In a family, there are $8$ men, $7$ women and $5$ children whose mean ages separately are respectively $24, 20$ and $6$ years. The mean age of the family is

17.1

years

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18.1

years

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19.1

years

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none of these

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Solution

The correct option is B $18.1$ years
Here we have three collections for which $A_{1} = 24, n_{1} = 8; A_{2} = 20, n_{2} = 7$ and $A_{3} = 6, n_{3} = 5$ .
Their combined mean is the required mean.
By the formula $A = \frac{n_{1} A_{1} + n_{2} A_{2} + n_{3} A_{3}}{n_{1} + n_{2} + n_{3}} = \frac{8 \times 24 + 7 \times 20 + 5 \times 6}{8 + 7 + 5}$
$= \frac{192 + 140 + 30}{20} = \frac{362}{20} = 18.1$
$∴$ The mean age of the family $= 18.1$ years

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In a family, there are 8 men, 7 women and 5 children whose mean ages separately are respectively 24,20 and 6 years. The mean age of the family is

The correct option is B 18.1 yearsHere we have three collections for which A1=24,n1=8;A2=20,n2=7 and A3=6,n3=5.Their combined mean is the required mean.By the formula A=n1A1+n2A2+n3A3n1+n2+n3=8×24+7×20+5×68+7+5=192+140+3020=36220=18.1∴ The mean age of the family =18.1 years

In a family, there are $8$ men, $7$ women and $5$ children whose mean ages separately are respectively $24, 20$ and $6$ years. The mean age of the family is