Question

noise levels at 5 airports were measured in decibels yielding the following data:132,126,135,147,133construct the 80% confidence interval for the mean noise level at such locations. assume the population is approximately normal.step 2 of 4 : calculate the sample standard deviation for the given sample data. round your answer to one decimal place.

Answer 1

The sample standard deviation noise levels at 5 airports is: 7.7

How to find standard deviation?

To calculate the sample standard deviation (s) for the given data set:

Data: 132, 126, 135, 147, 133

Find the mean (
`\( \bar{x} \)`):

`\[ \bar{x} = (132 + 126 + 135 + 147 + 133)/(5) \]`

`\[ \bar{x} = (673)/(5) \]`

`\[ \bar{x} = 134.6 \]`

Find the squared deviations from the mean for each data point:

`\[ (132 - 134.6)^2 = 6.76 \]`

`\[ (126 - 134.6)^2 = 73.96 \]`

`\[ (135 - 134.6)^2 = 0.16 \]`

`\[ (147 - 134.6)^2 = 154.76 \]`

`\[ (133 - 134.6)^2 = 2.56 \]`

Find the sum of the squared deviations:

`\[ 6.76 + 73.96 + 0.16 + 154.76 + 2.56 = 238.2 \]`

Divide the sum by n-1 (where n = number of data points):

`\[ s^2 = (238.2)/(5-1) = (238.2)/(4) = 59.55 \]`

Take the square root to find the sample standard deviation:

`\[ s = √(59.55) \approx 7.718 \]`

So, the sample standard deviation for the given data set is approximately 7.7 (rounded to one decimal place).