SOLUTION: Suppose that 13% of the voters in a state intend to vote for a certain candidate. What is the probability that a survey polling nine people reveals that three or fewer intend to vo

Question 596202: Suppose that 13% of the voters in a state intend to vote for a certain candidate. What is the probability that a survey polling nine people reveals that three or fewer intend to vote for that candidate? (Round your answer to 4 decimal places.)

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Use the binomial probability distribution

P(X = 0) = (9 ncr 0)*(0.13)^(0)*(1-0.13)^(9-0)
P(X = 0) = (9 ncr 0)*(0.13)^(0)*(0.87)^(9-0)
P(X = 0) = (1)*(0.13)^(0)*(0.87)^9
P(X = 0) = (1)*(1)*(0.285544154243029527)
P(X = 0) = 0.285544154243029527

P(X = 1) = (9 ncr 1)*(0.13)^(1)*(1-0.13)^(9-1)
P(X = 1) = (9 ncr 1)*(0.13)^(1)*(0.87)^(9-1)
P(X = 1) = (9)*(0.13)^(1)*(0.87)^8
P(X = 1) = (9)*(0.13)*(0.3282116715437121)
P(X = 1) = 0.384007655706143157

P(X = 2) = (9 ncr 2)*(0.13)^(2)*(1-0.13)^(9-2)
P(X = 2) = (9 ncr 2)*(0.13)^(2)*(0.87)^(9-2)
P(X = 2) = (36)*(0.13)^(2)*(0.87)^7
P(X = 2) = (36)*(0.0169)*(0.37725479487783)
P(X = 2) = 0.229521817203671772

P(X = 3) = (9 ncr 3)*(0.13)^(3)*(1-0.13)^(9-3)
P(X = 3) = (9 ncr 3)*(0.13)^(3)*(0.87)^(9-3)
P(X = 3) = (84)*(0.13)^(3)*(0.87)^6
P(X = 3) = (84)*(0.002197)*(0.433626201009)
P(X = 3) = 0.080024848143808932

In short,

P(X = 0)= 0.285544154243029527
P(X = 1)= 0.384007655706143157
P(X = 2)= 0.229521817203671772
P(X = 3)= 0.080024848143808932

So
P(X <= 3) = P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)
P(X <= 3) = 0.285544154243029527 + 0.384007655706143157 +0.229521817203671772 + 0.080024848143808932
P(X <= 3) = 0.979098475296653388

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Answer:

P(X <= 3) = 0.979098475296653388

Round this to four places to get P(X <= 3) = 0.9791

So the the probability that a survey polling nine people reveals that three or fewer intend to vote for that candidate is 0.9791
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