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Gaussian limit of binomial distribution

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In the attachment this problem is formulated in terms of noninteracting particles.

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Solution Summary

We extract the large N limit of the binomial distribution with p = 1/2 and then for general p. We derive the formulas for the mean and the standard deviation for the binomial distribution.

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documentclass[a4paper]{article}
usepackage{amsmath,amssymb}
newcommand{haak}[1]{left(#1right)}
newcommand{rhaak}[1]{left [#1right]}
newcommand{lhaak}[1]{left | #1right |}
newcommand{ahaak}[1]{left{#1right}}
newcommand{gem}[1]{leftlangle #1rightrangle}
newcommand{gemc}[2]{leftlangleleftlangleleft. #1right | #2
rightranglerightrangle}
newcommand{geml}[1]{leftlangle #1right.}
newcommand{gemr}[1]{left. #1rightrangle}
newcommand{haakl}[1]{left(#1right.}
newcommand{haakr}[1]{left.#1right)}
newcommand{rhaakl}[1]{left[#1right.}
newcommand{rhaakr}[1]{left.#1right]}
newcommand{lhaakl}[1]{left |#1right.}
newcommand{lhaakr}[1]{left.#1right |}
newcommand{ket}[1]{lhaakl{gemr{#1}}}
newcommand{bra}[1]{lhaakr{geml{#1}}}
newcommand{braket}[3]{gem{#1lhaak{#2}#3}}
newcommand{floor}[1]{leftlfloor #1rightrfloor}
newcommand{half}{frac{1}{2}}
newcommand{kwart}{frac{1}{4}}
renewcommand{imath}{text{i}}

begin{document}
section{Problem 1}
subsection{}
The probability that a given particle is in A is $P_{A}=half$. The probability that it is in B is thus also $P_{B}=half$. Let's pick $N_{A}$ particles and calculate the probability that these $N_{A}$ particles are in A. Since all the probabilities for the individual particles are independent, you can multiply these: You obtain:
begin{equation}label{p}
P = P_{A}^{N_{A}}times P_{B}^{N - N_{A}} = haak{half}^{N}
end{equation}

For any choice of the $N_{A}$ particles the probability is $P$. So, the probability that you have $N_{A}$ particles in A regardless ...

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