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(C) The total frequency is given as $120$. So,$17 + f_1 + 32 + f_2 + 19 = 120$.$f_1 + f_2 + 68 = 120 \implies f_1 + f_2 = 52$ (Equation $1$).The class

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$f_1 = 28, f_2 = 24$

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(C) The total frequency is given as $120$. So,$17 + f_1 + 32 + f_2 + 19 = 120$.$f_1 + f_2 + 68 = 120 \implies f_1 + f_2 = 52$ (Equation $1$).The class marks $(x_i)$ are $10, 30, 50, 70, 90$.The mean is $\frac{\sum f_i x_i}{\sum f_i} = 50$.$\frac{17(10) + f_1(30) + 32(50) + f_2(70) + 19(90)}{120} = 50$.$170 + 30f_1 + 1600 + 70f_2 + 1710 = 6000$.$30f_1 + 70f_2 + 3480 = 6000$.$30f_1 + 70f_2 = 2520 \implies 3f_1 + 7f_2 = 252$ (Equation $2$).From Equation $1$,$f_1 = 52 - f_2$. Substituting into Equation $2$:$3(52 - f_2) + 7f_2 = 252$.$156 - 3f_2 + 7f_2 = 252$.$4f_2 = 96 \implies f_2 = 24$.Then $f_1 = 52 - 24 = 28$.

Class	Frequency
$0 - 20$	$17$
$20 - 40$	$f_1$
$40 - 60$	$32$
$60 - 80$	$f_2$
$80 - 100$	$19$
Total	$120$

Set	Size	Mean	Variance
Observation $I$	$10$	$2$	$2$
Observation $II$	$n$	$3$	$1$

If the arithmetic mean of the following frequency distribution is $50$,find the values of $f_1$ and $f_2$.
Class Frequency
$0 - 20$ $17$
$20 - 40$ $f_1$
$40 - 60$ $32$
$60 - 80$ $f_2$
$80 - 100$ $19$
Total $120$

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If the arithmetic mean of the following frequency distribution is $50$,find the values of $f_1$ and $f_2$.ClassFrequency$0 - 20$$17$$20 - 40$$f_1$$40 - 60$$32$$60 - 80$$f_2$$80 - 100$$19$Total$120$