The mean deviation about the median of the di | vedclass

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

Question

(A) Given data in ascending order is $4, 4, 7, 12, 14, 15, 15, 23$. Here,the number of terms $n = 8$,which is even. So,the median is the average of th

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given data in ascending order is $4, 4, 7, 12, 14, 15, 15, 23$. Here,the number of terms $n = 8$,which is even. So,the median is the average of the $4^{th}$ and $5^{th}$ terms. Median $= \frac{12 + 14}{2} = \frac{26}{2} = 13$. Now,we calculate the mean deviation about the median using the formula $\frac{\sum |x_i - 	ext{Median}|}{n}$. $x_i$$|x_i - 13|$$4$$9$$4$$9$$7$$6$$12$$1$$14$$1$$15$$2$$15$$2$$23$$10$ Sum of absolute deviations $\sum |d_i| = 9 + 9 + 6 + 1 + 1 + 2 + 2 + 10 = 40$. Mean deviation about median $= \frac{40}{8} = 5$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq	$6$	$7$	$15$	$16$	$4$	$2$

$x_i$	$5, 15, 25, 35, 45$
$f_i$	$8, 48, 56, 32, 16$

The mean deviation about the median of the discrete data $12, 15, 7, 4, 4, 15, 23, 14$ is

Similar Questions

The mean deviation from the mean of the discrete data $2, 3, 5, 7, 11, 13, 17, 19, 22$ is

Calculate the mean deviation about the median for the following data:
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Freq $6$ $7$ $15$ $16$ $4$ $2$

The mean deviation of the following data about its mean is
$x_i$ $5, 15, 25, 35, 45$
$f_i$ $8, 48, 56, 32, 16$

Find the mean deviation about the median of the following distribution:
$ \begin{array}{|l|l|l|l|l|l|} \hline \text{Marks obtained} & 10 & 11 & 12 & 14 & 15 \\ \hline \text{Number of students} & 2 & 3 & 8 & 3 & 4 \\ \hline \end{array} $

If the mean deviation of the data $1, 1+d, 1+2d, \ldots, 1+100d$ $(d > 0)$ from their mean is $255$,then $d$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module