Battery Life

Solution

Answer: 95.4%

Let the battery lifetime be \[ X \sim N(100, 10^2). \] We want \[ P(80 \le X \le 120). \] We can use Z scores to Standardize: \[ Z = \frac{X-\mu}{\sigma}. \] For \(X=80\), \[ Z = \frac{80-100}{10} = -2. \] For \(X=120\), \[ Z = \frac{120-100}{10} = 2. \] Thus \[ P(80 \le X \le 120) = P(-2 \le Z \le 2). \] From the standard normal distribution table, \[ P(Z \le 2) \approx 0.9772 \] \[ P(Z \le -2) \approx 0.0228. \] Therefore \[ P(-2 \le Z \le 2) = 0.9772 - 0.0228 = 0.9544. \]