To find the probability that the sample mean number of TV sets in a U.S. household is between 2.5 and 3, we can again use the Central Limit Theorem. Here’s the step-by-step process:

1. State the known parameters:

   • Population mean ($\mu$) = 2.24
   • Population standard deviation ($\sigma$) = 1.2
   • Sample size ($n$) = 95

2. Calculate the standard error of the mean (SE): $SE = \frac{\sigma}{\sqrt{n}} = \frac{1.2}{\sqrt{95}} \approx 0.1231$

3. Find the Z-scores for the sample means 2.5 and 3: $Z_1 = \frac{2.5 - 2.24}{SE} = \frac{2.5 - 2.24}{0.1231} \approx 2.11$ $Z_2 = \frac{3 - 2.24}{SE} = \frac{3 - 2.24}{0.1231} \approx 6.17$

4. Determine the cumulative probabilities corresponding to these Z-scores:

   • Using the standard normal distribution table or a calculator:
     • $P(Z < 2.11) \approx 0.9826$
     • $P(Z < 6.17) \approx 1$ (since Z = 6.17 is extremely high, the cumulative probability is effectively 1)

5. Calculate the probability of the sample mean being between 2.5 and 3: $P(2.5 < \bar{x} < 3) = P(Z < 6.17) - P(Z < 2.11) = 1 - 0.9826 = 0.0174$

Therefore, the probability that the sample mean number of TV sets is between 2.5 and 3 is approximately 0.0174.

Would you like more details or have any questions?

Here are some questions you might want to ask:

1. How do you calculate Z-scores for different sample means?
2. What is the significance of extremely high or low Z-scores in normal distribution?
3. How can you use the Central Limit Theorem to approximate probabilities for sample means?
4. What are some other applications of the Central Limit Theorem in statistics?
5. How can you interpret the probability of sample means in real-world contexts?

Tip: When finding probabilities for ranges of sample means, always calculate the Z-scores for the endpoints of the range and then use the standard normal distribution to find the corresponding probabilities. Subtract these cumulative probabilities to find the probability within the desired range.