Calculate the mean deviation about the median | vedclass

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

Question

(A) The given data is not continuous. We convert it into a continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5

Vedclass · Accepted Answer

$7.35$

Vedclass · Answer

(A) The given data is not continuous. We convert it into a continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval.						Age			Frequency $(f_i)$			Cumulative Frequency $(c.f.)$			Mid-point $(x_i)$			$|x_i - 	ext{Median}|$			$f_i |x_i - 	ext{Median}|$							$15.5-20.5$			$5$			$5$			$18$			$20$			$100$							$20.5-25.5$			$6$			$11$			$23$			$15$			$90$							$25.5-30.5$			$12$			$23$			$28$			$10$			$120$							$30.5-35.5$			$14$			$37$			$33$			$5$			$70$							$35.5-40.5$			$26$			$63$			$38$			$0$			$0$							$40.5-45.5$			$12$			$75$			$43$			$5$			$60$							$45.5-50.5$			$16$			$91$			$48$			$10$			$160$							$50.5-55.5$			$9$			$100$			$53$			$15$			$135$							Total			$N=100$			-			-			-			$735$			Here,$N = 100$,so $\frac{N}{2} = 50$. The cumulative frequency just greater than $50$ is $63$,which corresponds to the class interval $35.5-40.5$.Median $= l + \frac{\frac{N}{2} - C}{f} 	imes h = 35.5 + \frac{50 - 37}{26} 	imes 5 = 35.5 + \frac{13 	imes 5}{26} = 35.5 + 2.5 = 38$.Mean Deviation about Median $= \frac{1}{N} \sum f_i |x_i - 	ext{Median}| = \frac{735}{100} = 7.35$.

Age (in years)	$16-20$	$21-25$	$26-30$	$31-35$	$36-40$	$41-45$	$46-50$	$51-55$
Number of persons	$5$	$6$	$12$	$14$	$26$	$12$	$16$	$9$

$x_i$	$2$	$4$	$6$	$8$	$10$
$f_i$	$1$	$2$	$3$	$2$	$1$

$x_i$	$2$	$4$	$5$	$7$	$9$
$f_i$	$2$	$4$	$10$	$8$	$6$

Calculate the mean deviation about the median age for the age distribution of $100$ persons given below:

Age (in years) $16-20$ $21-25$ $26-30$ $31-35$ $36-40$ $41-45$ $46-50$ $51-55$

Number of persons $5$ $6$ $12$ $14$ $26$ $12$ $16$ $9$

Similar Questions

Consider the following distribution:
$x_i$ $2$ $4$ $6$ $8$ $10$
$f_i$ $1$ $2$ $3$ $2$ $1$

The sum of the mean deviation from the mean and the mean deviation from the median of this distribution is:

The mean deviation about the mean for the following data is:
$x_i$ $2$ $4$ $5$ $7$ $9$
$f_i$ $2$ $4$ $10$ $8$ $6$

If $10$ is the mean deviation of $n$ observations $x_1, x_2, x_3, \ldots, x_n$,then the mean deviation of the observations $\frac{2x_1+5}{3}, \frac{2x_2+5}{3}, \frac{2x_3+5}{3}, \ldots, \frac{2x_n+5}{3}$ is

The mean deviation about the median of the numbers $2, 3, 5, 20, 17, 15, 13, 11, 9, 7$ is

For a frequency distribution,the mean deviation from the mean is computed by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Calculate the mean deviation about the median age for the age distribution of $100$ persons given below: Age (in years) $16-20$ $21-25$ $26-30$ $31-35$ $36-40$ $41-45$ $46-50$ $51-55$ Number of persons $5$ $6$ $12$ $14$ $26$ $12$ $16$ $9$