HW13Answers(4.3)

# HW13Answers(4.3) - Math331 Spring 2008 Instructor David...

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Math331, Spring 2008 Instructor: David Anderson Section 4.3 Homework Answers Homework: pgs. 157-158 #’s 1, 3, 4, 7, 8, 10. 1. F is given by F ( t ) = 0 t < 1 1 / 15 1 t < 2 3 / 15 2 t < 3 6 / 15 3 t < 4 10 / 15 4 t < 5 1 5 t . 3. We have R ( X ) = { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } . We may count the number of ways we can get each value and divide by 36, which is the number of elements in the sample space S = { ( a,b ) | a,b ∈ { 1 , 2 , 3 , 4 , 5 , 6 }} . We find p (2) = 1 / 36 ,p (3) = 2 / 36 ,p (4) = 3 / 36 ,p (5) = 4 / 36 ,p (5) = 5 / 36 ,p (7) = 6 / 36 ,p (8) = 5 / 36 ,p (9) = 4 / 36 ,p (10) = 3 / 36 ,p (11) = 2 / 36 ,p (12) = 1 / 36. 4. There are only jumps in the distribution function. Therefore, the probability mass function can be determined by the size of the jumps at each value. We therefore see that the range of the random variable is given by R ( X ) = {- 2 , 2 , 4 , 6 } . The probability mass function is given below p ( - 2) = 1 / 2 p (2) = 3 / 5 - 1 / 2 = . 1 p (4) = 8 / 9 - 3 / 5 = 13 / 45 p (6) = 1 - 8 / 9 = 1 / 9 . 7. (a) We need 1 = k * (1 + 2 + 3 + 4 + 5) = 15 k. Thus, k = 1 / 15. (b) We need 1 = k * (1 + 1 + 4 + 9) = 15 k. Thus, again, k = 1 / 15. (c) We need 1 = k summationdisplay i =1 (1 / 9) i = k 1 / 9 1 - 1 / 9 = k/ 8 . Thus, k = 8. 1

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(d) We need 1 = k n summationdisplay i =1 i = k n ( n + 1) 2 . Thus, k = 2 / ( n ( n + 1)). (e) We need 1 = k n summationdisplay i =1 i 2 = k n ( n + 1)(2 n + 1) 6 . Thus,
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