How to Solve Continuous Random Variables — Normal distribution — standardization Questions on Exam P
Sample Practice Problem
ID: #V67A7
Let \(T_{1},T_{2},\ldots\) be independent normal battery lifetimes with mean \(3\) hours and standard deviation \(1\) hour. Define \(S_{n}=\sum_{i=1}^nT_{i}\). Calculate the smallest positive integer \(n\) for which \(\Pr(S_{n}>32)\ge0.9772\).
📖 Worked Solution & Strategy
Let \(T_{i} \sim \operatorname{Normal}(3, 1)\) be independent battery lifetimes. Total lifetime \(S_{n} = \sum_{i=1}^n T_{i}\) is normal with mean \(\operatorname{E}[S_{n}] = 3n\) and variance \(\operatorname{Var}(S_{n}) = n(1^2) = n\). Standardizing gives \(\Pr(S_{n} > 32) = \Pr\left(\frac{S_{n} - 3n}{\sqrt n} > \frac{32 - 3n}{\sqrt n}\right) = 1 - \Phi\left(\frac{32 - 3n}{\sqrt n}\right) = \Phi\left(\frac{3n - 32}{\sqrt n}\right)\). The requirement \(\Pr(S_{n} > 32) \ge \Phi(2)\) simplifies to \(\frac{3n - 32}{\sqrt n} \ge 2\). Testing positive integers: at \(n = 13\), \(\frac{3(13) - 32}{\sqrt{13}} < 2\), which fails the requirement. At \(n = 14\), \(\frac{3(14) - 32}{\sqrt{14}} \ge 2\), which satisfies the inequality. Thus the smallest valid integer is \(n = 14\).