Let M denote the median of the following freq | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$208$

Vedclass · Answer

(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 $3$ $9$ $10$ $8$ $6$

Similar Questions

In a class of $100$ students,there are $70$ boys whose average marks in a subject are $75$. If the average marks of the complete class are $72$,then what are the average marks of the girls?

If the mean of the numbers $27, 31, 89, 107, 156$ is $82$,what is the mean of the numbers $130, 126, 68, 50, 1$?

The mean of the series $a, a + nd, a + 2nd$ is

The median of $10, 14, 11, 9, 8, 12, 6$ is

The mean of a set of numbers is $\bar x$. If $\lambda$ is subtracted from each number,what will be the mean of the new set?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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