# Let X_{1}, X_{2},...,X_{n} be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables. Find n such that Pr(X>2000)\geq 0.95.

Let $$X_{1}, X_{2},...,X_{n}$$ be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables.
Find n such that $$Pr(X>2000)\geq 0.95$$.

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Step 1
Given that $$\mu=100$$ and $$\sigma = 30$$
$$E(X)=E(\sum X_{i})=n\mu = 100n$$
$$\sigma(X)=\sigma(\sum X_{i})=\sqrt[30]{n}$$
Step 2
Consider,
$$P(X>2000)\geq 0.95$$
$$P(\frac{X-E(X)}{\sigma(X)}>\frac{2000-100n}{\sqrt[30]{n}})\geq 0.95$$
$$P(z>\frac{200-10n}{\sqrt[3]{n}})\geq 0.95$$
$$1-P(z\leq \frac{200-10n}{\sqrt[3]{n}})\geq 0.95$$
$$1-0.95\geq P(z\leq \frac{200-10n}{\sqrt[3]{n}})$$
$$P(z\leq \frac{200-10n}{\sqrt[3]{n}})\leq 0.05$$
$$\frac{200-10n}{\sqrt[3]{n}}\leq -1.645$$
$$200-10n\leq 4.935\sqrt{n}$$
$$4.935\sqrt{n}-10n+200\leq 0$$
$$\sqrt{n} \leq -4.23$$
$$n=17.89$$
$$\approx 18$$
Thus, $$n=18.$$