>>12538573Here
http://www.stat.yale.edu/~pollard/Courses/241.fall2005/notes2005/Bin.Normal.pdfhttps://www.math.utah.edu/~kenkel/normaldistributiontalk.pdfBasically, you start your approximation of the binomial mass function at a point where the number of successes is approximately the success rate times the number of trials. From this starting position, you take the ratios of neighboring terms of the binomial mass function. Using a logarithmic argument, and some more approximations, you show that the sequence of ratios is the exponential of an arithmetic sequence in terms of the total number of trials, the success rate, the failure rate, and number of slots from the start. Using the exponential identities, you can prove that the product of the successive ratios is the exponential of a sum of an arithmetic sequence. The arithmetic sequence has a closed form solution in terms of the square of the number of positions from the start, and in the denomiator; the number of trials, success rate, and failure rate.
This provides a closed form for the approximate ratio of the starting position, to other binomial mass function terms on its row, but it doesn’t give a good approximation of the starting position. Using the (aforementioned) choice of starting position as when the number of successes is equal to the number of trials times the success rate AND Stirling’s Approximation, you can get a good approximation for the starting binomial mass function in closed form (in terms of the number of trials, the success rate and the failure rate).
Converting this approximation of the mass function to a continuous function gets you the Gaussian function.
