a.? ?Show that b(x; n, 1 - p) = b(n - x; n, p). b. ?Show that B(x; n, 1 – p) = 1 – B ( n – x – 1; n, p). [Hint: At most x S’s is equivalent to at least ( n – x) F’s.] c. ?What do parts (a) and (b) imply about the necessity of including values of p greater than .5 in Appendix Table A.1?

Problem 63E Answer: Step1: We need to show that, a. Show that b(x; n, 1 - p) = b(n - x; n, p). b. Show that B(x; n, 1 – p) = 1 – B ( n – x – 1; n, p). [Hint: At most x S’s is equivalent to at least ( n – x) F’s.] c. What do parts (a) and (b) imply about the necessity of including values of p greater than .5 in Appendix Table A.1 Step2: The probability mass function of binomial distribution is given by p(x) = C p q n-, x = 0,1,2,...,n x With mean E(x) = np and Var(x) = np(1 - p) a). 1).Consider, n x nx b(x;n, 1 p)= ( )(1x p) [1 (1 p)] . = n! (1 p) (p) nx x!(nx)! = n! (p)nx (1 p) x x!(nx)! Hence, b(x;n, 1 p) = n! (p)nx (1 p) x ……..(1) x!(nx)! 2).Again consider, n nx n(nx) b(n x;n,p) = ( nx)(p) (1 p) n! nx x = (nx)![n(nx)]!) (1 p) n! nx x = (nx)!x!p) (1 p) n! nx x Hence, b(n x;n,p)= (nx)!x!p) (1 p) ……..(2) Therefore, From equation (1) and (2)...