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