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The problem you're working on involves a normal distribution with a **mean** of 190.3 minutes and a **standard deviation** of 19.8 minutes. The graph shows a normal distribution curve with points labeled \( U \), \( V \), and \( W \), which represent specific times on the x-axis, and you're tasked with determining the values of these points and the shaded area under the curve.

### Key information:
- **Mean** (\( \mu \)) = 190.3 minutes
- **Standard deviation** (\( \sigma \)) = 19.8 minutes
- \( V \) is at the peak of the curve, so \( V = \mu = 190.3 \) minutes.
- \( U \) and \( W \) are equidistant from \( V \), representing values one standard deviation away from the mean.

### Applying the **Empirical Rule**:
The empirical rule states that:
1. **68%** of data falls within **1 standard deviation** of the mean, so the area shaded between \( U \) and \( W \) would represent this 68%.
   - \( U = \mu - \sigma = 190.3 - 19.8 = 170.5 \) minutes
   - \( W = \mu + \sigma = 190.3 + 19.8 = 210.1 \) minutes

### Conclusion:
- The **percentage of the shaded area** under the curve is **68%**.
- The values are:
  - \( U = 170.5 \) minutes
  - \( V = 190.3 \) minutes
  - \( W = 210.1 \) minutes

Would you like more details or have any questions about this problem?

Here are 5 related questions you might find useful:
1. What percentage of data falls within two standard deviations of the mean in a normal distribution?
2. How does the standard deviation affect the shape of a normal distribution curve?
3. What would be the values of \( U \), \( V \), and \( W \) if the standard deviation were larger, say 30 minutes?
4. How is the empirical rule applied in other fields like finance or biology?
5. What is the relationship between the mean and the median in a normal distribution?

**Tip**: The empirical rule is most useful when data follows a bell-shaped, normal distribution. Always check if the data approximates this shape before applying the rule.

The lengths of movie files that are available for streaming are modeled using a normal distribution. The mean is 190.3 minutes, and the standard deviation is 19.8 minutes. Given points U, V, and W on the distribution curve, where V is at the peak, and U and W are symmetrically placed around V, determine the percentage of the shaded area and the values of U, V, and W.

Solve a problem involving normal distribution with a mean of 190.3 minutes and a standard deviation of 19.8 minutes. Learn how to use the Empirical Rule to find the values of U, V, and W, and calculate the percentage of the area shaded. Suitable for students in Grades 10-12.

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