The mean of the following frequency distribut | vedclass

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(A) Given total frequency $\sum f_i = 120$. Thus,$17 + f_1 + 32 + f_2 + 19 = 120 \implies f_1 + f_2 = 52$ (Equation $1$).The class marks $(x_i)$ are $

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$f_{1} = 28, f_{2} = 24$

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(A) Given total frequency $\sum f_i = 120$. Thus,$17 + f_1 + 32 + f_2 + 19 = 120 \implies f_1 + f_2 = 52$ (Equation $1$).The class marks $(x_i)$ are $20, 40, 60, 80, 100$. The sum $\sum f_i x_i = (17 	imes 20) + (f_1 	imes 40) + (32 	imes 60) + (f_2 	imes 80) + (19 	imes 100) = 340 + 40f_1 + 1920 + 80f_2 + 1900 = 4160 + 40f_1 + 80f_2$.Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 60 \implies \frac{4160 + 40f_1 + 80f_2}{120} = 60$.$4160 + 40f_1 + 80f_2 = 7200 \implies 40f_1 + 80f_2 = 3040 \implies f_1 + 2f_2 = 76$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(f_1 + 2f_2) - (f_1 + f_2) = 76 - 52 \implies f_2 = 24$.Substituting $f_2 = 24$ in Equation $1$: $f_1 + 24 = 52 \implies f_1 = 28$. Thus,$f_1 = 28$ and $f_2 = 24$.

Class	$10-30$	$30-50$	$50-70$	$70-90$	$90-110$
Frequency	$17$	$f_{1}$	$32$	$f_{2}$	$19$

Class	$0-20, 20-40, 40-60, 60-80, 80-100, 100-120$
Frequency	$10, f, 7, 6, 5, 4$

Mileage $(km/l)$	$10-12$	$12-14$	$14-16$	$16-18$
Number of cars	$7$	$12$	$18$	$13$

Class	$1-3$	$3-5$	$5-7$	$7-9$	$9-11$
Frequency	$6$	$3$	$8$	$2$	$1$

The mean of the following frequency distribution is $60$ and the total frequency is $120$. Find the missing frequencies $f_{1}$ and $f_{2}$.

Class $10-30$ $30-50$ $50-70$ $70-90$ $90-110$

Frequency $17$ $f_{1}$ $32$ $f_{2}$ $19$

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The mean of the following frequency distribution is $60$ and the total frequency is $120$. Find the missing frequencies $f_{1}$ and $f_{2}$. Class $10-30$ $30-50$ $50-70$ $70-90$ $90-110$ Frequency $17$ $f_{1}$ $32$ $f_{2}$ $19$