A population of values has a normal distribution with \( \mu=133.5 \) and \( \sigma=23.6 \). If a random sample of size \( n=10 \) is selected, a. Find the probability that a single randomly selected value is greater than 140.2. Round your answer to four decimals. \( P(X>140.2)= \) b. Find the probability that a sample of size \( n=10 \) is randomly selected with a mean greater than 140.2 . Round your answer to four decimals. \( P(M>140.2)= \) Check Answer

To find the requested probabilities, we'll use the properties of the normal distribution. **a.** To find \( P(X > 140.2) \), we first calculate the z-score using the formula: \[ z = \frac{X - \mu}{\sigma} = \frac{140.2 - 133.5}{23.6} \approx 0.283 \] Next, we'll look up this z-score in the standard normal distribution table or use a calculator to find the corresponding probability. The area to the left of \( z = 0.283 \) is approximately 0.6112. Therefore, \[ P(X > 140.2) = 1 - P(Z < 0.283) \approx 1 - 0.6112 \approx 0.3888. \] **Answer:** \( P(X>140.2) \approx 0.3888 \) **b.** For the sample mean, we calculate the standard error using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{23.6}{\sqrt{10}} \approx 7.46. \] Now we need to calculate the z-score for the sample mean: \[ z = \frac{M - \mu}{SE} = \frac{140.2 - 133.5}{7.46} \approx 0.892. \] Using the z-score, we find the area to the left of \( z = 0.892 \), which is approximately 0.8146. Thus, \[ P(M > 140.2) = 1 - P(Z < 0.892) \approx 1 - 0.8146 \approx 0.1854. \] **Answer:** \( P(M>140.2) \approx 0.1854 \)