P 45 X 50 P X 45 P X 46 P X 50 P X x C n x p x1 p n x n x n x p x 1 p n x P X

P 45 x 50 p x 45 p x 46 p x 50 p x x c n x p x1 p n x

This preview shows page 13 - 15 out of 15 pages.

P (45 X 50) = P ( X = 45) + P ( X = 46) + · · · + P ( X = 50) P ( X = x ) = C n x p x (1 p ) n x = n ! x !( n x )! p x (1 p ) n x P ( X = 45) = C 100 45 (0 . 4) 45 (0 . 6) 55 = 100! 45! × 55! (0 . 4) 45 (0 . 6) 55 P ( X = 46) , P ( X = 47) , P ( X = 48) , P ( X = 49) , P ( X = 50) 13
Image of page 13
Section 5.4: Normal Distribution Approximation for Binomial Distribution X Binomial( n, p ), E ( X ) = np , V ar ( X ) = np (1 p ) Result : If the number of trial n is large — such that np (1 p ) > 5 — then the distribution of the random variable Z = X E ( X ) V ar ( X ) = X np np (1 p ) is approximately a standard normal distribution. Z = X np np (1 p ) approx N (0 , 1) 14
Image of page 14
Z = X np np (1 p ) approx N (0 , 1) Example 5.8 Customer Visits Generated from Web Page Contacts, p. 202 Mary David makes the initial telephone contact with customers who have responded to an advisement on her company’s Web Page in an e ff ort to assess whether a follow-up visit to their hoomes is likely to be worthwhile. Her experience suggests that 40% of the initial contacts lead to follow-up visits. If she has 100 Web page contacts, what is the probability that between 45 and 50 home visits will result? 15
Image of page 15

You've reached the end of your free preview.

Want to read all 15 pages?

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

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors