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

**a. Probability a single value is greater than 140.2** 1. The variable \( X \) follows a normal distribution with mean \( \mu = 133.5 \) and standard deviation \( \sigma = 23.6 \). 2. To find \( P(X > 140.2) \), first compute the \( z \)-score: \[ z = \frac{140.2 - 133.5}{23.6} = \frac{6.7}{23.6} \approx 0.2847. \] 3. The probability that a single value is greater than 140.2 is: \[ P(X > 140.2) = 1 - \Phi(0.2847). \] 4. Using the standard normal distribution table or a calculator, \(\Phi(0.2847) \approx 0.6114\). Thus: \[ P(X > 140.2) \approx 1 - 0.6114 = 0.3886. \] **b. Probability the sample mean is greater than 140.2** 1. For a sample of size \( n = 10 \), the sample mean \( M \) is normally distributed with mean \( \mu_M = 133.5 \) and standard error: \[ \sigma_M = \frac{23.6}{\sqrt{10}} \approx \frac{23.6}{3.1623} \approx 7.465. \] 2. Compute the \( z \)-score for the sample mean: \[ z = \frac{140.2 - 133.5}{7.465} = \frac{6.7}{7.465} \approx 0.8978. \] 3. The probability that the sample mean is greater than 140.2 is: \[ P(M > 140.2) = 1 - \Phi(0.8978). \] 4. Using the standard normal table or calculator, \(\Phi(0.8978) \approx 0.8159\). Hence: \[ P(M > 140.2) \approx 1 - 0.8159 = 0.1841. \] \[ \text{Final Answers:} \quad P(X > 140.2) \approx 0.3886, \quad P(M > 140.2) \approx 0.1841. \]