To determine the percentage of values in a normal distribution between 19 and 29 using the 68-95-99.7 rule, let's follow these steps:

1. Identify the Mean and Standard Deviation:

   • Mean $(\mu) = 19$
   • Standard deviation $(\sigma) = 5$

2. Calculate the Z-Scores for 19 and 29:

   • Since the mean is 19, a score of 19 is exactly 0 standard deviations from the mean (within the middle of the distribution).
   • The score of 29 is 2 standard deviations above the mean: $Z = \frac{29 - 19}{5} = 2$

3. Apply the 68-95-99.7 Rule:

   • The 68-95-99.7 rule states:
     • 68% of values fall within 1 standard deviation of the mean.
     • 95% of values fall within 2 standard deviations of the mean.
     • 99.7% of values fall within 3 standard deviations of the mean.
   • Since 29 is 2 standard deviations above the mean, 95% of the values lie between $\mu - 2\sigma$ and $\mu + 2\sigma$.

4. Determine the Percentage Between 19 and 29:

   • Since 95% of the distribution lies within 2 standard deviations on both sides of the mean, half of that (or 47.5%) will lie between the mean and $\mu + 2\sigma$.
   • Therefore, the percentage of values between 19 and 29 is 47.5%.

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What percentage of values lie between 14 and 24 in this distribution?
2. How does the 68-95-99.7 rule help in understanding normal distributions?
3. What is the probability of a value falling below 9 in this distribution?
4. How would you calculate the percentage of values between two non-standardized points without the 68-95-99.7 rule?
5. What is the significance of the 99.7% in this rule?

Tip: The 68-95-99.7 rule is an excellent tool for estimating probabilities quickly in a normal distribution without calculations.