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(B) Given for the first set: $n_1 = 5$,$\bar{x}_1 = 8$,$\sigma_1^2 = 18$.Given for the second set: $n_2 = 3$,$\bar{x}_2 = 8$,$\sigma_2^2 = 24$.Combine

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$20.25$

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(B) Given for the first set: $n_1 = 5$,$\bar{x}_1 = 8$,$\sigma_1^2 = 18$.Given for the second set: $n_2 = 3$,$\bar{x}_2 = 8$,$\sigma_2^2 = 24$.Combined mean $\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{5 	imes 8 + 3 	imes 8}{5 + 3} = \frac{64}{8} = 8$.Since $\bar{x}_1 = \bar{x}_2 = \bar{x} = 8$,the deviations $D_1 = \bar{x}_1 - \bar{x} = 0$ and $D_2 = \bar{x}_2 - \bar{x} = 0$.The combined variance is given by $\sigma^2 = \frac{n_1(\sigma_1^2 + D_1^2) + n_2(\sigma_2^2 + D_2^2)}{n_1 + n_2}$.Substituting the values: $\sigma^2 = \frac{5(18 + 0^2) + 3(24 + 0^2)}{5 + 3} = \frac{90 + 72}{8} = \frac{162}{8} = 20.25$.

Class:	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq:	$\alpha$	$110$	$54$	$30$	$\beta$

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

One set containing five numbers has mean $8$ and variance $18$,and the second set containing $3$ numbers has mean $8$ and variance $24$. Then the variance of the combined set of numbers is

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