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To solve this problem, we need to determine the minimum number of hours a person needs to watch TV per day in order to be in the top 15% of the population. The key details provided are:

- The mean (μ) = 7.5 hours.
- The standard deviation (σ) = 1.8 hours.
- The population is normally distributed.
- We are looking for the threshold above which the top 15% fall.

### Steps:

1. **Use the z-score formula**:
   \[
   z = \frac{X - \mu}{\sigma}
   \]
   where \( X \) is the value we are solving for (the number of hours), \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

2. **Find the z-score that corresponds to the top 15%**:
   To find the critical z-value, we need to look up the z-score for the cumulative probability of \( 1 - 0.15 = 0.85 \). This gives the z-score at the 85th percentile.

3. **Using the z-table**:
   The z-value corresponding to a cumulative probability of 0.85 is approximately 1.036.

4. **Substitute the z-value into the formula** and solve for \( X \):
   \[
   1.036 = \frac{X - 7.5}{1.8}
   \]

   Solving for \( X \):
   \[
   X = 1.036 \times 1.8 + 7.5
   \]
   \[
   X \approx 9.3648
   \]

5. **Round to one decimal place**:
   \[
   X \approx 9.4 \, \text{hours}
   \]

Thus, at least **9.4 hours** of daily TV watching is necessary for a person to be eligible for the interview.

Would you like more details or have any questions?

### Related Questions:
1. What is the significance of a z-score in this context?
2. How would the result change if the standard deviation were smaller?
3. What if the psychologist wanted to interview the top 10% instead?
4. Can you calculate the probability of someone watching exactly 7.5 hours of TV?
5. How would you adjust the calculation if the population distribution were skewed?

### Tip:
Always round intermediate steps to at least four decimal places to maintain accuracy in your final result.

A newspaper article reported that people spend a mean of 7.5 hours per day watching TV, with a standard deviation of 1.8 hours. A psychologist would like to conduct interviews with the 15% of the population who spend the most time watching TV. She assumes that the daily time people spend watching TV is normally distributed. At least how many hours of daily TV watching are necessary for a person to be eligible for the interview? Round your answer to one decimal place.

Learn how to calculate the minimum hours of daily TV watching necessary for a person to be in the top 15% of the population using normal distribution and Z-scores. This problem involves key statistical concepts such as mean, standard deviation, and the Z-score formula, making it suitable for students in Grades 10-12.

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