A population of values has a normal distribution with \( \mu=54 \) and \( \sigma=79.5 \). If a random sample of size \( n=22 \) is selected, a. Find the probability that a single randomly selected value is greater than 103.2. Round your answer to four decimals. \( P(X>103.2)= \) b. Find the probability that a sample of size \( n=22 \) is randomly selected with a mean greater than 103.2. Round your answer to four decimals. \( P(M>103.2)= \)

To solve these problems, we need to understand the concepts of the normal distribution and the z-score. For (a): 1. First, we'll calculate the z-score for a single value of 103.2 using the formula: \[ z = \frac{X - \mu}{\sigma} = \frac{103.2 - 54}{79.5} = \frac{49.2}{79.5} \approx 0.6185 \] 2. Next, we can look up this z-score in a standard normal distribution table or use a calculator to find the corresponding probability. The probability that \( Z > 0.6185 \) is: \[ P(Z > 0.6185) = 1 - P(Z < 0.6185) \approx 1 - 0.7325 \approx 0.2675 \] So, \( P(X > 103.2) \approx 0.2675 \) (rounded to four decimals). For (b): 1. When looking for the probability that the sample mean is greater than 103.2, we use the standard error (SE) of the mean, which is calculated as: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{79.5}{\sqrt{22}} \approx \frac{79.5}{4.6904} \approx 16.93 \] 2. Then, we will find the z-score for the sample mean: \[ z = \frac{M - \mu}{SE} = \frac{103.2 - 54}{16.93} = \frac{49.2}{16.93} \approx 2.906 \] 3. Using the z-score, we look up 2.906 in a z-table or use a calculator: \[ P(Z > 2.906) = 1 - P(Z < 2.906) \approx 1 - 0.9982 \approx 0.0018 \] Thus, \( P(M > 103.2) \approx 0.0018 \) (rounded to four decimals). Final answers: a. \( P(X > 103.2) \approx 0.2675 \) b. \( P(M > 103.2) \approx 0.0018 \)