3. Let X ~ B(n, р), i..,P(X = k) = (")»*(1 – p)*-*for k € {0, 1, ...,n}. Assume that 0 P+i if k> пр— (1 — р).(a)You only have to give the details of the first case and note thatthe other two cases are similar.(b)Let t = np – (1 – p). If np is an integer, show that tsatisfiesпр — 1 (c)In general, np is not necessarily an integer, and it can beshown that t satisfies one of the following:i. [np] –1

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3. Let X ~ B(n, p), 1.e., P(X = k) = (:)»*'(1 – py-* for k e {0, 1, ...,n}. Assume that 0 < p < 1 (note the strict inequali- ties). In this problem, we will show that the peak of B(n, p) (i.e., the largest probability mass) occurs at k z np. In part (c), [·) denotes the floor function, where [a] is defined to be the largest integer less than or equal to a real number r. Denote P(X = k) by Pg. Show that for k € {0,1, -. · , (a) n– 1}, i. P < Pr+1 if k < np – (1 – p) ii. P = Pk+1 if k = np – (1 – p) iii. P > Pk+1 if k > np – (1 – p).

3. Let X ~ B(n, p), 1.e., P(X = k) = (:)»'(1 – py- for k e {0, 1, ...,n}. Assume that 0 < p < 1 (note the strict inequali- ties). In this problem, we will show that the peak of B(n, p) (i.e., the largest probability mass) occurs at k z np. In part (c), [·) denotes the floor function, where [a] is defined to be the largest integer less than or equal to a real number r. Denote P(X = k) by Pg. Show that for k € {0,1, -. · , (a) n– 1}, i. P < Pr+1 if k < np – (1 – p) ii. P = Pk+1 if k = np – (1 – p) iii. P > Pk+1 if k > np – (1 – p).

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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