A set of four observations has mean 1 and var | vedclass

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Question

(C) For the first set of observations: $n_1 = 4$,$\bar{x}_1 = 1$,and $\sigma_1^2 = 13$.Using the formula $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2$

Vedclass · Accepted Answer

$6.04$

Vedclass · Answer

(C) For the first set of observations: $n_1 = 4$,$\bar{x}_1 = 1$,and $\sigma_1^2 = 13$.Using the formula $\sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2$,we have $\frac{\sum x_1^2}{4} - (1)^2 = 13$,which gives $\sum x_1^2 = 4(14) = 56$.For the second set of observations: $n_2 = 6$,$\bar{x}_2 = 2$,and $\sigma_2^2 = 1$.Similarly,$\frac{\sum x_2^2}{6} - (2)^2 = 1$,which gives $\sum x_2^2 = 6(5) = 30$.Now,the combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{4(1) + 6(2)}{4 + 6} = \frac{16}{10} = 1.6$.The combined variance $\sigma^2 = \frac{\sum (x_1^2 + x_2^2)}{n_1 + n_2} - (\bar{x})^2 = \frac{56 + 30}{10} - (1.6)^2 = 8.6 - 2.56 = 6.04$.

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

$A$ 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:

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