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, \(n=18.\)