I need to find the values of UV and W in minutes and I need the percentage of total area shaded

To solve this problem, we need to determine the values of \( U \), \( V \), and \( W \) in minutes based on the given normal distribution parameters and the corresponding shaded areas. From the image, it's likely that the values represent certain time points on the curve, and we're interested in the percentage of the total area (under the normal curve) corresponding to the shaded region.

### Step 1: Identify the Mean and Standard Deviation
- **Mean (\(\mu\))**: 175 seconds
- **Standard Deviation (\(\sigma\))**: 25 seconds

### Step 2: Convert Seconds to Minutes
Since we need the values in minutes, first convert the mean and standard deviation from seconds to minutes.
- \( \mu = 175 \text{ seconds} = \frac{175}{60} \text{ minutes} \approx 2.92 \text{ minutes} \)
- \( \sigma = 25 \text{ seconds} = \frac{25}{60} \text{ minutes} \approx 0.42 \text{ minutes} \)

### Step 3: Determine the Values of \( U \), \( V \), and \( W \)
These values are typically set at specific standard deviations from the mean in a normal distribution. Assuming \( U \), \( V \), and \( W \) correspond to standard deviation points:

- \( U \) is likely \( \mu - \sigma \):  
  \( U = 2.92 - 0.42 \approx 2.50 \) minutes.
- \( V \) is likely \( \mu + \sigma \):  
  \( V = 2.92 + 0.42 \approx 3.34 \) minutes.
- \( W \) could be \( \mu \), or if it's meant to represent an area symmetric around the mean, this would just be \( \mu = 2.92 \) minutes.

### Step 4: Calculate the Percentage of Area Under the Curve
Assuming the standard interpretation:

- **Area between \( U \) and \( V \)**: This represents the area within 1 standard deviation from the mean, which is approximately **68%** of the total area under the curve.
- **Area outside \( U \) and \( V \)**: The area outside 1 standard deviation on either side of the mean would account for the remaining **32%** of the distribution, with **16%** on each tail.

### Final Answer:
- **\( U \)**: 2.50 minutes  
- **\( V \)**: 3.34 minutes  
- **\( W \)**: 2.92 minutes

The **percentage of the total area shaded** (likely the area between \( U \) and \( V \)) is **68%**.

Would you like further clarification on any part of this solution? Here are five related questions to consider:

1. What is the probability that a randomly selected time is more than 3.34 minutes?
2. How do you calculate the z-score for a specific value in this distribution?
3. What percentage of the data falls between 2.5 minutes and 3.5 minutes?
4. How would the curve change if the standard deviation were larger?
5. What is the interpretation of the shaded area in real-world terms?

**Tip**: Always double-check whether you're working with seconds or minutes to avoid unit conversion errors!

Learn how to find the values of UV and W in minutes using normal distribution parameters. Also, understand how to calculate the percentage of total area shaded under the curve. This problem involves applying statistical concepts such as the normal distribution formula and standard deviation calculation. Suitable for advanced high school or college-level students.

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