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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.
in Data Science & Statistics by Diamond (75,914 points) | 14 views

1 Answer

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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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