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Given $n = 10^4$ trials of independent Bernoulli experiments with $p = 0.6$, I'm trying to estimate the probability that the number of successes fall between $7901$ and $8100$.

I defined $S_n = \sum_{i=1}^{n}X_i$ where $X_i$ is the i-th trial. We then have $X_i \sim Be(p)$ and therefore $S_n \sim B(n, p)$. Since we know $E[X_i] = p = 0.6$ and $Var[X_i] = p(1-p) = 0.24$, I tried using the Central Limit Theorem:

$$ P(7901 \leq S_n \leq 8100) = P \left( \frac{7901 - n \cdot 0.6}{\sqrt{n \cdot 0.24}} \leq \frac{S_n - n \cdot 0.6}{\sqrt{n \cdot 0.24}} \leq \frac{8100 - n \cdot 0.6}{\sqrt{n \cdot 0.24}} \right) = P \left( \frac{1901}{\sqrt{2400}} \leq \frac{S_n - 6000}{\sqrt{2400}}\leq \frac{2100}{\sqrt{2400}} \right) $$

Since $\frac{S_n - 6000}{\sqrt{2400}} \xrightarrow[]{d} N(0,1)$, $$ P \left( \frac{1901}{\sqrt{2400}} \leq \frac{S_n - 6000}{\sqrt{2400}}\leq \frac{2100}{\sqrt{2400}} \right) = P \left( \frac{S_n - 6000}{\sqrt{2400}} \leq \frac{2100}{\sqrt{2400}} \right) - P \left( \frac{S_n - 6000}{\sqrt{2400}} \leq \frac{1900}{\sqrt{2400}}\right) \approx \phi \left( \frac{2100}{\sqrt{2400}} \right) - \phi \left( \frac{1900}{\sqrt{2400}} \right) $$

Where $\phi$ is the cumulative distribution function for the standard normal distribution. But this of course yields $$ \approx 1 - 1 = 0$$

I can't seem to pin down where I made a mistake, any hints/tips will be greatly appreciated. Thanks in advance.

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There's nothing wrong; you just haven't used enough precision. The correct answer is, rounded to four significant figures, $$2.434×10^{-329}$$ which tells you that it is negligible that between $7901$ and $8100$ heads will occur.

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  • $\begingroup$ (+1). In R, code ` diff(pbinom(c(7900,8100), 10^4, .6))` return $0$ (to 8 places). $\endgroup$
    – BruceET
    Dec 7, 2020 at 23:55

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