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$N$ independent trials are to be conducted, each with "success" probability $p .$ Let $X_{i}=1$ if trial $i$ is a success and $X_{i}=0$ if it is not. What is the distribution of the random variable $X=X_{1}+X_{2}+\ldots+X_{N} ?$ Express $P[a \leq X \leq b]$ as a sum (where $a \leq b$ and these are integers between 0 and $N$ ). Use the central limit theorem to provide an approximation to this probability. Compare your approximation with the limit theorem of De Moivre and Laplace on p1189 of Kreyszig.
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As $X$ is the total number of successes in $N$ independent trials $X$ is $\operatorname{Bin}(N, p)$. Thus
$$P[a \leq X \leq b]=\sum_{x=a}^{b}\left(\begin{array}{l} N \\ x \end{array}\right) p^{x}(1-p)^{N-x} .$$
As each $X_{i}$ has $E\left(X_{i}\right)=p$ and $\operatorname{Var}\left(X_{i}\right)=p(1-p)$ (you should confirm this) the central limit theorem says that $X$ is approximately Normal $(N p, N p(1-p))$. Let $\bar{a}=(a-N p) / \sqrt{N p(1-p)}$ and $\bar{b}=(b-N p) / \sqrt{N p(1-p)} .$ The approximation is $P[a \leq X \leq b] \approx P[\bar{a}<Z<\bar{b}]$ where $Z$ has the
standard Normal distribution. The reason for the small correction in the version of this result in Kreyszig is that $P[a \leq X \leq b]=P[a-\delta<X<b+\delta]$ for any $\delta \in(0,1)$ while the approximation varies with $\delta-$ the choice $\delta=0.5$ is arbitrary but generally sensible.
by Diamond (75,914 points)

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