Your company is auditing ABC Corp—an electronics products retailer. Because ABC

Question

Your company is auditing ABC Corp—an electronics products retailer. Because ABC has a habit of overcharging its customers, your focus is to check if the billing amounts on its invoices are correct. Assume that each invoice is for too high an amount with probability 0.06 and for too low an amount with probability 0.01 (so that the probability of a correct billing 0.93). Also, assume that the outcome for any invoice is independent of the outcome for other invoices.

(A) If you randomly select 200 invoices, what is the chance that you will find at least 15 invoices that overcharge the customer?

(B) What is the chance you won’t find any that undercharge a customer?

(C) Find an integer “k” such that the probability is at least 0.99 that you will find at least “k” invoices that overcharge the customer.

Continuing with the above question, suppose that when ABC overcharges a customer, the distribution of the amount overcharged (expressed as a percentage of the correct billing amount) is N(15%, 4%).

(D) What percentage of overbilled customers are charged at least 10% more than they should pay?

(E) What percentage of all customers are charged at least 10% more than they should pay?

(F) If you randomly select 200 invoices, what is the chance you will find at least 5 where the customer was overcharged by at least 10%?

Explanation / Answer

Part A

We are given

n = 200

p = 0.06

q = 1 – p = 1 – 0.06 = 0.94

We have to find P(X15)

Here, we have to use normal approximation to binomial distribution.

n*p = 200*0.06 = 12

n*q = 200*0.94 = 188

n*p and n*q > 5, so we can use normal approximation.

Mean = n*p = 200*0.06 = 12

SD = sqrt(n*p*q) = sqrt(200*0.06*0.94) = 3.358571

Now, we have subtract continuity correction 0.5 such that we have to find P(X>14.5)

P(X>14.5) = 1 – P(X<14.5)

Z = (X – mean) / SD

Z = (14.5 – 12) / 3.358571

Z = 0.744364

P(Z<0.744364) = 0.771672

P(X<14.5) = 0.771672

P(X>14.5) = 1 – P(X<14.5)

P(X>14.5) = 1 – 0.771672

P(X>14.5) = 0.228328

Required probability = 0.228328

There is a 22.83% chance that we will find at least 15 invoices that overcharge the customer.

Part B

Here, n = 200, p = 0.01,

n*p = 200*0.01 = 2 < 5

n*q = 200*0.99 = 198 > 5

Here, condition for using normal approximation is not met, so we don’t use normal approximation to binomial distribution. So, by using binomial distribution

P(X=x) = nCx*p^x*q^(n – x)

P(X=0) = 200C0*0.01^0*0.99^(200 – 0)

P(X=0) = 1*1*0.99^200

P(X=0) = 0.133979675

Required probability = 0.133979675

Part C

We are given

P(Xk) = 0.99

Z = (k – mean) / SD = 2.326348

We have

n = 200

p = 0.06

q = 1 – p = 1 – 0.06 = 0.94

Mean = n*p = 200*0.06 = 12

SD = sqrt(n*p*q) = sqrt(200*0.06*0.94) = 3.358571

(k – 12) / 3.358571 = 2.326348

(k – 12) = 2.326348*3.358571

(k – 12) = 7.813205

k = 7.813205 + 12

k = 19.81321

by adding continuity correction 0.5, k = 19.81321 + 0.5 = 20.31321

k = 20 approximately

Part D

We are given mean = 15, SD = 4 and X follow normal distribution.

We have to find P(X10)

P(X10) = 1 – P(X<10)

Z = (X – mean) / SD

Z = (10 – 15)/4 = -1.25

P(Z<-1.25) = 0.10565

P(X<10) = 0.10565

P(X10) = 1 – P(X<10)

P(X10) = 1 – 0.10565

P(X10) = 0.89435

Required percentage = 89.44%

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