Benford's LAw

Hi Struggling student :)

According to Benford's law, the probability that the first digit of the amount of a randomly chosen invoice is a 1 or a 2 is 0.477. You examine 77 invoices from a vendor and find that 25 have first digits 1 or 2. If Benford's law holds, the count of 1s and 2s will have the binomial distribution with n = 77 and p = 0.477. Too few 1s and 2s suggests fraud. What is the approximate probability of 25 or fewer 1s and 2s if the invoices follow Benford's law? (Use the normal approximation. Round your answer to four decimal places.)

Here is an example of what you want

http://www.mathwords.com/b/binomial_probability_formula.htm

\(\boxed{P(\text{k successes in n trials)}=\begin{pmatrix}n\\k\end{pmatrix}p^kq^{(n-k)}} \)

n=77

k=0 to 25

p=0.477

q=1-0.477 = 0.523

\(P(\text{k successes in 77 trials)}=\begin{pmatrix}77\\k\end{pmatrix}0.477^k*0.523^{(n-k)}\\~\\\ P(\text{25 or less successes in 77 trials)}=\displaystyle\sum_{k=0}^{25}\;\;\begin{pmatrix}77\\k\end{pmatrix}0.477^k*0.523^{(n-k)}\\~\\ \)

At this point in time I am going to go look for a binomial probability calculator on line.

I just googled 'binomial probability calculator' and this is the forst one that was on the list.

http://www.vassarstats.net/textbook/ch5apx.html

\(\begin{align}P(\text{25 or less successes in 77 trials)}&=\displaystyle\sum_{k=0}^{25}\;\;\begin{pmatrix}77\\k\end{pmatrix}0.477^k*0.523^{(n-k)}\\ \\&=0.0048 \end{align} \)

This would be a bit tedious to do without the probability calculator but it would just be a matter of adding the 26 elements of the sum.  :)

You know Struggling Student, it would be polite for you to acknowledge the answer that I have given you.