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To solve the problem in your image, we need to use the **68-95-99.7 rule** (also known as the **empirical rule**) to find the percentage of values that lie above 27, given a normal distribution with:

- Mean (\(\mu\)) = 22
- Standard deviation (\(\sigma\)) = 5

### Steps to solve:
1. **Find the z-score** for 27.
   The z-score is calculated using the formula:
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   Where:
   - \(x = 27\) (the value in question),
   - \(\mu = 22\) (the mean),
   - \(\sigma = 5\) (the standard deviation).

   Substituting the values:
   \[
   z = \frac{27 - 22}{5} = \frac{5}{5} = 1
   \]
   So, the z-score is 1.

2. **Use the empirical rule**:
   According to the 68-95-99.7 rule:
   - 68% of the data falls within 1 standard deviation of the mean.
   - This means 34% of the data falls between the mean and 1 standard deviation above the mean (because 68% is split evenly on both sides).

3. **Find the percentage of values above 27**:
   Since the z-score of 27 is 1, and 34% of the data falls between 22 and 27, the remaining area to the right (above 27) is half of the remaining 32% (because 68% is within 1 standard deviation).

   Therefore:
   \[
   50\% - 34\% = 16\%
   \]

### Answer: **16%** of the values lie above 27.

---

Would you like more details or have any other questions? Here are five related questions to deepen your understanding:

1. How is the z-score related to the standard normal distribution?
2. What if the standard deviation was 10 instead of 5? How would the percentage change?
3. Can you explain the 68-95-99.7 rule more thoroughly?
4. What is the probability that a value lies between 22 and 32 in this distribution?
5. How do you calculate the probability for values farther than 3 standard deviations from the mean?

**Tip**: The z-score is a useful tool in normal distributions as it standardizes values, making it easier to compare data across different normal distributions.

What percentage of values lie above 27 in a normal distribution with a mean of 22 and a standard deviation of 5 using the 68-95-99.7 rule?

Find the percentage of values above 27 in a normal distribution with a mean of 22 and a standard deviation of 5 using the empirical rule (68-95-99.7 rule). This problem involves the use of z-scores and the normal distribution curve.

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