Q 22. Benford鈥檚 law and fraud(a) Usi... [FREE SOLUTION]

91影视

The Practice of Statistics for AP Examination

Daren Starnes, Josh Tabor

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 837 pages

ISBN: 9781319113339

Chapter 6: Q 22. (page 369)

Benford鈥檚 law and fraud
(a) Using the graph from Exercise 21, calculate the standard deviation 蟽_Y. This gives us an idea of how much variation we鈥檇 expect in the employee鈥檚 expense records if he assumed that first digits from 1 to 9 were equally likely.
(b) The standard deviation of the first digits of randomly selected expense amounts that follow Benford鈥檚 law is $蟽_{X} = 2.46$ . Would using standard deviations be a good way to detect fraud? Explain your answer.

Short Answer

Expert verified

Part (a) The standard deviation is 2.5820.

Part (b) No

Step by step solution

Part (a) Step 1. Given information.

The given information is:

First Digit	$1 2 3 4 5 6 7 8 9$
Probability	$\frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9} \frac{1}{9}$

Part (a) Step 2. Find the standard deviation.

The expected value is:

渭 = 鈭� x P (X = x) = 1 脳 \frac{1}{9} + 2 脳 \frac{1}{9} + 3 脳 \frac{1}{9} + 4 脳 \frac{1}{9} + 5 脳 \frac{1}{9} + 6 脳 \frac{1}{9} + 7 脳 \frac{1}{9} + 8 脳 \frac{1}{9} + 9 脳 \frac{1}{9} = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9}{9} = \frac{45}{9} = 5

The standard deviation is:

$蟽^{2} = 鈭� {(x - 渭)}^{2} P (x) = {(1 - 5)}^{2} 脳 \frac{1}{9} + {(2 - 5)}^{2} 脳 \frac{1}{9} + {(3 - 5)}^{2} 脳 \frac{1}{9} + {(4 - 5)}^{2} 脳 \frac{1}{9} + {(5 - 5)}^{2} 脳 \frac{1}{9} + {(6 - 5)}^{2} 脳 \frac{1}{9} + {(7 - 5)}^{2} 脳 \frac{1}{9} + {(8 - 5)}^{2} 脳 \frac{1}{9} + {(9 - 5)}^{2} 脳 \frac{1}{9} = \frac{20}{3} 蟽 = \sqrt{蟽^{2}} = \sqrt{\frac{20}{3}} = \frac{2 \sqrt{15}}{3} 鈮� 2.5820$

Part (b) Step 1. Explanation.

The uniform distribution's standard deviation is 2.5820, while Benford's law's standard deviation is 2.46.

The two standard deviations are quite similar, so using the standard deviation to detect fraud is not a good idea because the two distributions have about the same standard deviation (the two distributions are difficult to distinguish by their standard deviation).