Question

"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"

Answer 1

Answer:

a

   $P(X = 4 ) = 0.1876$

b

   $P(X \le 4) = 0.8358$

Step-by-step 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

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

=>   $P(X \le 4) = 0.8358$