Calculate the mean deviation about the median | vedclass

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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	$5$	$6$	$12$	$14$	$26$	$12$	$16$	$9$

$\text{Size}$	$20$	$21$	$22$	$23$	$24$
$\text{Frequency}$	$6$	$4$	$5$	$1$	$4$

Marks	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
No of Girls	$6$	$8$	$14$	$16$	$4$	$2$

Class Interval	$0$–$2$	$2$–$4$	$4$–$6$	$6$–$8$	$8$–$10$
Frequency	$1$	$3$	$4$	$1$	$2$

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

Find the mean deviation about the mean of the distribution:
$\text{Size}$ $20$ $21$ $22$ $23$ $24$
$\text{Frequency}$ $6$ $4$ $5$ $1$ $4$

Let $\bar{X}$ and $M.D.$ be the mean and the mean deviation about $\bar{X}$ of $n$ observations $x_i,$ $i = 1, 2, \dots, n.$ If each of the observations is increased by $5,$ then the new mean and the mean deviation about the new mean,respectively,are

Find the mean deviation about the median for the following data:

Marks $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$

No of Girls $6$ $8$ $14$ $16$ $4$ $2$

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} $

The mean deviation about the mean for the following data is
Class Interval $0$–$2$ $2$–$4$ $4$–$6$ $6$–$8$ $8$–$10$
Frequency $1$ $3$ $4$ $1$ $2$

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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 $5$ $6$ $12$ $14$ $26$ $12$ $16$ $9$