2. (25 points) Our number system consists of tbe digits 0, 1, 2, 3, 4, 5, 6, 7,

Question

2. (25 points) Our number system consists of tbe digits 0, 1, 2, 3, 4, 5, 6, 7, 8,9. Because we do not write numbers such as 12 as 012, the first significant digit in any number must be 1, 2, 3, 4,5, 6, 7, 8,9. Consider entries of numbers in accounting records, tax returns, etc. We may think that each digit appears with equal frequency so that each of outcomes), bat this is not true. In 1881, Simon Newcomb discovered that first digits for mimbers in records do not oocur with equal frequency. The physicist Frank Benford discovered the same result in 1938. After studying lots and lots of data, he assigned probabilities of occurresce as shown in the table below. The probability law is now known as Benford's Lase and plays a major role in identifying fraudulent data on tax returns and accounting books. Probability 0.301 0.1760.125 0.097 0.079 0.067 008 0.01 0,046 Table 1: Benford's Law (a) (5 points) Verify that Benford's Law satisfies our second probability rule, namely P(S) = 1. (b) (5 points) Use Benford's Law to the determine the probability that a randomly selected entry in some accounting books has a first digit that is a 1 or a 2 (e) (5 points) Compute the probability that a randomly selected entry in some ac- counting books has a first digit that is 1 or 2 under the assumption that the digits are equally-likely How many more times is this eveat more lkely under Benford's Law? entry in some accounting books has a first digit that is at least 6, i.e. P(X 26). (d) (5 points) Use Benford's Law to determine the probability that a randomly selected (e) (5 points) Use the result of part d. and the Law of the Complement (Rule 4 in our list of probability rules) to find the probability that a randomly selected entry in some accounting books is less than 6, i.e., P(X

Explanation / Answer

a) here P(S) =P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)

=0.301+0.176+0.125+0.097+0.079+0.067+0.058+0.051+0.046 =1

therefore abvoe is a valid probability distribution

b)

frm Bernard law P(1 or 2) =P(X=1)+P(X=2) = =0.301+0.176 =0.477

c)

as each digit has equal probability therefore P(1 or 2) =2/9 =0.2222

this is (0.2222/0.477) =0.466 times (almost half) from what we get from Bernard law

d)

P(X>=6) =P(X=6)+P(X=7)+P(X=8)+P(X=9) =0.067+0.058+0.051+0.046 =0.222

e)P(X<6) = 1-P(X>=6)=1-0.222 =0.778

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