Let M denote the median of the following freq | vedclass

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(D) The total frequency $N = 3 + 9 + 10 + 8 + 6 = 36$. 						Class			Frequency			Cumulative Frequency							$0-4$			$3$			$3$							$4-8$			$9$			$12$

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$208$

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(D) The total frequency $N = 3 + 9 + 10 + 8 + 6 = 36$. 						Class			Frequency			Cumulative Frequency							$0-4$			$3$			$3$							$4-8$			$9$			$12$							$8-12$			$10$			$22$							$12-16$			$8$			$30$							$16-20$			$6$			$36$			Since $\frac{N}{2} = \frac{36}{2} = 18$,the median class is $8-12$. The median formula is $M = l + \left( \frac{\frac{N}{2} - C}{f} ight) 	imes h$,where $l = 8$,$C = 12$,$f = 10$,and $h = 4$. $M = 8 + \left( \frac{18 - 12}{10} ight) 	imes 4 = 8 + \left( \frac{6}{10} ight) 	imes 4 = 8 + 2.4 = 10.4$. Therefore,$20M = 20 	imes 10.4 = 208$.

Let $M$ denote the median of the following frequency distribution. Then $20M$ is equal to:

Class $0-4$ $4-8$ $8-12$ $12-16$ $16-20$

Frequency $3$ $9$ $10$ $8$ $6$

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Class	$0-4$	$4-8$	$8-12$	$12-16$	$16-20$
Frequency	$3$	$9$	$10$	$8$	$6$

Let $M$ denote the median of the following frequency distribution. Then $20M$ is equal to: Class $0-4$ $4-8$ $8-12$ $12-16$ $16-20$ Frequency $3$ $9$ $10$ $8$ $6$