Question

Benford’s law and fraud Refer to Exercise 13. It might also be possible to detect an employee’s fake expense records by looking at the variability in the first digits of those expense amounts.

(a) Calculate the standard deviation σY. This gives us an idea of how much variation we’d expect in the employee’s expense records if he assumed that first digits from $1$ to $9$ were equally likely.

(b) Now calculate the standard deviation of first digits that follow Benford’s law (Exercise 5). Would using standard deviations be a good way to detect fraud? Explain.

Short answer

a). The standard deviation is the square root of the variance is$\sigma \approx 2.5820$.

b). The standard deviation is the square root of the variance is $\sigma \approx 2.4618$.

Step-by-step solution

Part (a) Step 1: Given Information 

Given in the question is to refer exercise 13.

The histogram depicts the probability distribution.

Part (a) Step 2: Explanation

The provided distribution is symmetric (the graph to the left of $5$is a mirror image of the graph to the right of $5$), and the mean is located in the middle, which is $5$.

The expected value of the squared variation from the mean is the variance:

$$\begin{gathered} {\sigma }^2=\sum (x-\mu {)}^{2}P(x) \\ =(1-5{)}^2\times \frac{1}{9}+(2-5{)}^2\times \frac{1}{9}+(3-5{)}^2\times \frac{1}{9}+(4-5{)}^2\times \frac{1}{9}+(5-5{)}^2\times \frac{1}{9}+(6-5{)}^2\times \frac{1}{9}+(7-5{)}^2\times \frac{1}{9}+(8-5{)}^2\times \frac{1}{9}+(9-5{)}^2\times \frac{1}{9} \\ =\frac{20}{3} \\ \approx 6.6667 \end{gathered}$$

The square root of the variance is the standard deviation:

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{6.6667} \\ \approx 2.5820 \end{gathered}$$

Part (b) Step 1: Given Information 

The following histogram shows the probability distribution

Part (b) Step 2: Explanation 

The expected value is calculated by multiplying each possibility by its probability:

$$\begin{gathered} E\left(X\right)=\sum xP\left(x\right) \\ =1\times 0.301+2\times 0.176+3\times 0.125+4\times 0.097+5\times 0.079+6\times 0.067+7\times 0.058+8\times 0.051+9\times 0.046 \\ =3.441 \end{gathered}$$

The expected value of the squared variation from the mean is the variance:

$$\begin{gathered} {\sigma }^2=\sum (x-\mu {)}^{2}P(x) \\ =(1-3.441{)}^2\times 0.301+(2-3.441{)}^2\times 0.176+(3-3.441{)}^2\times 0.125+(4-3.441)+(5-3.441{)}^2\times 0.079+(6-3.441{)}^2\times 0.067+(7-3.441{)}^2\times 0.058+(8-3.441{)}^2\times 0.051+(9-3.441) \end{gathered}$$

The square root of the variance is the standard deviation:

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{6.060519} \\ \approx 2.4618 \end{gathered}$$