Broadcom (NASDAQ:AVGO) Shares Gap Up Following Better-Than-Expected Earnings

Let’s break down the problem step by step. We are given: - The total number of students: \( n = 100 \) - The mean score: \( \bar{x} = 40 \) - The standard deviation: \( s = 10 \) We are asked to find the number of students who scored between 30 and 50. --- ### Step 1: Standardize the scores We convert the scores to z-scores using the formula: \[ z = \frac{x - \bar{x}}{s} \] For \( x = 30 \): \[ z_{30} = \frac{30 - 40}{10} = \frac{-10}{10} = -1 \] For \( x = 50 \): \[ z_{50} = \frac{50 - 40}{10} = \frac{10}{10} = 1 \] So, we are looking for the proportion of students with z-scores between -1 and 1. --- ### Step 2: Use the empirical rule The empirical rule (68-95-99.7 rule) states that approximately 68% of the data lies within one standard deviation of the mean in a normal distribution. So, the proportion of students scoring between 30 and 50 is approximately 68%. --- ### Step 3: Calculate the number of students \[ \text{Number of students} = 0.68 \times 100 = 68 \] --- ### Step 4: Check the options The options given are: - A) 68 - B) 95 - C) 99 - D) 100 So, the answer should be **A) 68**. --- **Final answer:** \[ \boxed{A} \]