Chapter 6.1, Problem 21E

(a)

To determine

Probability for the event $Y>6$ , proportion of first digits in the employee’s expense amounts should be greater than 6 and use of the information to detect a fake expense report.

Answer to Problem 21E

Probability,

  $P\left(Y>6\right)\approx 0.3333$

According to Benford’s law,

Probability,

  $P\left(Y>6\right)=0.155$

If the first digit proportion greater than 6 is closer to 0.3333 instead of 0.155, then the expense report is false.

Explanation of Solution

Given information:

Histogram representing the first digit Y (from 0 to 9) of a randomly selected expense amount:

Calculations:

Two events are mutually exclusive, if the events cannot occur at the same time.

Addition rule for mutually exclusive events:

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

We have

The probability for each event is $\frac{1}{9}$ .

For $Y=7$ :

  $P\left(Y=7\right)=\frac{1}{9}$

For $Y=8$ :

  $P\left(Y=8\right)=\frac{1}{9}$

For $Y=9$ :

  $P\left(Y=9\right)=\frac{1}{9}$

We know that

It is not possible for the first digit to be two different digits at the same time.

Thus,

The events are mutually exclusive.

Then

Apply the addition rule for mutually exclusive events:

  $\begin{array}{l}P\left(Y>6\right)=P\left(Y=7\right)+P\left(Y=8\right)+P\left(Y=9\right)\\ =\frac{1}{9}+\frac{1}{9}+\frac{1}{9}\\ =\frac{1+1+1}{9}\\ =\frac{3}{9}\\ =\frac{1}{3}\\ \approx 0.3333\end{array}$

According to Benford’s law,

Probabilities from the table,

We have

For $Y=7$ :

  $P\left(Y=7\right)=0.058$

For $Y=8$ :

  $P\left(Y=8\right)=0.051$

For $Y=9$ :

  $P\left(Y=9\right)=0.046$

We know that

It is not possible for the first digit to be two different digits at the same time.

Thus,

The events are mutually exclusive.

Then

Apply the addition rule for mutually exclusive events:

  $\begin{array}{l}P\left(Y>6\right)=P\left(Y=7\right)+P\left(Y=8\right)+P\left(Y=9\right)\\ =0.058+0.051+0.046\\ =0.155\end{array}$

If the first digit proportion greater than 6 is closer to 0.3333 instead of 0.155, then the expense report is false.

(b)

To determine

Mean of the random variable Y is located at the solid red line in the figure.

Answer to Problem 21E

The distribution is symmetric and the mean lies in the middle of the distribution, which means mean needs to lie at 5.

Explanation of Solution

Given information:

Histogram representing the first digit Y (from 0 to 9) of a randomly selected expense amount:

Calculations:

The expected value (or mean) is the sum of the product of each possibility x with its probability $P\left(X=x\right)$ .

Thus,

  $\begin{array}{l}\mu ={\displaystyle \sum xP\left(X=x\right)}\\ =1\times \frac{1}{9}+2\times \frac{1}{9}+3\times \frac{1}{9}+4\times \frac{1}{9}+5\times \frac{1}{9}+6\times \frac{1}{9}+7\times \frac{1}{9}+8\times \frac{1}{9}+9\times \frac{1}{9}\\ =\frac{1+2+3+4+5+6+7+8+9}{9}\\ =\frac{45}{9}\\ =5\end{array}$

Now,

Note that

The expected value is 5 in the histogram.

We came to know that

The distribution is symmetric and the mean lies exactly in the middle of the distribution.

This implies

The mean needs to lie at 5.

(c)

To determine

Information according to Benford’s law is used to detect a fake expense report.

Answer to Problem 21E

If the mean of all first digits in an expense report is closer to 5 instead of 3.441, then the expense report is false.

Explanation of Solution

Given information:

Histogram representing the first digit Y (from 0 to 9) of a randomly selected expense amount:

According to Benford’s law,

Expected value (or mean),

  ${\mu }_X=3.441$

Calculations:

From Part (b),

We have

The mean of uniform distribution is 5.

In this Part,

The mean according to Benford’s law is 3.441.

If the mean of all digits in an expense report is closer to 5 instead of 3.441, then the expense report is false.

Thus,

If we obtain an expense report with a mean of first digits close to 5 instead of 3.441,

Then

We will assume that the expense report is fake.