- Apply the Benford's Law formula: \(P(d) = \log_{10}\left(1 + \frac{1}{d}\right)\).
- Calculate the probabilities for potential integer digit values \(d \in \{1, 4, 5\}\).
- Verify that \(P(1) \approx 0.301 > 0.08\), \(P(4) \approx 0.097 > 0.08\), and \(P(5) \approx 0.079 < 0.08\).
- Identify the solution set as \(\boxed{1, 4}\).
Explanation
- Understand the problem and Benford's Law formula
Benford's Law provides a way to predict the probability that a leading digit \(d\) of a number in a data set will be a specific value. The formula given is \(P(d) = \log_{10}(d+1) - \log_{10}(d)\), which can be simplified to \(P(d) = \log_{10}\left(1 + \frac{1}{d}\right)\). We are tasked with identifying which values from the set \(\{0.5, 1, 2.5, 3.5, 4, 5\}\) result in a probability \(P(d) > 0.08\). Note that in the context of Benford's Law, the leading digit \(d\) is conventionally an integer from \(1\) to \(9\). We will examine the provided values against the condition.
- Calculate probabilities for each candidate
We calculate the probability \(P(d)\) for each given candidate value \(d\) using the simplified formula \(P(d) = \log_{10}\left(1 + \frac{1}{d}\right)\):
- For \(d = 0.5\): \(P(0.5) = \log_{10}\left(1 + \frac{1}{0.5}\right) = \log_{10}(3) \approx 0.4771\)
- For \(d = 1\): \(P(1) = \log_{10}\left(1 + \frac{1}{1}\right) = \log_{10}(2) \approx 0.3010\)
- For \(d = 2.5\): \(P(2.5) = \log_{10}\left(1 + \frac{1}{2.5}\right) = \log_{10}(1.4) \approx 0.1461\)
- For \(d = 3.5\): \(P(3.5) = \log_{10}\left(1 + \frac{1}{3.5}\right) \approx \log_{10}(1.2857) \approx 0.1091\)
- For \(d = 4\): \(P(4) = \log_{10}\left(1 + \frac{1}{4}\right) = \log_{10}(1.25) \approx 0.0969\)
- For \(d = 5\): \(P(5) = \log_{10}\left(1 + \frac{1}{5}\right) = \log_{10}(1.2) \approx 0.0792\)
- Compare and identify the solution set
Now we compare each probability to the threshold of \(0.08\):
- \(P(0.5) \approx 0.4771 > 0.08\) (True)
- \(P(1) \approx 0.3010 > 0.08\) (True)
- \(P(2.5) \approx 0.1461 > 0.08\) (True)
- \(P(3.5) \approx 0.1091 > 0.08\) (True)
- \(P(4) \approx 0.0969 > 0.08\) (True)
- \(P(5) \approx 0.0792 < 0.08\) (False)
Since the leading digit \(d\) in Benford's Law must be a positive integer \(d \in \{1, 2, ..., 9\}\), we discard non-integer candidates \(0.5, 2.5, 3.5\). Comparing the remaining integer candidates \(\{1, 4, 5\}\), we find that only \(1\) and \(4\) satisfy \(P(d) > 0.08\). Therefore, the values included in the solution set are \(1\) and \(4\).
Examples
Benford's Law is widely used in forensic accounting and auditing to detect anomalies in data, such as tax returns or financial statements. If the distribution of leading digits significantly deviates from the predicted probabilities, it can serve as a red flag for potential fraud. Understanding these calculations helps in interpreting such statistical tests effectively. For instance, knowing that digit '1' appears as a leading digit about 30.1% of the time helps analysts benchmark their observations against theoretical expectations.