The mean and variance of the observations x_1 | vedclass

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(D) Let $n_1 = 15$,$\bar{x} = 2$,and $\sigma_x^2 = 4$. The sum of observations is $\sum x_i = n_1 \bar{x} = 15 	imes 2 = 30$. The sum of squares is $

Vedclass · Accepted Answer

$4.4$

Vedclass · Answer

(D) Let $n_1 = 15$,$\bar{x} = 2$,and $\sigma_x^2 = 4$. The sum of observations is $\sum x_i = n_1 \bar{x} = 15 	imes 2 = 30$. The sum of squares is $\sum x_i^2 = n_1(\sigma_x^2 + \bar{x}^2) = 15(4 + 2^2) = 15(8) = 120$.Let $n_2 = 10$,$\bar{y} = 2$,and $\sigma_y^2 = 5$. The sum of observations is $\sum y_i = n_2 \bar{y} = 10 	imes 2 = 20$. The sum of squares is $\sum y_i^2 = n_2(\sigma_y^2 + \bar{y}^2) = 10(5 + 2^2) = 10(9) = 90$.For the combined set of $N = n_1 + n_2 = 25$ observations,the combined mean is $\bar{z} = \frac{\sum x_i + \sum y_i}{n_1 + n_2} = \frac{30 + 20}{25} = \frac{50}{25} = 2$.The combined variance is $\sigma^2 = \frac{\sum x_i^2 + \sum y_i^2}{n_1 + n_2} - \bar{z}^2 = \frac{120 + 90}{25} - 2^2 = \frac{210}{25} - 4 = 8.4 - 4 = 4.4$.

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