Question

Calculate the mean deviation of the following income groups of five and seven members from their medians.

IIncome in ₹	HIncome in ₹
4000	3800
4200	4000
4400	4200
4600	4400
4800	4600
4800
5800

Answer 1

Here, we will first find the median of the group I using the formula. Then by using the mean deviation formula and the obtained median, we will find the mean deviation of the group. We will follow the same process to find the mean deviation of group H. Mean deviation is defined as the average deviation from the mean of the given data values.

Formula Used:
We will use the following formulas:

1. If the number of terms $n$ is odd, then the median is given by ${\left( {\dfrac{{n + 1}}{2}} \right)^{th}}$ term.
2. Mean deviation is given by the formula $M.D. = \dfrac{1}{n}\sum\limits_{i = 1}^n {\left| {{d_i}} \right|}$ where ${d_i} = {x_i} - M$ where ${x_i}$ is the given data value and $M$ is the median

Complete step by step solution:
We will calculate the mean deviation of the income group I from the median.
The Income in Rupees of Income group I with five members are given as 4000, 4200, 4400, 4600, 4800.
The given data is arranged in ascending order.
Since the number of terms $n$ is odd, therefore
Median $= {\left( {\dfrac{{5 + 1}}{2}} \right)^{th}}$
Adding the terms in the numerator, we get
$\Rightarrow$ Median $= \dfrac{6}{2} = {3^{rd}}$ value
Therefore, the median is ${\rm{Rs}}.4400$.
Mean deviation is given by the formula $M.D. = \dfrac{1}{n}\sum\limits_{i = 1}^n {\left| {{d_i}} \right|}$ where ${d_i} = {x_i} - M$

IIncome in ₹| \left| {{d_i}} \right| = \left| {{x_i} - M} \right|$$$$\left| {{d_i}} \right| = \left| {{x_i} - 4400} \right|
---|---
$4000$ | $\left| {{d_i}} \right| = \left| {4000 - 4400} \right| = 400$
$4200$ | $\left| {{d_i}} \right| = \left| {4200 - 4400} \right| = 200$
$4400$ | $\left| {{d_i}} \right| = \left| {4400 - 4400} \right| = 0$
$4600$ | $\left| {{d_i}} \right| = \left| {4600 - 4400} \right| = 200$
$4800$ | $\left| {{d_i}} \right| = \left| {4800 - 4400} \right| = 400$
Total| $\sum {{d_i} = 1200}$

Mean deviation $M.D. = \dfrac{1}{5}\left( {1200} \right)$
$\Rightarrow$ Mean deviation $M.D. = 240$
We will calculate the mean deviation of the income group H from the median.
The Income in Rupees of Income group I with five members are given as 3800, 4000, 4200, 4400, 4600, 4800, 5800
The given data is arranged in ascending order.
Since the number of terms $n$ is odd, therefore
Median $= {\left( {\dfrac{{7 + 1}}{2}} \right)^{th}}$
$\Rightarrow$ Median $= \dfrac{8}{2} = {4^{th}}$ value
Therefore, the median is ${\rm{Rs}}.4400$.
Mean deviation is given by the formula $M.D. = \dfrac{1}{n}\sum\limits_{i = 1}^n {\left| {{d_i}} \right|}$ where ${d_i} = {x_i} - M$

HIncome in ₹| \left| {{d_i}} \right| = \left| {{x_i} - M} \right|$$$$\left| {{d_i}} \right| = \left| {{x_i} - 4400} \right|
---|---
$3800$ | $\left| {{d_i}} \right| = \left| {3800 - 4400} \right| = 600$
$4000$ | $\left| {{d_i}} \right| = \left| {4000 - 4400} \right| = 400$
$4200$ | $\left| {{d_i}} \right| = \left| {4200 - 4400} \right| = 200$
$4400$ | $\left| {{d_i}} \right| = \left| {4400 - 4400} \right| = 0$
$4600$ | $\left| {{d_i}} \right| = \left| {4600 - 4400} \right| = 200$
$4800$ | $\left| {{d_i}} \right| = \left| {4800 - 4400} \right| = 400$
$5800$ | $\left| {{d_i}} \right| = \left| {5800 - 4400} \right| = 1400$
Total| $\sum {{d_i} = 3200}$

Mean deviation $M.D. = \dfrac{1}{7}\left( {3200} \right)$
Dividing 3200 by 7, we get
$\Rightarrow$ Mean deviation $M.D. = 457.143$

Therefore, the mean deviation of Income Group I is 240 and the mean deviation of Income Group H is $457.143$.

Note:
Here, We will find the mean of the given data values and subtract it from the given data values to find the average deviation. We should remember that here the mean deviation is from the median of the data values. Instead of finding mean we will find the median of the data values which is the middle term of the data values to find the average deviation from the median and also the mean deviation cannot be negative. We should also remember while finding the median the data should be arranged in the ascending order.