Motorcycle Noise \( A \) recent study of 49 motorcycles found that the average noise pollution in decibels is 90 . If the population standard deviation is 4.2 decibels, find the \( 91 \% \) confidence interval for the mean. Round answers to the nearest whole decibel. Round intermediate answers to two decimal places. We are \( 91 \% \) confident that the mean noise pollution level is between

To find the \( 91\% \) confidence interval for the mean noise pollution level of motorcycles, we can use the formula for the confidence interval for the mean when the population standard deviation is known: \[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \(\bar{x}\) is the sample mean, - \(z\) is the z-score corresponding to the desired confidence level, - \(\sigma\) is the population standard deviation, - \(n\) is the sample size. Given: - \(\bar{x} = 90\) decibels, - \(\sigma = 4.2\) decibels, - \(n = 49\). ### Step 1: Find the z-score for \( 91\% \) confidence level To find the z-score for a \( 91\% \) confidence level, we need to find the critical value that leaves \( 4.5\% \) in each tail (since \( 100\% - 91\% = 9\% \) and half of that is \( 4.5\% \)). Using a z-table or calculator, the z-score for \( 0.945 \) (which is \( 1 - 0.045 \)) is approximately \( 1.695 \). ### Step 2: Calculate the standard error The standard error (SE) is calculated as follows: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{4.2}{\sqrt{49}} = \frac{4.2}{7} = 0.6 \] ### Step 3: Calculate the margin of error Now, we can calculate the margin of error (ME): \[ ME = z \cdot SE = 1.695 \cdot 0.6 \approx 1.017 \] ### Step 4: Calculate the confidence interval Now we can calculate the confidence interval: \[ \text{Lower limit} = \bar{x} - ME = 90 - 1.017 \approx 88.98 \quad \text{(round to 89)} \] \[ \text{Upper limit} = \bar{x} + ME = 90 + 1.017 \approx 91.02 \quad \text{(round to 91)} \] ### Conclusion We are \( 91\% \) confident that the mean noise pollution level is between **89** and **91** decibels.