Assessment 1: Fix the Problem! Directions: Read and analyze the given problems below. Write your solutions and final answer on a separate sheet of paper. 1. ILAW Manufacturing company produces bulbs that last a mean of 900 hours with a standard deviation of 110 hours. What is the probability that the mean lifetime of a random sample of 15 of these bulbs is less than 850 hours? 2. A school principal claims that grade 11 students have a mean grade of 86 with a standard deviation of 4 . Suppose that the distribution is approximately normal. What is the probability that a randomly selected grade will be greater than 82 but less than 90 ?

### Problem 1 1. The sample mean, \(\bar{X}\), is approximately normally distributed with mean \(\mu = 900\) hours and standard error \[ \sigma_{\bar{X}} = \frac{110}{\sqrt{15}}. \] Calculate the standard error: \[ \sigma_{\bar{X}} = \frac{110}{\sqrt{15}} \approx \frac{110}{3.873} \approx 28.4. \] 2. We seek the probability that \(\bar{X} < 850\). Standardize using the \(Z\)-score: \[ Z = \frac{850 - 900}{28.4} = \frac{-50}{28.4} \approx -1.76. \] 3. Using the standard normal distribution, find: \[ P(Z < -1.76) \approx 0.0392. \] Final answer for Problem 1: \[ \boxed{0.0392} \] --- ### Problem 2 1. The grades are approximately normally distributed with mean \(\mu = 86\) and standard deviation \(\sigma = 4\). 2. We want to find the probability that a randomly selected grade is between 82 and 90. Standardize both endpoints. 3. For \(x = 82\): \[ Z = \frac{82 - 86}{4} = \frac{-4}{4} = -1. \] 4. For \(x = 90\): \[ Z = \frac{90 - 86}{4} = \frac{4}{4} = 1. \] 5. The probability is: \[ P(82 < x < 90) = P(-1 < Z < 1). \] From the standard normal table or known results: \[ P(-1 < Z < 1) \approx 0.6826. \] Final answer for Problem 2: \[ \boxed{0.6826} \]