{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Asst8-suggested - Statistics 427 Homework Solutions to...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Statistics 427: Homework # ! Solutions to Suggested Problems Note : This page contains solutions for 4.59 and 4.67; the following pages contain solutions for 4.65, 4.68, 5.3 and 5.13. 4.59 X is Gamma with ! = 1 and " = 1 a. E(X) = 1 1 = ! b. 1 1 = = ! " c. P(X ! 4 ) = 982 . 1 1 4 ) 4 )( 1 ( = ! = ! ! ! e e d. P(2 ! X ! 5) = [ ] 129 . 1 1 5 2 ) 2 )( 1 ( ) 5 )( 1 ( = ! = ! ! ! ! ! ! ! e e e e 4.67 Suppose μ = 24 and " 2 = 144 # $% = 24, $% 2 = 144 # % = 6, $ = 4 a. P(12 ! X ! 24 ) = F(4;4) – F(2;4) = .424 b . P(X ! 24) = F(4;4) = .567, so while the mean is 24, the median is less than 24. (Note that P(X ! μ ~ ) = .5) The fact that the median is less than the mean is caused by the positive skew of the gamma distribution. c . We want a value of X for which F(X; 4) =.99. From Table A.4, we see F(10;4)=.990. So with % = 6, the 99 th percentile = 6(10)=60. d . We want a value of X for which F(X; 4) =.995. Based on Table A.4, F(11; 4) = .995, so t = 6(11)=66. Thus at 66 weeks, only .5% of all transistors would still be operating.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
4.65 We are given that X Gamma( α = 7 , β = 1). (a) P ( X 5) = F (5; α = 7) = 0 . 238 (from table) (b) P ( X < 5) = P ( X 5) = 0 . 238 (c) P ( X > 8) = 1 - P ( X 8) = 1 - F (8; α = 7) = 1 - 0 . 687 = 0 . 313 (d) P (3 X 8) = P ( X 8) - P ( X < 3) = F (8; α = 7) - F (3; α = 7) = 0 . 687 - 0 . 034 = 0 . 653 (e) P (3 < X < 8) = P (3 X 8) = 0 . 653 (f) P ( X < 4 or
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern