The mean of the following frequency distribut | vedclass

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(D) Given total frequency $N = 60$.Sum of frequencies: $6 + f_1 + 20 + 15 + f_2 + 4 = 60$$45 + f_1 + f_2 = 60 \implies f_1 + f_2 = 15$  ---(Equation $

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$8, 7$

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(D) Given total frequency $N = 60$.Sum of frequencies: $6 + f_1 + 20 + 15 + f_2 + 4 = 60$$45 + f_1 + f_2 = 60 \implies f_1 + f_2 = 15$  ---(Equation $1$)Calculate midpoint $(x_i)$ and $f_i x_i$:Class $0-10$: $x_i = 5, f_i x_i = 30$Class $10-20$: $x_i = 15, f_i x_i = 15f_1$Class $20-30$: $x_i = 25, f_i x_i = 500$Class $30-40$: $x_i = 35, f_i x_i = 525$Class $40-50$: $x_i = 45, f_i x_i = 45f_2$Class $50-60$: $x_i = 55, f_i x_i = 220$Sum $\sum f_i x_i = 30 + 15f_1 + 500 + 525 + 45f_2 + 220 = 1275 + 15f_1 + 45f_2$Mean $\bar{x} = \frac{\sum f_i x_i}{N} = 28.5$$\frac{1275 + 15f_1 + 45f_2}{60} = 28.5$$1275 + 15f_1 + 45f_2 = 1710$$15f_1 + 45f_2 = 435$Divide by $15$: $f_1 + 3f_2 = 29$ ---(Equation $2$)Subtract Equation $1$ from Equation $2$:$(f_1 + 3f_2) - (f_1 + f_2) = 29 - 15$$2f_2 = 14 \implies f_2 = 7$Substitute $f_2 = 7$ in Equation $1$: $f_1 + 7 = 15 \implies f_1 = 8$.Thus,$f_1 = 8$ and $f_2 = 7$.

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

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$2$	$6$	$8$	$16$	$20$	$18$	$16$	$14$

The mean of the following frequency distribution is $28.5$ and the total frequency is $60$. Find the missing frequencies $f_{1}$ and $f_{2}$.

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$

Frequency $6$ $f_1$ $20$ $15$ $f_2$ $4$

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The mean of the following frequency distribution is $28.5$ and the total frequency is $60$. Find the missing frequencies $f_{1}$ and $f_{2}$. Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ Frequency $6$ $f_1$ $20$ $15$ $f_2$ $4$