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?

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:

[tex]$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\}\\\\[/tex]

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

[tex]$c. what is the probability that 0 customers exceed their credit limits?\\$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$\\\\[/tex][tex]$e) what is the probability that 2 customers will exceed their credit limits?\\

$f(2)=\binom{22}{2}0.08^20.92^{20}=0.744$\\\\[/tex]