander and Yolanda each receive exactly 5 calls? The number of calls Xander receives in an hour is Poisson distributed with mean 6, and therefore parameter 6. The number of calls Yolanda receives in an hour is Poisson distributed with mean 10, and therefore parameter 10.
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