Ages of Proofreaders At a large publishing company, the mean age of proofreaders is 36.2 years and the standard deviation is 3.7 years. Assume the variable is normally distributed. Use 9 The Standard Normal Distribution Table. Round intermediate \( z \)-value calculations to two decimal places and the final answers to at least four decimal places. Part 1 of 2 If a proofreader from the company is randomly selected, find the probability that his or her age will be between 33.5 and 35 years. \[ P(33.5

Ask by Fuentes Valdez. in the United States

Mar 11,2025

To find the probability that the mean age of a sample of 25 proofreaders will be between 33.5 and 35 years, we first calculate the standard error (SE) of the mean. The formula for the SE is given by \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation (3.7 years) and \( n \) is the sample size (25). Calculating the SE gives: \[ SE = \frac{3.7}{\sqrt{25}} = \frac{3.7}{5} = 0.74. \] Next, we need to convert the ages 33.5 and 35 into z-scores using the mean age of 36.2 along with the calculated SE. The z-scores are computed as follows: For 33.5 years: \[ z_1 = \frac{33.5 - 36.2}{0.74} \approx \frac{-2.7}{0.74} \approx -3.65. \] For 35 years: \[ z_2 = \frac{35 - 36.2}{0.74} \approx \frac{-1.2}{0.74} \approx -1.62. \] Now we can look up these z-scores in the Standard Normal Distribution Table or use a calculator to find: - \( P(Z < -3.65) \approx 0.0001 \). - \( P(Z < -1.62) \approx 0.0526 \). Finally, to find the probability that the sample mean age is between 33.5 and 35 years: \[ P(33.5 < \bar{X} < 35) = P(Z < -1.62) - P(Z < -3.65) \approx 0.0526 - 0.0001 = 0.0525. \] So, rounding to four decimal places, \[ P(33.5 < \bar{X} < 35) \approx 0.0525. \]