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.