To solve this, we are given:

1. IQ scores are normally distributed.
2. Standard deviation ($\sigma$) = 18.
3. 58% of individuals have IQ scores greater than 100.

We need to find the mean IQ score ($\mu$) for this distribution.

Solution:

Since 58% of individuals have an IQ greater than 100, 42% (or 0.42) have an IQ of 100 or less. This corresponds to the 42nd percentile.

1. Find the z-score corresponding to the 42nd percentile: Using the standard normal distribution table or a calculator, we find that the z-score for 42% is approximately $-0.2019$.

2. Set up the z-score equation: The z-score formula is: $z = \frac{X - \mu}{\sigma}$ where $X = 100$, $\sigma = 18$, and $z = -0.2019$.

3. Solve for $\mu$: $-0.2019 = \frac{100 - \mu}{18}$ $\mu = 100 + (0.2019 \times 18)$ $\mu \approx 100 + 3.6342$ $\mu \approx 103.6$

Answer:

The mean IQ score for this region is approximately 103.6.

Would you like more details on any step, or do you have further questions?

1. How do we interpret percentiles in a normal distribution?
2. How can we calculate z-scores without a table?
3. Why is the standard deviation important in this context?
4. What if the percentile had been different?
5. Can this approach be used for other distributions?

Tip: Always double-check your z-score values to ensure accuracy in solving normal distribution problems.