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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
$42$
$20.25$
$18$
None of these
Solution
(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 \times 8 + 3 \times 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}$ = $20.25$.
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