Answer :
To find the variance of the given data set, follow these steps:
1. List the Data and the Mean:
The data values are: 198, 190, 245, 211, 193, 193.
The mean ([tex]\(\bar{x}\)[/tex]) is given as 205.
2. Calculate the Deviations from the Mean:
For each data point, subtract the mean and find the deviation:
- [tex]\(198 - 205 = -7\)[/tex]
- [tex]\(190 - 205 = -15\)[/tex]
- [tex]\(245 - 205 = 40\)[/tex]
- [tex]\(211 - 205 = 6\)[/tex]
- [tex]\(193 - 205 = -12\)[/tex]
- [tex]\(193 - 205 = -12\)[/tex]
3. Square the Deviations:
- [tex]\((-7)^2 = 49\)[/tex]
- [tex]\((-15)^2 = 225\)[/tex]
- [tex]\(40^2 = 1600\)[/tex]
- [tex]\(6^2 = 36\)[/tex]
- [tex]\((-12)^2 = 144\)[/tex]
- [tex]\((-12)^2 = 144\)[/tex]
4. Sum of Squared Deviations:
Add up all the squared deviations:
[tex]\[
49 + 225 + 1600 + 36 + 144 + 144 = 2198
\][/tex]
5. Calculate the Variance:
Divide the sum of squared deviations by the number of data points (6 in this case):
[tex]\[
\text{Variance} = \frac{2198}{6} \approx 366.33
\][/tex]
Therefore, the variance of the data is approximately [tex]\(366.33\)[/tex].
1. List the Data and the Mean:
The data values are: 198, 190, 245, 211, 193, 193.
The mean ([tex]\(\bar{x}\)[/tex]) is given as 205.
2. Calculate the Deviations from the Mean:
For each data point, subtract the mean and find the deviation:
- [tex]\(198 - 205 = -7\)[/tex]
- [tex]\(190 - 205 = -15\)[/tex]
- [tex]\(245 - 205 = 40\)[/tex]
- [tex]\(211 - 205 = 6\)[/tex]
- [tex]\(193 - 205 = -12\)[/tex]
- [tex]\(193 - 205 = -12\)[/tex]
3. Square the Deviations:
- [tex]\((-7)^2 = 49\)[/tex]
- [tex]\((-15)^2 = 225\)[/tex]
- [tex]\(40^2 = 1600\)[/tex]
- [tex]\(6^2 = 36\)[/tex]
- [tex]\((-12)^2 = 144\)[/tex]
- [tex]\((-12)^2 = 144\)[/tex]
4. Sum of Squared Deviations:
Add up all the squared deviations:
[tex]\[
49 + 225 + 1600 + 36 + 144 + 144 = 2198
\][/tex]
5. Calculate the Variance:
Divide the sum of squared deviations by the number of data points (6 in this case):
[tex]\[
\text{Variance} = \frac{2198}{6} \approx 366.33
\][/tex]
Therefore, the variance of the data is approximately [tex]\(366.33\)[/tex].
