Questions: Noise levels at 3 concerts were measured in decibels yielding the following data: 124,134,134 Construct the 98% confidence interval for the mean noise level at such locations. Assume the population is approximately normal. Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place.

Solution Steps

The mean \( \mu \) of the noise levels is calculated as follows:

\[ \mu = \frac{\sum x_i}{n} = \frac{124 + 134 + 134}{3} = \frac{392}{3} \approx 130.7 \]

The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} \]

First, we find \( (x_i - \mu)^2 \) for each \( x_i \):

• For \( x_1 = 124 \): \( (124 - 130.7)^2 = 44.89 \)
• For \( x_2 = 134 \): \( (134 - 130.7)^2 = 10.89 \)
• For \( x_3 = 134 \): \( (134 - 130.7)^2 = 10.89 \)

Now, summing these values:

\[ \sum (x_i - \mu)^2 = 44.89 + 10.89 + 10.89 = 66.67 \]

Thus, the variance is:

\[ \sigma^2 = \frac{66.67}{3-1} = \frac{66.67}{2} = 33.335 \approx 33.3 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{33.3} \approx 5.8 \]

Final Answer