The mean of the following frequency distribut | vedclass

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(B) To find the mean,we first calculate the class mark $(x_i)$ for each interval: $11-13: x_1 = 12$$13-15: x_2 = 14$$15-17: x_3 = 16$$17-19: x_4 = 18$

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$8$

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(B) To find the mean,we first calculate the class mark $(x_i)$ for each interval: $11-13: x_1 = 12$$13-15: x_2 = 14$$15-17: x_3 = 16$$17-19: x_4 = 18$$19-21: x_5 = 20$$21-23: x_6 = 22$$23-25: x_7 = 24$Now,calculate $f_i x_i$:$3 	imes 12 = 36$$6 	imes 14 = 84$$9 	imes 16 = 144$$13 	imes 18 = 234$$f 	imes 20 = 20f$$5 	imes 22 = 110$$4 	imes 24 = 96$Sum of frequencies $\sum f_i = 3+6+9+13+f+5+4 = 40+f$Sum of products $\sum f_i x_i = 36+84+144+234+20f+110+96 = 704+20f$Given mean $\bar{x} = 18$,we use the formula $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$:$18 = \frac{704+20f}{40+f}$$18(40+f) = 704+20f$$720+18f = 704+20f$$720-704 = 20f-18f$$16 = 2f$$f = 8$

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

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$6$	$x$	$17$	$y$	$8$

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

Class	$0-100$	$100-200$	$200-300$	$300-400$	$400-500$	$500-600$
Frequency	$12$	$18$	$27$	$20$	$17$	$6$

The mean of the following frequency distribution is $18$. Find the missing frequency $f$.

Class $11-13$ $13-15$ $15-17$ $17-19$ $19-21$ $21-23$ $23-25$

Frequency $3$ $6$ $9$ $13$ $f$ $5$ $4$

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The mean of the following frequency distribution is $18$. Find the missing frequency $f$. Class $11-13$ $13-15$ $15-17$ $17-19$ $19-21$ $21-23$ $23-25$ Frequency $3$ $6$ $9$ $13$ $f$ $5$ $4$