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Find out the missing frequencies of the following series if the AM is 35 and the number of items is 100:
Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
Frequency | 5 | 10 | ? | 4 | 20 | 3 | ? |
Given :
A.M. of the series $= 35$.
The total number of items $= 100$.
To do :
We have to find the missing frequencies.
Solution :
Let the missing frequencies be $f_1$ for the class interval 20-30 and $f_2$ for the class interval 60-70.
This implies,
$5+10+f_1+4+20+3+f_2=100$
$f_1+f_2=100-42$
$f_1+f_2=58$
$\frac{\sum f_{i}x_{i}}{\sum f_{i}} = 100$
$\frac{5 \times 5 + 15 \times 10 + 25 \times f_1 + 35 \times 4 + 45 \times 20 + 55 \times 3 + 65 \times f_2}{100} = 35$
$\frac{25 + 150 + 140 + 900 + 165 + 25f_1 + 65f_2}{100} = 35$
$1380 + 25f_1 + 65f_2 = 35\times 100$
$ 25f_1 + 65f_2 = 3500 - 1380$
$5 (5f_1 + 13f_2) = 2120$
$5f_1 + 13f_2 = \frac{2120}{5} = 424$
$5f_1 + 13f_2 =424$......................................(i)
$f_1 + f_2 = 58$
Multiply 5 on both sides,
$5f_1 + 5f_2 = 58 \times 5$
$5f_1 + 5f_2 = 290$......................................(ii)
Subtract (ii) from (i)
$5f_1 + 13f_2 =424$ $(-)$
$5f_1 + 5f_2 = 290$
________________
$8f_2 = 134$
$f_2 = \frac{134}{8} \approxeq 17$
$f_1 + 17 = 58$
$f_1 = 58 - 17 = 41$
Therefore, the missing frequencies are 41 and 17 respectively.