# Estimating Binomial random variable with Normal distribution

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.

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