To solve this problem, we need to find the IQ score corresponding to a cumulative probability (area under the curve) of 0.10. The normal distribution provided has:

• Mean ($\mu$) = 100
• Standard deviation ($\sigma$) = 15
• The shaded area = 0.10

Approach:

1. Standard normal distribution (Z-score): The given normal distribution can be converted to the standard normal distribution using: $Z = \frac{X - \mu}{\sigma}$ where $X$ is the IQ score, $\mu$ is the mean, and $\sigma$ is the standard deviation.

2. Finding the Z-score: The Z-score corresponding to the cumulative area of 0.10 can be looked up in a Z-table or calculated using a calculator. For a left-tailed area of 0.10, the Z-score is approximately -1.28.

3. Converting Z-score to IQ score: Now, convert the Z-score back to the original scale of the IQ test using the formula: $X = Z \cdot \sigma + \mu$ Substituting the values: $X = (-1.28) \cdot 15 + 100 = 80.8$

Conclusion:

The IQ score corresponding to the shaded area of 0.10 is 80.8, so the correct answer is:

$\boxed{80.8}$

Would you like further clarification or more details on any step?

Here are 5 related questions for practice:

1. What IQ score corresponds to a cumulative probability of 0.05 in this distribution?
2. How do we find the Z-score for a given cumulative area in a normal distribution?
3. If the shaded area was 0.50, what IQ score would you expect?
4. How would the IQ score change if the standard deviation were smaller, say 10 instead of 15?
5. What is the probability of an IQ score greater than 119.2?

Tip: Always remember to convert between the Z-score and actual data points when working with normal distributions!