A sample of 35 different payroll departments found that employees worked an average of 240.6 days a year. If the population standard deviation is 18.8 days, find the \( 90 \% \) confidence interval for the average number of days \( \mu \) worked by all employees who are paid through payroll departments. \( 230.9<\mu<250.3 \) \( 232.4<\mu<248.8 \) \( 236.8<\mu<244.4 \) \( 235.4<\mu<245.8 \)

To calculate the \(90\%\) confidence interval for the average number of days \( \mu \) worked by employees, we can use the formula for the confidence interval of the mean when the population standard deviation is known. The confidence interval is given by: \[ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \( \bar{x} \) is the sample mean (240.6 days) - \( z \) is the z-score corresponding to the desired confidence level (\( z \approx 1.645 \) for \(90\%\)) - \( \sigma \) is the population standard deviation (18.8 days) - \( n \) is the sample size (35) Calculating the standard error: \[ SE = \frac{18.8}{\sqrt{35}} \approx 3.17 \] Now we compute the margin of error: \[ ME = 1.645 \times 3.17 \approx 5.22 \] Finally, we can find the confidence interval: \[ 240.6 \pm 5.22 \implies (235.38, 245.82) \] Thus, rounding to one decimal place, the \(90\%\) confidence interval is approximately: \[ (235.4 < \mu < 245.8) \] So, the correct answer is \( 235.4 < \mu < 245.8 \).