Question

atunim- im un Question 4 Consider that the probability of a customer making a purchase of an item is 30 or will not make a purchase is ,70, this is assume for all customers. Let also assume that there are 3 customers that made purchases denoted by X1, X2, and X3. a. If customer X2 make two purchases what will be the probability of purchasing these items, using binomial probability function (4 points) 6. compute the expected value for the binomial distribution (4 points) c. compute the variance and the standard deviation (4 points)

Answer 1

Solution:

Solution :- Here we have to giver that the probability of a customer making purchese of an item is 0.30 de probability of sucess = p = 0.30 and not purchesing probability is 0-70 ée. Cl-P) = q = 0.70 There are 3 Customer ie 0-3 Now we know theet Pielf binomical distribution is P(x=x) = mx på C1-P)^2 a) Now we have to find probability of customer - X₂ make two purches i e. x=24 ..PLX22) = 36, (0.30) C1-0-82952 = 3! * 0.0g * 0.70 (3-2)!! = 3 * 0.09 * 0.70 :| P(X2=2) = 0.189) Therefore probability of Customer Xz - make purchases 2 is 0.189. b) Now mean= Expected veelue of binomical distribution is = np lie. F= E(X)= np : E(X) = 3* (0:30) TE(X) = o.g

C) Now variance of binomical distribution is varex) = npq = np C1-P) i. varex) = Varcx) = 3* 0.30* 070 0.63 and standard devicetion - S.D - V variance ;; SD = V0-63 S.D= 0.7937 Thank you!! please like it..