One set containing five numbers has mean 8 an | vedclass

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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}$ = $20.25$.

Classes	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
${f_i}$	$3$	$7$	$12$	$15$	$8$	$3$	$2$

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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