For the given frequency table the arithmetic mean was found to be nearly Rs $28.07.$
Income Rs No of workers
15 8
20 12
25 f1
30 16
35 f2
40 10
If the total number of workers is $75$ , then the missing frequencies.

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Solution

$\sum f i = 8 + 12 + f_{1} + 16 + f_{2} + 10 = 46 + f_{1} + f_{2} = 75$
So, $f_{1} + f_{2} = 29 ⟶ (I)$
Now, $¯ ¯ ¯ x = 28.07 = \frac{\sum f i x i}{\sum f i}$
$¯ ¯ ¯ x = \frac{(15 \times 8) + (20 \times 12) + (25 f_{1}) + (30 \times 16) + (35 f_{2}) + (40 \times 10)}{75}$
$120 + 240 + 25 f_{1} + 480 + 35 f_{2} + 400 = 2105.25$
${5 f}_{1} + {7 f}_{2} = 173.05 ⟶ (I I)$
On solving $(I)$ and $(I I)$ , we get
$f_{1} = 14.975 \approx 15$ and $f_{2} = 14.025 \approx 14$

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Similar questions

The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers	A	B	C	D	E	F	G	H	I	J
Daily Income (in Rs)	120	150	180	200	250	300	220	350	370	260

Weekly wages (Rs.)	40-43	43-46	46-49	49-52	52-55
Number of workers:	31	58	60	?	27

50.

120.

f_{1}, f_{2}

\begin{matrix} Class & 0 - 20 & 20 - 40 & 40 - 60 & 60 - 80 & 80 - 100 Frequency & 17 & f_{1} & 32 & f_{2} & 19 \end{matrix}

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