Question 481082: In a recent study 90 percent of the homes in the United States were found to have color TVs. In a sample of nine homes, what is the probability that:
a. All nine have color TVs?
b. Less than five have color TVs?
c. More than five have color TVs?
d. At least seven homes have color TVs?
I am confused I was told this problem is Binomial but I can't seem to solve it can someone please help me out please. Thank you very much.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In a recent study 90 percent of the homes in the United States were found to have color TVs. In a sample of nine homes, what is the probability that:
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Binomial Problem with n = 9 and p = 0.9
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a. All nine have color TVs?
P(x=9) = 9C9(0.9)^9*(0.1)^0 = 0.9^9 = 0.3874
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b. Less than five have color TVs?
P(0<= x <=4) = binomcdf(9,0.9,4) = 0.0008909
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c. More than five have color TVs?
P(6<= x <=9) = 1 - binomcdf(9,0.9,5) = 0.9917
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d. At least seven homes have color TVs?
P(7<= x <=9) = 1-binomcdf(9,0.9,6) = 0.9470
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I used a TI-84 to get the cumulative probabilities.
Cheers,
Stan H.
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