hw8sol - ISyE 2027 Fall 2011 Hmwk 8 Solution 1 1(a Binomial...

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

View Full Document Right Arrow Icon
ISyE 2027 Fall 2011 Hmwk 8 Solution November 17, 2011 1. (a) Binomial with parameter ( 5, 1 4 ) ; (b) 5 4 ; (c) 15 16 ; (d) ( 5 i ) ( 1 4 ) i ( 5 - i j ) ( 1 2 ) j . (e) mean 5, variance 0. 2. P (- k 6 Z 6 k ) = 0.682 if k = 1, 0.956 if k = 2 0.997 if k = 3 3. (a) 0.682, (b) 0.956, (c) 0.997, (d) 0.022, (e) 0.021, (f) 0.044. 4. (a) 0.975, (b) 0.05, (c) 0.2195, (d) 0.2195. (e) 1. 5. Let Z be a standard normal random variable. I would not expect the nor- mal approximation to be great since X is discrete, rather skewed (asym- metric), and the sample size is not particularly large. (a) P { Y 6 3.5 } , P { Y 6 2.5 } , P { 2.5 < Y 6 3.5 } , P { 2.5 < Y 6 4.5 } . (b) P { X 6 3 } = 3 X k = 0 P { X = k } = 3 X k = 0 15 k ( 1 / 5 ) k ( 4 / 5 ) 15 - k 0.6482 P { Y 6 3.5 } = P Y - 3 2 15 / 5 6 3.5 - 3 2 15 / 5 = P { Z 6 15 / 12 } .6266 P { X < 3 } = 2 X k = 0 P { X = k } = 2 X k = 0 15 k ( 1 / 5 ) k ( 4 / 5 ) 15 - k 0.3980 1
Image of page 1

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

View Full Document Right Arrow Icon
P { Y 6 2.5 } = P Y - 3 2 15 / 5 6 2.5 - 3 2 15 / 5 = P { Z 6 - 15 / 12 } 0.3734 P { X = 3 } = 15 k ( 1 / 5 ) k ( 4 / 5 ) 15 - k 0.2501 P { 2.5 < Y 6 3.5 } = P 2.5 - 3 2 15 / 5 < Y - 3 2 15 / 5 6 3.5 - 3 2 15 / 5 = P { - 15 / 12 < Z 6 15 / 12 } .2531 P { 2 < X 6 4 } = 4 X k = 3 P { X = k } = 4 X k = 3 15 k ( 1 / 5 ) k ( 4 / 5 ) 15 - k 0.4377 P { 2.5 < Y 6 4.5 } = P 2.5 - 3 2 15 / 5 < Y - 3 2 15 / 5 6 4.5 - 3 2 15 / 5 = P { - 15 / 12 < Z 6 15 / 4 } .4601 6. Note that E [ X ] = μ = np = 20 and StDev [ X ] = σ = np ( 1 - p ) = 4 . I will
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