According to Benford's Law, which of the following would be most likely to occur $$ 30.10 \% $$ of the time as a first digit? O 8 8 2

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The digit most likely to occur $$ 30.10 \% $$ of the time as a first digit is 1. According to Benford's Law, the distribution of first digits in many naturally occurring datasets is not uniform. Instead, smaller digits occur more frequently as the leading digit. Specifically, the probability $$ P(d) $$ of a digit $$ d $$ being the first digit is given by the formula: $$ P(d) = \log_{10}(d + 1) - \log_{10}(d) = \log_{10}\left(\frac{d + 1}{d}\right) $$ Calculating the probabilities for the digits 1 through 9: - For $$ d = 1 $$: $$ P(1) = \log_{10}\left(\frac{2}{1}\right) \approx 0.3010 \text{ or } 30.10\% $$ - For $$ d = 2 $$: $$ P(2) = \log_{10}\left(\frac{3}{2}\right) \approx 0.1761 \text{ or } 17.61\% $$ - For $$ d = 8 $$: $$ P(8) = \log_{10}\left(\frac{9}{8}\right) \approx 0.0458 \text{ or } 4.58\% $$ From these calculations, we see that the digit that occurs approximately $$ 30.10\% $$ of the time as a first digit is **1**. Therefore, neither 2 nor 8 would be the correct answer for $$ 30.10\% $$. The correct answer is **1**.

According to Benford's Law, which of the following would be most likely to occur \( 30.10 \% \) of the time as a first digit? O 8 8 2

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According to Benford's Law, which of the following would be most likely to occur \( 30.10 \% \) of the time as a first digit? O 8 8 2

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