Q.
The mean of following frequency table is $50$.
Class
Frequency
$0 -20$
$17$
$20-40$
$f_1$
$40-60$
$32$
$60-80$
$f_2$
$80 - 100$
$19$
Total
$120$
The missing frequencies are
| Class | Frequency |
|---|---|
| $0 -20$ | $17$ |
| $20-40$ | $f_1$ |
| $40-60$ | $32$ |
| $60-80$ | $f_2$ |
| $80 - 100$ | $19$ |
| Total | $120$ |
Statistics
Solution:
We have,
Mid point
$(x)$
Frequency
$(f)$
$fx$
$10$
$17$
$170$
$30$
$f_1$
$30\,f_1$
$50$
$32$
$1600$
$70$
$f_2$
$70\,f_2$
$90$
$19$
$1710$
Total
$120$
$30\,f_1+70\,f_2+3480$
Now, $\bar{x} = \frac{1}{120}\,\Sigma fx$
$\Rightarrow 50 = \frac{1}{120} \times \left(30\,f_{1}+70\,f_{2}+3480\right)$
$\Rightarrow 600 = 3\,f_{1}+7\,f_{2}+348$
$\Rightarrow 3\,f_{1}+7\,f_{2}=252\quad\ldots\left(i\right)$
Also, $f_{1} + f_{2} + 68 = 120$
$\Rightarrow f_{1} + f_{2} = 52\quad\ldots\left(ii\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $f_{1} = 28$, $f_{2} = 24$
| Mid point $(x)$ |
Frequency $(f)$ |
$fx$ |
|---|---|---|
| $10$ | $17$ | $170$ |
| $30$ | $f_1$ | $30\,f_1$ |
| $50$ | $32$ | $1600$ |
| $70$ | $f_2$ | $70\,f_2$ |
| $90$ | $19$ | $1710$ |
| Total | $120$ | $30\,f_1+70\,f_2+3480$ |