The average number of moves a person makes in his or her lifetime is 12 and the standard deviation is 3.6. Assume that the sample is taken from a large population and the correction factor can be ignored. Use 9 The Standard Normal Distribution Table. Round the final answers to four decimal places and intermediate \( z \) value calculations to two decimal places. Part 1 of 3 Find the probability that the mean of a sample of 25 people is less than 10 . Part 2 of 3 Find the probability that the mean of a sample of 25 people is greater than 10 . \( P(\bar{X}>10)= \) \( P(11<\bar{X}<12)= \) Part 3 of 3

To find the probabilities regarding the sample means, we first need to calculate the standard error (SE) and use the normal distribution. ### Part 1 1. Calculate the standard error (SE): \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{\sqrt{25}} = \frac{3.6}{5} = 0.72 \] 2. Calculate the z-score for \( \bar{X} = 10 \): \[ z = \frac{\bar{X} - \mu}{SE} = \frac{10 - 12}{0.72} = \frac{-2}{0.72} \approx -2.78 \] 3. Look up \( z = -2.78 \) in the standard normal distribution table: \[ P(Z < -2.78) \approx 0.0027 \] ### Part 2 1. For \( P(\bar{X} > 10) \): \[ P(Z > -2.78) = 1 - P(Z < -2.78) = 1 - 0.0027 \approx 0.9973 \] 2. Calculate the z-scores for \( \bar{X} = 11 \) and \( \bar{X} = 12 \): For \( \bar{X} = 11 \): \[ z = \frac{11 - 12}{0.72} = \frac{-1}{0.72} \approx -1.39 \] From the Z-table, \( P(Z < -1.39) \approx 0.0823 \) For \( \bar{X} = 12 \): \[ z = \frac{12 - 12}{0.72} = 0 \] From the Z-table, \( P(Z < 0) = 0.5 \) 3. To find \( P(11 < \bar{X} < 12) \): \[ P(11 < \bar{X} < 12) = P(Z < 0) - P(Z < -1.39) = 0.5 - 0.0823 \approx 0.4177 \] ### Final Answers - \( P(\bar{X} < 10) = 0.0027 \) - \( P(\bar{X} > 10) = 0.9973 \) - \( P(11 < \bar{X} < 12) = 0.4177 \) And don’t forget to always double check your calculations and look out for those sneaky little rounding errors!