Answer :
To find the variance, we follow these steps.
1. Compute the difference between each data point and the mean, then square the difference. Given the data points and the mean $ \bar{x} = 205 $, we have:
- For $x_1 = 198$:
$$
(198 - 205)^2 = (-7)^2 = 49
$$
- For $x_2 = 190$:
$$
(190 - 205)^2 = (-15)^2 = 225
$$
- For $x_3 = 245$:
$$
(245 - 205)^2 = (40)^2 = 1600
$$
- For $x_4 = 211$:
$$
(211 - 205)^2 = (6)^2 = 36
$$
- For $x_5 = 193$:
$$
(193 - 205)^2 = (-12)^2 = 144
$$
- For $x_6 = 193$:
$$
(193 - 205)^2 = (-12)^2 = 144
$$
2. Sum the squared differences:
$$
49 + 225 + 1600 + 36 + 144 + 144 = 2198
$$
3. Since we are finding the population variance, we divide the sum by the number of data points $N = 6$:
$$
\sigma^2 = \frac{2198}{6} \approx 366.3333\ldots
$$
4. Rounding to the nearest tenth, the variance is:
$$
\sigma^2 \approx 366.3
$$
Thus, the variance of the data is $\boxed{366.3}$.
1. Compute the difference between each data point and the mean, then square the difference. Given the data points and the mean $ \bar{x} = 205 $, we have:
- For $x_1 = 198$:
$$
(198 - 205)^2 = (-7)^2 = 49
$$
- For $x_2 = 190$:
$$
(190 - 205)^2 = (-15)^2 = 225
$$
- For $x_3 = 245$:
$$
(245 - 205)^2 = (40)^2 = 1600
$$
- For $x_4 = 211$:
$$
(211 - 205)^2 = (6)^2 = 36
$$
- For $x_5 = 193$:
$$
(193 - 205)^2 = (-12)^2 = 144
$$
- For $x_6 = 193$:
$$
(193 - 205)^2 = (-12)^2 = 144
$$
2. Sum the squared differences:
$$
49 + 225 + 1600 + 36 + 144 + 144 = 2198
$$
3. Since we are finding the population variance, we divide the sum by the number of data points $N = 6$:
$$
\sigma^2 = \frac{2198}{6} \approx 366.3333\ldots
$$
4. Rounding to the nearest tenth, the variance is:
$$
\sigma^2 \approx 366.3
$$
Thus, the variance of the data is $\boxed{366.3}$.
