Question

Using Benford's law According to Benford's law (Exercise 15, page 377), the probability that the first digit of the amount of a randomly chosen invoice is an 8 or a 9 is 0.097. Suppose you examine randomly selected invoices from a vendor until you find one whose amount begins with an 8 or a 9 .

a. How many invoices do you expect to examine before finding one that begins with an 8 or 9 ?

b. In fact, the first invoice you find with an amount that starts with an 8 or 9 is the 40 th invoice. Does this result provide convincing evidence that the invoice amounts are not genuine? Calculate an appropriate probability to support your answer.

Short answer

(a)it is expected that to examine about $10.3093$invoices until get the first success which is an invoice that begins with an 8 or 9 .

(b)It is observed that the probability of getting the first success on the ${40}^{\text{th}}$invoice or after it is small (less than $0.05$), which shows that the event is not likely to get and therefore there is enough convincing proof that the invoice amounts are not genuine.

Step-by-step solution

Part (a) Step 1: Given Information

Given,$p=0.097$

Formula used:

The expected value

$\mu =\frac{1}{p}$

Part (a) Step 2: Simplification

The number of independent trials required until first success is distributed geometrically.

A geometric distribution's expected value is

$\mu =\frac{1}{p}=\frac{1}{0.097}=10.3039$

Hence, it is expected that to examine about $10.3093$invoices until you achieve your first success, which is an invoice starting with an 8 or 9.

Part (b) Step 1: Given Information

Given$p=0.097$

Formuale to be used

Geometric probability:

$P(X=k)=q^{k-1}p=(1-p{)}^{k-1}p$

Addition rule:

$P(A\cup B)=P\left(A\text{or}B\right)=P\left(A\right)+P\left(B\right)$

Complement rule:

$P\left(A°\right)=P(notA)=1-P\left(A\right)$

Part (b) Step 2: Simplification

Consider,$p=0.097$

Compute the binomial probability definition at $k=1,2$, and $3\dots 40$:

$P(X=1)=(1-0.097{)}^{1-1}(0.097)=0.097$

Similarly

localid="1654759250726" $$\begin{gathered} P(X=2)=0.0876 \\ P(X=3)=0.0791 \\ P(X=39)=0.0020 \\ P(X\le 39)=P(X=1)+P(X=2)+P(X=3)+\dots +P(X=39) \\ =0.097+0.0876+0.0791+\dots +0.0020 \\ =0.9813 \end{gathered}$$

Using the complement rule:

$$\begin{gathered} P(X\ge 40)=1-P(X\le 39) \\ =1-0.9813 \\ =0.0187=1.87\% \end{gathered}$$

Using Ti83/84- calculator

It has been discovered that the likelihood of achieving the first success on the ${40}^{\text{th}}$invoice or after it is small (less than $0.05$), This demonstrates that the occurrence is unlikely to occur, and hence there is sufficient convincing evidence that the invoice amounts are not genuine.