<PRE><B>Question 342697</B>
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?
Ans: 0.9^9 = 0.3874
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b. Less than five have color TVs?
Ans: binomcdf(9,0.9,4) = 0.0008909...
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c. More than five have color TVs?
Ans: 1 - binomcdf(9,0.9,5) = 1 - 0.0083 = 0.9917
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d. At least seven homes have color TVs?
Ans: 1 - binomcdf(9,0.9,6) = 1-0.0530 = 0.9470
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Cheers,
Stan H.
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