Question

The monthly income (in rupees) of $7$ households in a village are:$1200,{\text{ }}1500,{\text{ }}1400,{\text{ }}1000,{\text{ }}1000,{\text{ }}1600,{\text{ }}10000.$
${\text{(i)}}$ Find the median income of the households.
${\text{(ii)}}$ If one more household with monthly income of $1500$ is added, what will the median income be?

Answer 1

Here we will first arrange all the terms in ascending order and then find the median using the below mentioned median formulas.
Finally we get the required answer.

Formula used: ${\text{Median = }}{\left[ {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right]^{th}}{\text{term if n is odd,}}$
${\text{or, = }}\left[ {\dfrac{{{{\left[ {\dfrac{{\text{n}}}{{\text{2}}}} \right]}^{th}}{\text{term + }}{{\left[ {\dfrac{{\text{n}}}{{\text{2}}} + 1} \right]}^{th}}{\text{term}}}}{{\text{2}}}} \right]{\text{ if n is even}}$

Complete step-by-step solution:
First we have to find out ${\text{(i)}}$,
Since $n$ represents the total number of terms, in the question stated as we can see that the total number of terms is $7$, this means that $n = 7$.
Now to find the median,
So we have to arrange all the terms given in the question in ascending order from left to right,
Now the terms could be written as:
$1000,{\text{ }}1000,{\text{ }}1200,{\text{ }}1400,{\text{ }}1500,{\text{ }}1600,{\text{ }}10000$
Since all the terms are written in ascending order,
Now, we can calculate the median from the formula.
Since we know $n = 7$ this is an odd number the formula to calculate the median is:
${\text{Median = }}{\left[ {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right]^{th}}{\text{ term}}$
Now on substituting the value of $n = 7$ we get:
\Rightarrow$$${\text{Median = }}{\left[ {\dfrac{{{\text{7 + 1}}}}{{\text{2}}}} \right]^{th}}{\text{ term}}$$ On adding the numerator term and we get, \Rightarrow{\text{Median = }}{\left[ {\dfrac{8}{{\text{2}}}} \right]^{th}}{\text{ term}}$$ Let us divide the term and we get: $\Rightarrow{\text{Median = }}{{\text{4}}^{{\text{th}}}}{\text{ term}} Here we can write the ${4^{th}}$ term in the distribution is$1400$, **Hence, the Median is $1400$.** Now we have to find out ${\text{(ii)}}$ One more household with amount $1500$ is to be added in the distribution, therefore the distribution arranged in ascending order from left to right becomes: {\text{1000, 1000, 1200, 1400, 1500, 1500, 1600, 10000}} Since there are $8$ terms in the distributionn = 8 Since $n = 8$ which is an even number the formula to calculate the median is: {\text{Median = }}\left[ {\dfrac{{{{\left[ {\dfrac{{\text{n}}}{{\text{2}}}} \right]}^{{\text{th}}}}{\text{ term + }}{{\left[ {\dfrac{{\text{n}}}{{\text{2}}}{\text{ + 1}}} \right]}^{{\text{th}}}}{\text{ term}}}}{{\text{2}}}} \right]{\text{ }} On substituting $n = 8$ we get: $\Rightarrow$$${\text{Median = }}\left[ {\dfrac{{{{\left[ {\dfrac{{\text{8}}}{{\text{2}}}} \right]}^{{\text{th}}}}{\text{ term + }}{{\left[ {\dfrac{{\text{8}}}{{\text{2}}}{\text{ + 1}}} \right]}^{{\text{th}}}}{\text{ term}}}}{{\text{2}}}} \right]{\text{ }}
Taking LCM we get,
\Rightarrow$$${\text{Median = }}\left[ {\dfrac{{{{\left[ {\dfrac{{\text{8}}}{{\text{2}}}} \right]}^{{\text{th}}}}{\text{ term + }}{{\left[ {\dfrac{{{\text{8 + 2}}}}{{\text{2}}}} \right]}^{{\text{th}}}}{\text{ term}}}}{{\text{2}}}} \right]{\text{ }}$$ Let us add the terms and we get, \Rightarrow$$${\text{Median = }}\left[ {\dfrac{{{{\left[ {\dfrac{{\text{8}}}{{\text{2}}}} \right]}^{{\text{th}}}}{\text{ term + }}{{\left[ {\dfrac{{10}}{{\text{2}}}} \right]}^{{\text{th}}}}{\text{ term}}}}{{\text{2}}}} \right]{\text{ }}$$
Let us divided the terms and we get,
$\Rightarrow$ ${\text{Median = }}\dfrac{{{{\text{4}}^{{\text{th}}}}{\text{ term + }}{{\text{5}}^{{\text{th}}}}{\text{ term}}}}{{\text{2}}}$
Now on substituting the ${4^{th}}$ and ${5^{th}}$ terms from the distribution we get:
$\Rightarrow$ ${\text{Median}} = \dfrac{{1400 + 1500}}{2}$
On adding the numerator, we get:
$\Rightarrow$ ${\text{Median = }}\dfrac{{{\text{2900}}}}{{\text{2}}}$
On dividing we get:
$\Rightarrow$ ${\text{Median = 1450}}$
$\therefore$ The median of the $8$ households is $1450$.

Note: Median should not apply to qualitative data.
The given values should be only group and also computation in ordering.
Ratio and ordinal scale can be represented as median.