SOLUTION: Please help!! I have tried solving this problem but I never ger the right answer and it is driving me insane. It is known that 30% of all customers of a major credit card pay their

Algebra ->  Probability-and-statistics -> SOLUTION: Please help!! I have tried solving this problem but I never ger the right answer and it is driving me insane. It is known that 30% of all customers of a major credit card pay their      Log On


   



Question 478387: Please help!! I have tried solving this problem but I never ger the right answer and it is driving me insane. It is known that 30% of all customers of a major credit card pay their bills in full before any interest charges are incurred. Suppose a simple random sample of 150 credit card holders is selected. What is the probability that 30 or fewer customers pay their account balances in full before any interest charges are incurred? What is the probability that between 40 and 50 customers pay their accounts balances in full before any interest charges are incurred? What is the probability that more than 60 customers pay their accounts balances in full before any interest charges are incurred?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
It is known that 30% of all customers of a major credit card pay their bills in full before any interest charges are incurred. Suppose a simple random sample of 150 credit card holders is selected.
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Binomial Problem with n = 150 and p = 0.3
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What is the probability that 30 or fewer customers pay their account balances in full before any interest charges are incurred?
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P(0<= x <=30) = 150C30(0.3)^30*(0.7)^120 = binomcdf(150,0.3,30) = 0.0038
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What is the probability that between 40 and 50 customers pay their accounts balances in full before any interest charges are incurred?
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P(40<= x <=50) = binomcdf(150,0.3,50) - binomcdf(150,0.3,39) = 0.6750
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What is the probability that more than 60 customers pay their accounts balances in full before any interest charges are incurred?
P(61<= x <=150) = 1 - binomcdf(150,0.3,60) = 0.0035
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Cheers,
Stan H.
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