16. A normal distribution has a mean of \( \mu=58 \) and a standard deviation of \( \sigma=12 \). a. What is the probability of randomly selecting a score less than \( X=52 \) ? b. What is the probability of selecting a sample of \( n=9 \) scores with a mean less than \( M=52 \) ? c. What is the probability of selecting a sample of \( n=16 \) scores with a mean less than \( M=52 \) ?

To find the probabilities associated with your questions, we can use the properties of the normal distribution and z-scores. 1. For part a, you want to find the probability of selecting a score less than \( X = 52 \). We first convert the score into a z-score using the formula: \[ z = \frac{X - \mu}{\sigma} = \frac{52 - 58}{12} = \frac{-6}{12} = -0.5 \] Next, we look up the z-score of -0.5 in the standard normal distribution table or use a calculator. The probability associated with \( z = -0.5 \) is approximately 0.3085. Therefore, \( P(X < 52) \approx 0.3085 \) or 30.85%. 2. For part b, to find the probability of selecting a sample of \( n = 9 \) scores with a mean less than \( M = 52 \), we need to use the standard error of the mean (SEM): \[ SEM = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{9}} = \frac{12}{3} = 4 \] Now, we convert the sample mean \( M = 52 \) into a z-score with the new mean \( \mu = 58 \): \[ z = \frac{M - \mu}{SEM} = \frac{52 - 58}{4} = \frac{-6}{4} = -1.5 \] The probability corresponding to \( z = -1.5 \) is approximately 0.0668. Thus, \( P(M < 52) \approx 0.0668 \) or 6.68%. 3. For part c, using a sample size of \( n = 16 \), we again calculate the SEM: \[ SEM = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{16}} = \frac{12}{4} = 3 \] Again, we convert \( M = 52 \) into a z-score: \[ z = \frac{M - \mu}{SEM} = \frac{52 - 58}{3} = \frac{-6}{3} = -2 \] The probability corresponding to \( z = -2 \) is approximately 0.0228. Therefore, \( P(M < 52) \approx 0.0228 \) or 2.28%. In summary: - \( P(X < 52) \approx 30.85\% \) - \( P(M < 52 \ text{ for } n=9) \approx 6.68\% \) - \( P(M < 52 \ text{ for } n=16) \approx 2.28\% \)