The mean of the following data is 55.5 and th | vedclass

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(A) Given total frequency $N = 20$.Sum of frequencies: $3 + x + 7 + 7 + y = 20 \implies x + y + 17 = 20 \implies x + y = 3$ (Equation $1$).The class m

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$x=2, y=1$

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(A) Given total frequency $N = 20$.Sum of frequencies: $3 + x + 7 + 7 + y = 20 \implies x + y + 17 = 20 \implies x + y = 3$ (Equation $1$).The class marks $(x_i)$ are $35, 45, 55, 65, 75$.The sum of $f_i x_i = (3 	imes 35) + (x 	imes 45) + (7 	imes 55) + (7 	imes 65) + (y 	imes 75) = 105 + 45x + 385 + 455 + 75y = 945 + 45x + 75y$.Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = 55.5$.$\frac{945 + 45x + 75y}{20} = 55.5 \implies 945 + 45x + 75y = 1110 \implies 45x + 75y = 165$.Dividing by $15$: $3x + 5y = 11$ (Equation $2$).From Equation $1$,$x = 3 - y$. Substituting into Equation $2$: $3(3 - y) + 5y = 11 \implies 9 - 3y + 5y = 11 \implies 2y = 2 \implies y = 1$.Then $x = 3 - 1 = 2$.Thus,$x = 2$ and $y = 1$.

Class	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$3$	$x$	$7$	$7$	$y$

Marks	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Number of students	$2$	$4$	$7$	$6$	$1$

Class	$4-7$	$8-11$	$12-15$	$16-19$
Frequency	$5$	$4$	$9$	$10$

Number of pages written per day	$16-18$	$19-21$	$22-24$	$25-27$	$28-30$
Number of days	$1$	$3$	$4$	$9$	$13$

The mean of the following data is $55.5$ and the total frequency is $20$. Find the missing frequencies $x$ and $y$.
Class $30-40$ $40-50$ $50-60$ $60-70$ $70-80$
Frequency $3$ $x$ $7$ $7$ $y$

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The mean of the following data is $55.5$ and the total frequency is $20$. Find the missing frequencies $x$ and $y$.Class$30-40$$40-50$$50-60$$60-70$$70-80$Frequency$3$$x$$7$$7$$y$