Let's solve this problem step-by-step:

Step 1: Calculate the z-scores

To convert the values into z-scores, we'll use the formula: $z = \frac{X - \mu}{\sigma}$ where:

• $X$ is the data point
• $\mu$ is the mean
• $\sigma$ is the standard deviation

Given:

• Mean, $\mu = 120$
• Standard deviation, $\sigma = 6$

Let's find the z-scores for 115 and 125.

For $X = 115$:

$z = \frac{115 - 120}{6} = \frac{-5}{6} \approx -0.83$

For $X = 125$:

$z = \frac{125 - 120}{6} = \frac{5}{6} \approx 0.83$

Step 2: Use the z-score table to find the percentage of values between these z-scores

Using the z-score table:

• The z-score of $-0.83$ corresponds to a percentile of approximately 20.34%.
• The z-score of $0.83$ corresponds to a percentile of approximately 79.66%.

To find the percentage of people with readings between 115 and 125, subtract the percentile for $-0.83$ from that of $0.83$: $79.66\% - 20.34\% = 59.32\%$

Conclusion

The percentage of people with readings between 115 and 125 is approximately 59.32%.

Would you like more details or have any questions?

Here are some related questions that might help deepen your understanding:

1. What is a z-score and why is it useful in statistics?
2. How do you interpret a z-score in terms of standard deviations from the mean?
3. How does the area under the normal distribution curve relate to percentiles?
4. What are some examples of other real-life scenarios where z-scores might be applied?
5. How do you use a z-score table to find probabilities for values in a normal distribution?

Tip: Understanding z-scores helps in standardizing different datasets, allowing for comparisons across different scales or units.