Question

When a customer places an order with a certain​ company's on-line supply​ store, a computerized accounting information system​ (AIS) automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is 0.08. Suppose​ that, on a given​ day, 22 customers place orders. Assume that the number of customers that the AIS detects as having exceeded their credit limit is distributed as a binomial random variable. Complete parts​ (a) through​ (d) below. a) what is the mean of customers exceeding their credit limits?

b) what is the standard deviation of the customers?

c) what is the probability that 0 customers exceed their credit limits?

d) what is the proability that 1 customer will exceed their credit limits?

e) what is the probability that 2 customers will exceed their credit limits?

Answer 1

a. 1.76 customers

b. 0.6192 customers

c. 0.1600

d. 0.7442

Explanation:

Suppose that p = 0.08 is the probability that a particular client exceeds his credit, and take the number of Bernoulli-type attempts as n = 22$ $. As the number of customers detected by AIS as having exceeded their credit limits is a binomial model, you have to:

`$f(x)=\binom{n}{x}p^x(1-p)^(n-x) = \binom{22}{x}0.08^x0.92^(22-x) $ for $ x \in\left \{ 1,2,3,4,...,n\right\}\\\\`

`$$a. what is the mean of customers exceeding their credit limits?\\$\mu=np=22*0.08= 1.76$  costumers\\\\b. what is the standard deviation of the customers?$\\sd=√(np(1-p)))=√(22*0.08*0.92) = √(0.6192)=0.78689$Costumers\\\\`

`$$c. what is the probability that 0 customers exceed their credit limits?\\$f(0)=\binom{22}{0}0.08^00.92^(22)=0.1600$\\\\d) what is the proability that 1 customer will exceed their credit limits?\\$f(1)=\binom{22}{1}0.08^10.92^(21)=0.3055$\\\\`
`$$e) what is the probability that 2 customers will exceed their credit limits?\\</p><p>$f(2)=\binom{22}{2}0.08^20.92^(20)=0.744$\\\\`