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Question

(b) Here ${n_1} = 5$, ${\bar x_1} = 8$, $\sigma _1^2 = 18$, ${n_2} = 3$ ${\bar x_2} = 8$, $\sigma _2^2 = 24$

$\bar x = $ combined mean $ = \frac{{5

Vedclass · Accepted Answer

$20.25$

Vedclass · Answer

(b) Here ${n_1} = 5$, ${\bar x_1} = 8$, $\sigma _1^2 = 18$, ${n_2} = 3$ ${\bar x_2} = 8$, $\sigma _2^2 = 24$

$\bar x = $ combined mean $ = \frac{{5 	imes 8 + 3 	imes 8}}{{5 + 3}}$ $ = \frac{{64}}{8} = 8$

Combined variance $ = \frac{{{n_1}(\sigma _1^2 + D_1^2) + {n_2}(\sigma _2^2 + D_2^2)}}{{{n_1} + {n_2}}}$,

where ${D_1} = {\bar x_1} - \bar x$, ${D_2} = {\bar x_2} - \bar x$

Now, ${D_1} = 8 - 8;\,\,{D_2} = 8 - 8 = 0$

Combined variance $ = \frac{{5(18) + 3(24)}}{{5 + 3}}$ $ = \frac{{90 + 72}}{8}$ $ = \frac{{162}}{8}$ = $20.25$.

class	0-10	10-20	20-30	30-40
Freq	1	3	4	2

Marks	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
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Group $B$	$10$	$20$	$30$	$25$	$43$	$15$	$7$

$X_i$	$0$	$1$	$2$	$3$	$4$	$5$
$f_i$	$k+2$	$2k$	$K^{2}-1$	$K^{2}-1$	$K^{2}-1$	$k-3$

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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