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I got a very small probability in my homework so are there any mistakes?

Benford's law says that if we have a collection of numbers, the first digit has the distribution $\mathbb{P}(D=k)=\log_{10}(1+1/k).$ Take a collection of 1200 numbers. Approximate by using normal distribution that the probability that those numbers has first digit 1, 2 or 3 at least 680 cases.

My approach:

One can compute easily that $\sum_{k=1}^9\log_{10}\left (1+\frac{1}{k}\right )=1.$ Benford's law says that the probability that the first digit is 1, 2 or 3 is

$\sum_{i=1}^3\log_{10}(1+1/i)\approx 0.778$.

Now $1200\cdot 0.778>10$ and $1200(1-0.778)>10$, so it is reasonable to approximate binomial distribution by a normal distribution. Now $\operatorname{Bin}(1200,0.778)$ is approximately $N(1200\cdot 0.778,1200\cdot 0.778(1-0.778))=N(933.6,207.3)$.

Therefore, $P(X\leq 680.5)\approx P\left (Z\leq \frac{680.5-1200\cdot 0.778}{\sqrt{1200\cdot 0.778\cdot 0.222}}\right )\approx P(Z<-17.6)$. Now $\Phi(-17.6)\approx 0.$

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  • $\begingroup$ The sum up to 5 (not 3) gives 0.778. You need to correct your question in one way or another (either 3 is wrong or 0.778 is). Note also you were asked a question about "at least 680". Are you sure you calculate the right quantity? $\endgroup$
    – Glen_b
    Commented Feb 4, 2013 at 23:29
  • $\begingroup$ @Glen_b The mistake I made was that I computed the sum up to 5 instead up to 3. $\endgroup$
    – Jaakko Seppälä
    Commented Feb 8, 2013 at 21:19

1 Answer 1

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Are you sure $\sum_{i=1}^3\log_{10}(1+1/i)\approx 0.778$?

> sum(log10(1+1/(1:3)))
[1] 0.60206
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