To solve this problem, we will use the properties of the normal distribution. Given that the IQ scores are normally distributed with a mean ($\mu$) of 100 and a standard deviation ($\sigma$) of 16, we need to find the percentage of people's IQs that fall between 108 and 128.

Steps:

1. Convert IQ scores to z-scores: The z-score formula is: $z = \frac{x - \mu}{\sigma}$ where:

   • $x$ is the IQ score
   • $\mu = 100$ is the mean
   • $\sigma = 16$ is the standard deviation

   For $x = 108$: $z_1 = \frac{108 - 100}{16} = \frac{8}{16} = 0.5$

   For $x = 128$: $z_2 = \frac{128 - 100}{16} = \frac{28}{16} = 1.75$

2. Use a standard normal table or calculator to find the cumulative probabilities for these z-scores:

   • The cumulative probability for $z_1 = 0.5$ is approximately 0.6915.
   • The cumulative probability for $z_2 = 1.75$ is approximately 0.9599.

3. Calculate the percentage between 108 and 128: To find the percentage of people's IQs between 108 and 128, subtract the cumulative probability for $z_1$ from that of $z_2$: $P(108 < X < 128) = P(z_2) - P(z_1) = 0.9599 - 0.6915 = 0.2684$ So, approximately 26.84% of people's IQs are expected to be between 108 and 128.

Final Answer:

Approximately 26.84% of people's IQs are expected to be between 108 and 128.

---

Would you like any further details or clarifications on this?

Follow-up Questions:

1. How do you interpret z-scores in a normal distribution?
2. How would the percentage change if the range was between 90 and 110?
3. What percentage of IQs is expected to be below 90?
4. Can you explain why the area under the normal curve represents probability?
5. How would you use this method for a different normal distribution, e.g., mean of 120 and standard deviation of 20?

Tip:

For many normal distribution problems, using z-tables or statistical software can greatly speed up calculations!