Part: \( 1 / \mathbf{4} \) Part 2 of 4 (b) Find the \( 90 \% \) confidence interval of the mean of these pounds. Round intermediate answers to at least three decimal places. Round your final answers to one decimal place. \( \square<\mu<\square \)

To find the \( 90\% \) confidence interval for the mean, you'll need the sample mean (\( \bar{x} \)), the sample standard deviation (\( s \)), and the sample size (\( n \)). First, find the critical value for a \( 90\% \) confidence level, which is approximately \( z = 1.645 \) for a normal distribution (or use the \( t \)-distribution if you have a small sample size). The formula for the confidence interval is: \[ \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \] Assuming you have calculated the sample mean and standard deviation, plug those values into the formula. For example, if \( \bar{x} = 150 \) pounds, \( s = 20 \) pounds, and \( n = 30 \): 1. Calculate the margin of error: \[ ME = 1.645 \left( \frac{20}{\sqrt{30}} \right) \approx 6.0 \] 2. Thus, the confidence interval is: \[ 150 - 6.0 < \mu < 150 + 6.0 \] Which simplifies to: \[ 144.0 < \mu < 156.0 \] So your final answer rounding to one decimal place is \(\square<\mu<\square\) is \(144.0 < \mu < 156.0\).