MCQ
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
- A$42$
- ✓$20.25$
- C$18$
- DNone of these
$\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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