Question 909190: 7% of the customers did not receive delivery
If a sample of 50 customers is selected on a given day, assuming that the trials are independent of each other with one outcome, what is the probability that.
a. fewer than 3 customers would receive a free newspaper?
b. 2, 3, or 4 customers would receive a free newspaper?
c. more than 5 customers would receive a free newspaper?
I believe n=50 P=.07 and X=3 for part A
I have tried this problem multiple times using the Binomial probability formula, but all my answers seem incorrect.
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! X~binomial(n=50, p=.07)
A) P[X<3] = P[X<=2] = P[X=0] + P[X=1] + P[X=2] = (50 choose 0)(.07)^0(.93)^50 + (50 choose 1)(.07)^1(.93)^49 + (50 choose 2)(.07)^2(.93)^48 = .3108
B) P[X=2] + P[X=3] + P[X=4] = (50 choose 2)(.07)^2(.93)^48 + (50 choose 3)(.07)^3(.93)^47 + (50 choose 4)(.07)^4(.93)^46 = .6025
C) P[X>5] = 1-P[X<=5] (using the complement).
1-(P[X=0] + P[X=1] + (P[X=2] + P[X=3] + P[X=4])+ P[X=5])
1-(P[X=0] + P[X=1] + B + P[X=5]) where B is the answer from part B.
1-((50 choose 0)(.07)^0(.93)^50 + (50 choose 1)(.07)^1(.93)^49 + .6025 + (50 choose 5)(.07)^5(.93)^45) = .1351
Hope this helps!
Devin [swincher4391@yahoo.com]
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