Answer: 91.04%

## Explanation To solve this problem, we need to calculate the percentage of values between 32 and 116 in a normal distribution with mean μ = 80 and standard deviation σ = 24. ### Step 1: Calculate Z-scores For the lower bound (32): \[ Z_1 = \frac{32 - 80}{24} = \frac{-48}{24} = -2 \] For the upper bound (116): \[ Z_2 = \frac{116 - 80}{24} = \frac{36}{24} = 1.5 \] ### Step 2: Find probabilities using Z-table - P(Z < -2) = 0.0228 (2.28%) - P(Z < 1.5) = 0.9332 (93.32%) ### Step 3: Calculate the probability between the bounds \[ P(32 < X < 116) = P(Z < 1.5) - P(Z < -2) = 0.9332 - 0.0228 = 0.9104 \] This equals 91.04% of the distribution. ### Verification - The range from 32 to 116 covers from μ - 2σ to μ + 1.5σ - In a normal distribution: - About 95.44% of data falls within μ ± 2σ - About 86.64% of data falls within μ ± 1.5σ - Our calculated value of 91.04% is reasonable and matches option D.

Author: LeetQuiz .