SOLUTION: If 60% of all households own VCRs, what is the probability that 11 households are selected at random, more than 7 will own VCRs? With explanation please. Thanks

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Question 913776: If 60% of all households own VCRs, what is the probability that 11 households are selected at random, more than 7 will own VCRs? With explanation please. Thanks
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
p(owns VCR) = .60, n = 11
P(x > 7) =1- binomcdf(11, .60, 7) = 1 - .7037 = .2963 Using TI
0r
By hand...
Using P+%28x%29=+highlight_green%28nCx%29%28p%5Ex%29%28q%29%5E%28n-x%29+
p and q are the probabilities of success and failure respectively.
In this case p = .6 & q are = .4, n = 11
nCx+=+%28n%21%29%2Fx%21%28n+-+x%29%21%29
P(x > 7) = P(8) + P(9) + P(10) + P(11)
P(x > 7) =
P(x >7) = .1774 + .0887 + .0266 + .0036 = .2963