The mean of the following frequency distribut | vedclass

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(C) To find the mean,we first calculate the class mark $(x_i)$ for each interval:$10-20: x_1 = 15$$20-30: x_2 = 25$$30-40: x_3 = 35$$40-50: x_4 = 45$$

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$7$

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(C) To find the mean,we first calculate the class mark $(x_i)$ for each interval:$10-20: x_1 = 15$$20-30: x_2 = 25$$30-40: x_3 = 35$$40-50: x_4 = 45$$50-60: x_5 = 55$$60-70: x_6 = 65$$70-80: x_7 = 75$The sum of frequencies $\sum f_i = 5 + 3 + 4 + f + 2 + 6 + 13 = 33 + f$.The sum of products $\sum f_i x_i = (5 	imes 15) + (3 	imes 25) + (4 	imes 35) + (f 	imes 45) + (2 	imes 55) + (6 	imes 65) + (13 	imes 75)$$= 75 + 75 + 140 + 45f + 110 + 390 + 975 = 1765 + 45f$.Given mean $\bar{x} = 52$,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$:$52 = \frac{1765 + 45f}{33 + f}$$52(33 + f) = 1765 + 45f$$1716 + 52f = 1765 + 45f$$52f - 45f = 1765 - 1716$$7f = 49$$f = 7$.

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$5$	$3$	$4$	$f$	$2$	$6$	$13$

Class	$11-13$	$13-15$	$15-17$	$17-19$	$19-21$	$21-23$	$23-25$
Frequency	$3$	$6$	$9$	$13$	$f$	$5$	$4$

Number of wickets	$20$-$60$	$60$-$100$	$100$-$150$	$150$-$250$	$250$-$350$	$350$-$450$
Number of bowlers	$7$	$5$	$16$	$12$	$2$	$3$

The mean of the following frequency distribution is $52$. Find the missing frequency $f$.

Class $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Frequency $5$ $3$ $4$ $f$ $2$ $6$ $13$

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The mean of the following frequency distribution is $52$. Find the missing frequency $f$. Class $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $5$ $3$ $4$ $f$ $2$ $6$ $13$