According to Benford's law, the probability that the first digit of the amount of a randomly chosen invoice is a 1 or a 2 is 0.477. You examine 81 invoices from a vendor and find that 25 have first digits 1 or 2. If Benford's law holds, the count of 1s and 2s will have the binomial distribution with n = 81 and p = 0.477. Too few 1s and 2s suggests fraud. What is the approximate probability of 25 or fewer 1s and 2s if the invoices follow Benford's law? (Use the normal approximation. Round your answer to four decimal places.)

Fia11223
Oct 24, 2017

#1**0 **

It has been a long time since I have done questions like this but this is what I think.

n=81 p=0.477 q=1-p = 0.523

expected number of 1s and 2s in 81 invoices is np = 81*0.477=38.637

\(\bar x=38.637\)

\(\sigma ^2= npq\\ \sigma ^2= 81*0.477*0.523\\ \sigma ^2= 20.2072\\ \sigma = 4.4952\\\)

Now there were 25 ones and twos. This is below the expected mean of 38.637

38.637-25=13.637 It is below by this many in actual number but how many standard deviations below it is?

13.637/4.4952 = 3.0336803701726286

Mmm it is 3.0337 standard deviations below the expected mean.

I think the probability of there being 25 or less ones or twos is 0.0012 or 0.12%

http://onlinestatbook.com/2/calculators/normal_dist.html

So it seems likely there is some fraud happening here.

I used this site to help me muddle through this question.

Accept my answer at your peril :)

Melody
Oct 24, 2017