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.