A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to :

To find the variance of the combined set of observations, we use the formula for combined variance.

$$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2}$$

$$\bar{x} = \frac{4(1) + 6(2)}{4 + 6} = \frac{4 + 12}{10} = 1.6$$

We need the squared difference between each set's mean and the combined mean:

• $$d_1^2 = (\bar{x}_1 - \bar{x})^2 = (1 - 1.6)^2 = (-0.6)^2 = 0.36$$
• $$d_2^2 = (\bar{x}_2 - \bar{x})^2 = (2 - 1.6)^2 = (0.4)^2 = 0.16$$

The formula for combined variance is:

$$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$$

Substitute the values:

$$\sigma^2 = \frac{4(13 + 0.36) + 6(1 + 0.16)}{10}$$

$$\sigma^2 = \frac{4(13.36) + 6(1.16)}{10}$$

$$\sigma^2 = \frac{53.44 + 6.96}{10}$$

$$\sigma^2 = \frac{60.4}{10} = 6.04$$

Correct Option: C (6.04)