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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 116e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 3 - Problem 116e

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# The mode of a discrete random variable X with pmf p(x) is

ISBN: 9780321629111 32

## Solution for problem 116E Chapter 3

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

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Problem 116E

The mode of a discrete random variable X with pmf p(x) is that value x* for which p(x) is largest (the most probable x value). a. Let X ?Bin(n, p). By considering the ratio b( x +1; n , p) / b (x; n , p), show that b(x; n, p) increases with x as long as x < np – ( 1 – p). Conclude that the mode x* is the integer satisfying (n + 1)p - 1? x* ? (n + 1) p. b. ?Show that if X has a Poisson distribution with parameter ?, the mode is the largest integer less than ?. If ? is an integer, show that both ? – 1 and ? are modes.

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Answer: Step1: * Given the mode of a discrete random variable X with the pmf p(x) is that value X for which p(x) is largest. Step2: a). Let X follows binomial distribution with parameters n,p. That is, X ~ B(n,p) The pmf of binomial distribution is n x nx B(x;n,p) = ( x p (1 p) ; x = 0,1,2,... b(x +1;n,p) By considering the ratio b(x;n,p) Now we have to show that b(x;n,p) increases with ‘x’ as long as x < np - (1- p) and also conclude that the mode x* is the integer satisfying (n + 1)p - 1 x* (n + 1) p. n x+1 n(x+1) b(x +1;n,p) (x+1) p (1p) Now, b(x;n,p) = ( ) p (1p) nx x x+1 nx1 (x+1)!(nx1)!(1p) = p (1p)x x!(nx)! b(x +1;n,p) p = nx × >1. b(x;n,p) x+1 1p Multiplying the equation np xp = x(1p)+1(1p) >1 = np - xp > x(1-p) + (1-p) = np-xp > x-xp + (1-p) = x < np - (1-p) So, from x < np - (1-p) ; B(x;n,p) increasing x. Step3: b). To show that if X has a poisson distribution with parameter , and the mode is the largest integer less than . if is an integer to show that both - 1 and are modes. Then, the pmf of poisson distribution is x P(x,)= ex! ,x = 1,2,3… ( here and are same) x+1 P(x+1,) (x+1)! Now, = x P(x+) e x!, = x+1 > 1 = > x + 1 P(x+1,) = x < 1 P(x+) From which the state conclusion follow that, is not integer then, mode is largest integer less than . If is integer, then P(,) = P(-1,). So, both and -1 are mode.

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##### ISBN: 9780321629111

This full solution covers the following key subjects: mode, show, Integer, largest, bin. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. The full step-by-step solution to problem: 116E from chapter: 3 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. The answer to “The mode of a discrete random variable X with pmf p(x) is that value x* for which p(x) is largest (the most probable x value). a. Let X ?Bin(n, p). By considering the ratio b( x +1; n , p) / b (x; n , p), show that b(x; n, p) increases with x as long as x < np – ( 1 – p). Conclude that the mode x* is the integer satisfying (n + 1)p - 1? x* ? (n + 1) p. b. ?Show that if X has a Poisson distribution with parameter ?, the mode is the largest integer less than ?. If ? is an integer, show that both ? – 1 and ? are modes.” is broken down into a number of easy to follow steps, and 121 words. Since the solution to 116E from 3 chapter was answered, more than 434 students have viewed the full step-by-step answer.

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