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Asst8-solved - Statistics 427: Homework #8 Solutions to...

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Unformatted text preview: Statistics 427: Homework #8 Solutions to Assigned Problems 4.60 [8 pts (3,3,2)] (a) We are told that X Ex( ) where = . 01386. The probability that the distance is at most 100km is P ( X 100) = Z 100 e- w dw =- e- w 100 =- e- 100(0 . 01386) + 1 = 0 . 7499 . We can use the cdf, F X ( x ) = 1- e- x to directly compute these values. The probability that the distance is at most 200km is P ( X 200) = F X (200) = 1- e- 200( . 01386) = 0 . 9375 , and the probability that the distance is betwee 100 and 200 km is P (100 X 200) = F X (200)- F X (100) = 0 . 9375- . 7499 = 0 . 1876 . (b) The mean is E[ X ] = 1 / = 72 . 15 = . The variance is V [ X ] = 1 / 2 and so the standard deviation is = 1 / = 72 . 15 = . So the probability that X exceeds the mean distance by more than two standard deviations is P ( X + 2 ]) = P ( X + 2 ) = P ( X 3 ) = 1- P ( X 3 ) = 1- F X (3 ) = 1- [1- exp {- 3 } ] = exp {- 3 / } = exp {- 3 } = 0 . 0498 (c) The median distance is the value of that solves P ( X ) = 0 . 5. Using the cdf, we have F X ( ) = 0 . 5 and so 1- exp {- } = 0 . 5, so 0 . 5 = exp {- } . Taking the natural logarithm, we have- ln(2) =- , which leaves us with = - 1 ln(2). Using = . 01386, this shows that the median is 50 . 01. 1 4.63 [8 pts (2,6)] Let X be a continuous random variable that represents call duration. We are given that X Exp( ). (a) If the probability that a person makes short telephone calls (less than 10 minutes) is large, then the person will save money by using the calling plan that charges a flat rate of 10 cents per minute. If the person usually makes long telephone calls, then the person will save money by using the plan that charges 99 cents for all calls up to 20 minutes. (b) Under the first plan, if a call lasts for x minutes the total cost is h 1 ( x ) = x/ 10 dollars. Under the second plan, the cost is h 2 ( x ) = . 99 x 20 . 99 + ( x- 20) / 10 x > 20 ....
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Asst8-solved - Statistics 427: Homework #8 Solutions to...

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