Answer :
To calculate the sample standard deviation for the given data set of body temperatures, follow these steps:
1. List the Data Set:
The body temperatures are as follows:
98.1, 98.0, 98.4, 97.2, 99.2, 97.7, 98.2, 96.5, 97.1, 97.9, 96.6, 97.8, 97.8, 99.2, 97.0, 97.9, 97.1, 99.2, 97.1, 97.7.
2. Calculate the Mean:
To find the mean, add all the temperatures together and divide by the number of values.
[tex]\[
\text{Mean} = \frac{98.1 + 98.0 + 98.4 + \ldots + 97.7}{20} = 97.785
\][/tex]
3. Calculate Each Deviation:
Subtract the mean from each data point to find the deviation of each.
4. Square Each Deviation:
After finding each deviation, square these values.
5. Calculate the Variance:
Add up all the squared deviations and divide by the number of data points minus one (because it's a sample, not a population):
[tex]\[
\text{Variance} = \frac{\sum (\text{Each value} - \text{Mean})^2}{n - 1} = 0.6403
\][/tex]
6. Calculate the Standard Deviation:
The sample standard deviation is the square root of the variance:
[tex]\[
\text{Standard Deviation} = \sqrt{0.6403} \approx 0.800
\][/tex]
Therefore, the sample standard deviation for the body temperatures is approximately 0.800.
1. List the Data Set:
The body temperatures are as follows:
98.1, 98.0, 98.4, 97.2, 99.2, 97.7, 98.2, 96.5, 97.1, 97.9, 96.6, 97.8, 97.8, 99.2, 97.0, 97.9, 97.1, 99.2, 97.1, 97.7.
2. Calculate the Mean:
To find the mean, add all the temperatures together and divide by the number of values.
[tex]\[
\text{Mean} = \frac{98.1 + 98.0 + 98.4 + \ldots + 97.7}{20} = 97.785
\][/tex]
3. Calculate Each Deviation:
Subtract the mean from each data point to find the deviation of each.
4. Square Each Deviation:
After finding each deviation, square these values.
5. Calculate the Variance:
Add up all the squared deviations and divide by the number of data points minus one (because it's a sample, not a population):
[tex]\[
\text{Variance} = \frac{\sum (\text{Each value} - \text{Mean})^2}{n - 1} = 0.6403
\][/tex]
6. Calculate the Standard Deviation:
The sample standard deviation is the square root of the variance:
[tex]\[
\text{Standard Deviation} = \sqrt{0.6403} \approx 0.800
\][/tex]
Therefore, the sample standard deviation for the body temperatures is approximately 0.800.
