Print 9. When a customer places an order with a certain company\'s on-line suppl

Question

Print 9. 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 he credit limit. Past records indicate that the probability of customers exceeding their credit limit is 0.07. Suppose that, on a given day, 18 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 are the mean and standard deviation of the number of customers exceeding their credit limits? The mean number of customers exceeding their credit limits is Round to four decimal places as needed.) customers The standard deviation is (Round to three decimal places as needed.) b. What is the probability that 0 customers will exceed their limits? (Round to four decimal places as needed.) C. What is the probability that 1 customer will exceed his or her limit? (Round to four decimal places as needed.) d. What is the probability that 2 or more customers will exceed their limits? (Round to four decimal places as needed.)

Explanation / Answer

a)

Here n (number of customers) = 18

p denotes the probability of the customer exceeding their credit limit = 0.07

p = 0.07

This is the binomial distribution

mean of binomial distribution = n*p = 18*0.07 = 1.26

So mean number of the customers exceeding their credit limit is = 1.26

Variance of binomial distribution = n*p*(1-p) = 18*0.07*(1-0.07) = 1.1718

Standard deviation = square root ( variance ) = square root (1.1718) = 1.082497113 = 1.082(rounded upto 3 decimal places )

The standard deviation is 1.082 coustomers.

b) Conside X :customer exceeding their credit limit

we know the binomial pdf as P( X= x) = [n!/(x!*(n-x)!)] * px * ( 1- P)n-x

we have to find P (X= 0 ) =  [18!/(0!*(18-0)!)] * (0.07)0 * ( 1- 0.07)18 - 0

= (0.93)18 = 0.270827695 = 0.2708

probability that 0 customers will exceed their limit = 0.2708

(C)  

In C part we have to find P ( X= 1 ) = [18!/(1!*(18-1)!)] * (0.07)1 * ( 1- 0.07)18 - 1

= [18!/ 17!] * 0.07*(0.93)17 = 18 *0.07 * (0.93)17 = 0.366927845= 0.3669

Probability that 1 customer wiil exceed their limit = 0.3669.

(d) In d part we have to find

P (X >= 2 )

we find this probablity as

P ( X >= 2 ) = 1 - P (X = 0 ) - P ( X = 1 )

= 1 - 0.270827695- 0.366927845 = 0.362244459 =0.3622 ( we get this answer when we used P ( X= 0 ) and P( X= 1 ) without any rounding )

If we used P ( X = 0 ) and P ( X = 1 ) with rounding up to four decimal

then we get P ( X > = 2 ) = 1 - 0.2708 - 0.3669 = 0.3623

Please check both the answer one by one  

So P ( X <= 2 ) = 0.3622 ( we get this answer when we used P ( X= 0 ) and P( X= 1 ) without any rounding )

or

P ( X <= 2 ) = 0.3623 ( we get this answer when we used P ( X= 0 ) and P( X= 1 ) with rounding upto 4 decimal)

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