Question

f(31–43 10.320.72 543 Computing Binomial Probabilities If X is a binomial random variable with parameters n and p, the probability distribution of Xis given by f(k) = P(X=k) = (pkan* for k =0, 1. , .,where q=1-p. Example: Suppose n = 5 and p = 0.3. Then q = 1 - p = 0.7, f(k)= 10.3)* (0.75% f(0)=C6 20.3)%0.7)-1-1-(0.16807)-0.16807. f(1)=( )(0,3)(0.7) 10.3)(0.2401) - 5(0.07203)0.36015 0 0.1681 f(2)=(3 10.3)2(0.73 (0.09)(0.343) – 10(0.03087)-0.30870 1 0.3602 0.027)(0.49) =10(0.01323)-0.132302 0.3087 f(4)=( )(0.3)*(0.7) (0.0081)(0.7) -5(0.00567)=0.0284 3 0.1323 f(5)–(10.3%(0.7)° 54321 0.00243) 1-1(0.00243)=0.00243 4 0.0284 5 0.0024 P(X s 1) = P(X = 0 or 1) = f(0) + f(1) = 0.1681 +0.3602 = 0.5283 P(X> 3) - P(X= 4 or 5) = f(4) + f(5) - 0.0284 +0.0024 -0.0308 P(2 s Xs 4) = P(X=2 or 3 or 4) = f(2) + f(3) + f(4) = 0.3087 + 0.1323 +0.0284 = 0.4694 Problem: Let X be a binomial random variable with parameters n=6 and p = 0.4. (1) Calculate the probability distribution of X (to 4 decimal places) xl f(x) (a) f(0) 0 (b) f(1)- 32.1 54:32 4321 5-4-321 1 (c) f(2) 2 3 4 5 (d) f(3)= (e) f(4) (1) f(5)- (g) f(6) (2) Complete the table and use it to calculate the following probabilities: (a) P(Xs 2)- 6 (b) P(X25)= (c) P(25 Xs 5) =

Answer 1

n = 6, p = 0.4

1)

(a) f(0) = P(X = 0) = $0.6^{6}$ = 0.0467

(b) f(1) = P(X = 1) = $\binom{6}{1}*0.4*0.6^{5}$ = 0.1866

(c) f(2) = $\binom{6}{2}*0.4^{2}*0.6^{4}$ = 0.3110

(d) f(3) = = 0.2765

+ 0.4 + 0.6% 3

(e) f(4) = $\binom{6}{4}*0.4^{4}*0.6^{2}$ = 0.1382

(f) f(5) = $\binom{6}{5}*0.4^{5}*0.6$ = 0.0369

(g) f(6) = $0.4^{6}$ = 0.0041

2)

(a)

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5443

(b) P(X ≥ 5) = f(5) + f(6) = 0.0410

(c) P(2 ≤ X ≤ 5) = 0.7626