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Show that E(X) = np when X is a binomial random variable.

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 64E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 64E

Show that E(X) = np when X is a binomial random variable. [Hint: First express E(X) as a sum with lower limit x = 1.Then factor out np, let y = x - 1 so that the sum is from y = 0 to y = n - 1, and show that the sum equals 1.]

Step-by-Step Solution:
Step 1 of 3

Problem 64E Answer: Step1: Let “X” be random variable it follows binomial distribution with parameters “p” and “n”. The probability mass function of binomial distribution is given by p(x) = C p q n-, x = 0,1,2,...,n x We need to prove mean E(x) = np. Step2: Consider, n E(X) = xP(X = x) x=0 n n x nx = x[( )px(1 p) ] x=0 n = x[ n! p (1 p) nx] x=0 x!(nx)! For x = 0 term in the sum will be 0,hence we can change the index on the sum n = x[ n! p (1 p) nx] x=1 x!(nx)! n n! x nx = [ (x1)!(nx)! (1 p) ] x=1 Move the index back to zero n1 n! x nx = [ x![n(x+1)]! (1 p) ] x=0 n1 = np [ (n1)! p (1 p) nx] x![n1x)]! x=0 = np{B(n 1, p)} The sum of pmf is 1 = np Therefore,E(X) = np Hence the proof.

Step 2 of 3

Chapter 3, Problem 64E is Solved
Step 3 of 3

Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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Show that E(X) = np when X is a binomial random variable.