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"The manager for State Bank and Trust has recently examined the credit card account balances for the customers of her bank and found that 20% have an outstanding balance at the credit card limit. Suppose the manager randomly selects 15 customers and finds 4 that have balances at the limit. Assume that the properties of the binomial distribution apply.a.What is the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards?b.What is the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card"

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Answer:

a


P(X = 4 ) = 0.1876

b


P(X \le 4)  =  0.8358

Explanation:

From the question we are told that

The proportion that has outstanding balance is p = 0.20

The sample size is n = 15

Given that the properties of the binomial distribution apply, for a randomly selected number(X) of credit card


X \ \ ~ Bin (n , p )

Generally the probability of finding 4 customers in a sample of 15 who have "maxed out" their credit cards is mathematically represented as


P(X = 4 ) = ^nC_4 * p^4 * (1 - p)^(n-4)

=>
P(X = 4 ) = ^(15)C_4 * (0.20)^4 * (1 - 0.20)^(15-4)

Here C stand for combination

=>
P(X = 4 ) = 0.1876

Generally the probability that 4 or fewer customers in the sample will have balances at the limit of the credit card is mathematically represented as


P(X \le 4) = [ ^(15)C_0 * (0.20)^0 * (1 - 0.20)^(15-0)]+[ ^(15)C_1 * (0.20)^1 * (1 - 0.20)^(15-1)]+\cdots+[ ^(15)C_4 * (0.20)^4 * (1 - 0.20)^(15-4)]

=>
P(X \le 4)  =  0.8358

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