To find the indicated IQ score based on a cumulative probability, we need to use the standard normal distribution (Z-distribution) and convert the given probability into an IQ score using the mean and standard deviation.

Steps:

1. Identify the probability: From the problem, the cumulative probability is given as 0.2969.

2. Convert the probability to a Z-score: Using a Z-table or an inverse cumulative distribution function, find the Z-score corresponding to 0.2969.

   • The Z-score for a cumulative probability of 0.2969 is approximately -0.52.

3. Use the Z-score formula: The formula to convert a Z-score to an IQ score is:

   $\text{IQ score} = \mu + Z \cdot \sigma$ where:

   • $\mu = 100$ (mean IQ)
   • $\sigma = 15$ (standard deviation)
   • $Z = -0.52$ (Z-score)

4. Substitute values:

   $\text{IQ score} = 100 + (-0.52) \cdot 15$ $\text{IQ score} = 100 - 7.8 = 92.2$

5. Round to the nearest whole number: The indicated IQ score is 92.

---

Would you like further details on this calculation, or do you have any other questions?

Here are 5 relative questions to expand on this concept:

1. How is a Z-score related to the normal distribution?
2. What other applications are there for Z-scores in real-world problems?
3. How can cumulative probabilities be interpreted in terms of percentiles?
4. Why is the mean of IQ scores typically set at 100?
5. What is the significance of the standard deviation in IQ scoring?

Tip: When rounding, always consider whether the number after the decimal point is 5 or greater to determine if it should round up.