Question

Mean deviation from the median of the data \[90,100,125,115,110\] is?
${\text{(A) 10}}$
${\text{(B) 20}}$
${\text{(C) 30}}$
${\text{(D) None of these}}$

Answer 1

Hint: Here we have to calculate the deviation from the medium; we will first calculate the median and find the deviation from it. We will make use of the below mentioned formulas to solve. Finally we get the required answer.

Formula used: ${\text{Mean = }}\dfrac{{{\text{sum of terms}}}}{{{\text{number of terms}}}}$
\[{\text{Median = }}{\left[ {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right]^{th}}{\text{ term if n is odd,}}\]
\[{\text{Median = }}\left[ {\dfrac{{{{\left[ {\dfrac{{\text{n}}}{{\text{2}}}} \right]}^{th}}{\text{ term + }}{{\left[ {\dfrac{{\text{n}}}{{\text{2}}} + 1} \right]}^{th}}{\text{ term}}}}{{\text{2}}}} \right]{\text{ if n is even}}{\text{.}}\]

Complete step-by-step solution:
To calculate the deviation, we first find the median of the given data
\[90,100,125,115,110\]
To calculate the median, we have to first arrange the given data into ascending order, the data after being sorted in ascending order is:
$90,100,110,115,125$
Now the total number of terms in the given distribution is $5$ which is an odd number therefore the formula to find the median is:
\[{\text{Median = }}{\left[ {\dfrac{{{\text{n + 1}}}}{{\text{2}}}} \right]^{th}}{\text{ term}}\]
Now $n$is the total number of terms in the distribution therefore we know $n = 5$.
After substituting the values in the formula, we get:
\[{\text{Median = }}{\left[ {\dfrac{{{\text{5 + 1}}}}{{\text{2}}}} \right]^{th}}{\text{ term}}\]
On simplifying we get:
\[{\text{Median = }}{\left[ 3 \right]^{rd}}{\text{ term}}\]
Now the ${3^{rd}}$ term in the distribution is $110$ therefore the median of the distribution is $110$.
Now we will find the deviation from the median of all the values.
The deviation from median can be represented from the following table:
We know the deviation from median is $|{x_i} - Median|$ where ${x_i}$ is the ${i^{th}}$ term in the distribution

Value ${x_i}$	Deviation from median $|{x_i} - Median|$
\[90\]	\[20\]
\[100\]	\[10\]
\[125\]	\[15\]
\[115\]	\[5\]
\[110\]	\[0\]

Now the mean deviation from median will be the mean of all the deviation of means which could be done as:
$\sum {\dfrac{{|{x_i} - Median|}}{n}} $
On substituting the values, we get:
${\text{Mean deviation from median}} = \dfrac{{20 + 10 + 15 + 5 + 0}}{5}$
On simplifying we get:
${\text{Mean deviation from median}} = \dfrac{{50}}{5}$
On further simplifying we get:
${\text{Mean deviation from median}} = 10$,

Therefore, the correct option is ${\text{(A)}}$ which is $10$.

Note: Apart from the mean deviation from median which was solved in this question, there also exists mean absolute deviation and mean deviation from mode.
The formula for mean deviation is:
\[{\text{Mean deviation = }}\dfrac{{\sum {{\text{|}}{{\text{x}}_{\text{i}}}{\text{ - Mean|}}} }}{{\text{n}}}\]
Also, the formula for mean deviation from mode is:
\[{\text{Mean deviation from mode = }}\dfrac{{\sum {{\text{|}}{{\text{x}}_{\text{i}}}{\text{ - Mode|}}} }}{{\text{n}}}\]
Where ${x_i}$ are the terms in the distribution and $n$ is the total number of terms in the distribution