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.

vazelinahS 2021-06-04 Answered
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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Expert Answer

unessodopunsep
Answered 2021-06-05 Author has 3701 answers

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, \(n=18.\)

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