Binomial Approximation. Calculate $n$

Question

I have the next question: If it is known that $P(X \geq300)=0.3$ and $X \sim \mathop{Bin}(n,0.2)$ How can one estimate $n$?

Answer 1

Use the Normal approximation. Our binomial random variable $X$ has mean $(0.2)n$ and variance $(n)(0.2)(0.8)$, so standard deviation $0.4\sqrt{n}$.

Looking "backwards" in the table of the standard normal, we find that the probability that a standard normal $Z$ is $\ge z$ is $0.3$ for $z\approx 0.525$. But $$\Pr(X\ge 300)\approx \Pr\left(Z\ge \frac{300-0.2n}{0.4\sqrt{n}}\right).$$
Thus we want to choose $n$ so that $$\frac{300-0.2n}{0.4\sqrt{n}}\approx 0.525.$$ Let us pretend we have equality. Then we want to solve the equation $300-0.2n=(0.4)(0.525)\sqrt{n}$, or equivalently $$0.2n +(0.4)(0.525)\sqrt{n}-300=0.$$ Make the substitution $x=\sqrt{n}$. We get a quadratic equation in $x$. Solve for $x$, using the Quadratic Formula, and square the result to estimate the appropriate $n$.