A race director is preparing for an upcoming marathon and estimates that the mean time to finish is 326 minutes. Assume that the times are normally distributed, with a standard deviation of 50 minutes. Use this table or the ALEKS calculator to find the percentage of times that are longer than 334 minutes. For your intermediate computations, use four or more decimal places. Give your final answer to two decimal places (for example \( 98.23 \% \).

To find the percentage of marathon finish times longer than 334 minutes, we can use the properties of the normal distribution. **Given:** - Mean (\(\mu\)) = 326 minutes - Standard deviation (\(\sigma\)) = 50 minutes - We want to find \( P(X > 334) \) **Step 1: Convert the raw score to a Z-score.** The Z-score formula is: \[ Z = \frac{X - \mu}{\sigma} \] Plugging in the values: \[ Z = \frac{334 - 326}{50} = \frac{8}{50} = 0.16 \] **Step 2: Find the probability corresponding to the Z-score.** Using standard normal distribution tables or a calculator: \[ P(Z > 0.16) = 1 - P(Z \leq 0.16) \] \[ P(Z \leq 0.16) \approx 0.5636 \] \[ P(Z > 0.16) = 1 - 0.5636 = 0.4364 \] **Step 3: Convert the probability to a percentage.** \[ 0.4364 \times 100 = 43.64\% \] **Final Answer:** Approximately **43.64%** of the marathon finish times are longer than 334 minutes.