Find the mean of the following frequency distribution is $57.6$ and the total number of observations is $50$ . Find the missing frequencies $f_{1}$ and $f_{2}$ .
Class $0 - 20$ $20 - 40$ $40 - 60$ $60 - 80$ $80 - 100$ $100 - 120$ Total
Frequency $7$ $f_{1}$ $12$ $f_{2}$ $8$ $5$ $50$

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Solution

Step 1 Finding mean:
The given table can be completed as below.
Class Frequency $(f_{i})$ Mid values $(x_{i})$ $(f_{i} x_{i})$
$0 - 20$ $7$ $10$ $70$
$20 - 40$ $f_{1}$ $30$ $30 f_{1}$
$40 - 60$ $12$ $50$ $600$
$60 - 80$ $f_{2}$ $70$ $1260 - 70 f_{1}$
$80 - 100$ $8$ $90$ $720$
$100 - 120$ $5$ $110$ $550$
Total $\sum_{} f_{i} = 50$ $\sum_{} (f_{i} x_{i}) = 3200 - 40 f_{1}$
The formula for mean is $M e a n (\bar{x}) = \frac{\sum_{} (f_{i} x_{i})}{\sum_{} f_{i}}$ .
where $\sum_{} f_{i}$ is the total frequency and $x_{i}$ is the mid-values of the class-intervals.
Mean is given as $57.6$ .
The sum of frequencies is given as $50$ . So,
$\Rightarrow 7 + f_{1} + 12 + f_{2} + 8 + 5 = 50 \Rightarrow f_{1} + f_{2} = 18 \Rightarrow f_{2} = 18 - f_{1}$
Step 2 Finding values of $f_{1}$ and $f_{2}$ :
Apply the mean formula and substitute values,
$\Rightarrow 57.6 = \frac{3200 - 40 f_{1}}{50} \Rightarrow 40 f_{1} = 320 ∴ f_{1} = 8$
Find the value of $f_{2}$ by putting the value of $f_{1}$ in $f_{2} = 18 - f_{1}$ .
$\Rightarrow f_{2} = 18 - f_{1} ∴ f_{2} = 10$
Hence, the missing frequencies are $f_{1} = 8$ and $f_{2} = 10$ .

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Class	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	$100 - 120$	Total
Frequency	$7$	$f_{1}$	$12$	$f_{2}$	$8$	$5$	$50$

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$0 - 20$	$7$	$10$	$70$
$20 - 40$	$f_{1}$	$30$	$30 f_{1}$
$40 - 60$	$12$	$50$	$600$
$60 - 80$	$f_{2}$	$70$	$1260 - 70 f_{1}$
$80 - 100$	$8$	$90$	$720$
$100 - 120$	$5$	$110$	$550$
Total	$\sum_{} f_{i} = 50$		$\sum_{} (f_{i} x_{i}) = 3200 - 40 f_{1}$

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Find the mean of the following frequency distribution is $57.6$ and the total number of observations is $50$ . Find the missing frequencies $f_{1}$ and $f_{2}$ .
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