TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household was 2.24 . Assume the standard deviation is 1.2 . A sample of 95 households is drawn. Part 1 of 5 (a) What is the probability that the sample mean number of TV sets is greater than 2 ? Round your answer to at least four decimal places. The probability that the sample mean number of TV sets is greater than 2 is 0.9744 . Part: \( \mathbf{1} / \mathbf{5} \). Part 2 of 5 (b) What is the probability that the sample mean number of TV sets is between 2.5 and 3 ? Round your answer to at least four decimal places. The probability that the sample mean number of TV sets is between 2.5 and 3 is S.

Alright, I need to solve this probability problem involving the number of TV sets in U.S. households. The given information is: - Mean (μ) = 2.24 - Standard deviation (σ) = 1.2 - Sample size (n) = 95 **Part 1: Probability that the sample mean is greater than 2** First, I need to find the probability that the sample mean number of TV sets is greater than 2. To do this, I'll use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normally distributed if the sample size is large enough (n ≥ 30). Since n = 95, this condition is satisfied. The formula for the z-score in this context is: \[ z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \] Where: - \(\bar{x}\) is the sample mean - \(\mu\) is the population mean - \(\sigma\) is the population standard deviation - \(n\) is the sample size Plugging in the values: \[ z = \frac{2 - 2.24}{1.2 / \sqrt{95}} = \frac{-0.24}{1.2 / 9.7468} = \frac{-0.24}{0.123} \approx -1.95 \] Now, I need to find \( P(Z > -1.95) \). Using standard normal distribution tables or a calculator, the cumulative probability up to z = -1.95 is approximately 0.0256. Therefore, the probability that the sample mean is greater than 2 is: \[ P(Z > -1.95) = 1 - 0.0256 = 0.9744 \] So, the probability is 0.9744. **Part 2: Probability that the sample mean is between 2.5 and 3** Next, I need to find the probability that the sample mean number of TV sets is between 2.5 and 3. Again, I'll use the z-score formula: \[ z_1 = \frac{2.5 - 2.24}{1.2 / \sqrt{95}} = \frac{0.26}{0.123} \approx 2.11 \] \[ z_2 = \frac{3 - 2.24}{1.2 / \sqrt{95}} = \frac{0.76}{0.123} \approx 6.16 \] Now, I need to find \( P(2.11 < Z < 6.16) \). Using standard normal distribution tables: - \( P(Z < 2.11) \approx 0.9826 \) - \( P(Z < 6.16) \approx 1 \) (since 6.16 is far in the right tail) Therefore: \[ P(2.11 < Z < 6.16) = P(Z < 6.16) - P(Z < 2.11) = 1 - 0.9826 = 0.0174 \] So, the probability that the sample mean number of TV sets is between 2.5 and 3 is approximately 0.0174.