If the mean of two samples of sizes 20 and 30 | vedclass

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(D) Given: $n_1 = 20, \bar{x}_1 = 25, \sigma_1^2 = 9$ and $n_2 = 30, \bar{x}_2 = 10, \sigma_2^2 = 16$. Combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n

Vedclass · Accepted Answer

$67.2$

Vedclass · Answer

(D) Given: $n_1 = 20, \bar{x}_1 = 25, \sigma_1^2 = 9$ and $n_2 = 30, \bar{x}_2 = 10, \sigma_2^2 = 16$. Combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{20 	imes 25 + 30 	imes 10}{20 + 30} = \frac{500 + 300}{50} = \frac{800}{50} = 16$. Let $d_1 = \bar{x}_1 - \bar{x} = 25 - 16 = 9$ and $d_2 = \bar{x}_2 - \bar{x} = 10 - 16 = -6$. Combined variance $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$. $\sigma^2 = \frac{20(9 + 9^2) + 30(16 + (-6)^2)}{20 + 30} = \frac{20(9 + 81) + 30(16 + 36)}{50} = \frac{20(90) + 30(52)}{50} = \frac{1800 + 1560}{50} = \frac{3360}{50} = 67.2$.

Marks:	$2$	$3$	$5$	$7$
Frequency:	$(x+1)^2$	$2x-5$	$x^2-3x$	$x$

Items of expenditure	Number of families
$Education$	$150$
$Food and clothing$	$400$
$House rent$	$40$
$Electricity$	$250$
$Miscellaneous$	$160$

If the mean of two samples of sizes $20$ and $30$ are $25$ and $10$ respectively,and their variances are $9$ and $16$ respectively,then their combined variance is:

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