Question

Find the missing frequencies $f_1$ and $f_2$, if the mean of the frequency table is $62.8$ and the sum of all frequencies is $50$

Class	Frequency
0-20	5
20-40	$f_1$
40-60	10
60-80	$f_2$
80-100	7
100-120	8
Total	50

• A. $f_1=12,f_2=8$
• B. $f_1=18,f_2=2$
• C. $f_1=6,f_2=6$
• D. $f_1=8,f_2=12$

Answer 1

Answer: (D)

$f_1=8,f_2=12$

$ci$	$f_i$	$x_i$	$f_i{x}_i$
$0-20$	$5$	$10$	$50$
$20-40$	$f_1$	$30$	$30f_1$
$40-60$	$10$	$50$	$500$
$60-80$	$f_2$	$70$	$70f_2$
$80-100$	$7$	$90$	$630$
$100-120$	$8$	$110$	$880$

$N=\Sigma f_i=50$ ...(given)

$N=\Sigma f_i=30+f_1+f_2$

$\Rightarrow 50=30+f_1+f_2$

$\Rightarrow f_1=20-f_2$ ...(1)

$\Sigma f_i{x}_i=2060+30f_1+70f_2$

Mean $\overline{x}=\frac{\Sigma f_i{x}_i}{N}$

$\therefore 62.8=\frac{2060+30f_1+70f_2}{50}$

$\therefore 3140=2060+30(20-f_2)+70f_2$ ...from(1)

$\therefore 3140=2060+600-30f_2+70f_2$

$\therefore 3140=2660+40f_2$

$\therefore f_2=12$

substituting $f_2$ in (1), we get

$f_1=20-12$

$\therefore f_1=8$

Hence, option $D$ is correct.