Question

Find the median for the data set 22, 45, 56, 56,45,123,122,56,103,56\[\]
A.103\[\]
B. 102\[\]
C.122\[\]
D.56\[\]

Answer 1

Hint: We arrange the given data values in the question in ascending order and find the number of data values $n$ by counting. If $n$ is odd we find the median as the data, a value at ${{\left( \dfrac{n+1}{2} \right)}^{\text{th}}}$ position in ascending order and if $n$ is even then we take the average of data values at ${{\left( \dfrac{n}{2} \right)}^{\text{th}}}$ and ${{\left( \dfrac{n}{2}+1 \right)}^{\text{th}}}$ position.

Complete step-by-step solution:
We know that a median of a data set is any value such that at most half of the data set is less than the proposed median and at most half is greater than the proposed median. If there are $n$ data values say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ arranged in ascending order and $n$ is odd then median is the data, a value at ${{\left( \dfrac{n+1}{2} \right)}^{\text{th}}}$ position as \[m={({x}_{(\dfrac{n+1}{2})})}\]
If $n$ is even , then the median is average of two middle values at ${{\left( \dfrac{n}{2} \right)}^{\text{th}}}$ and ${{\left( \dfrac{n}{2}+1 \right)}^{\text{th}}}$ position.
\[m=\dfrac{1}{2}\left( {{x}_{\dfrac{n}{2}}}+{{x}_{\dfrac{n}{2}+1}} \right)\]
Let us observe the data set given in the question 22, 45, 56, 56,45,123,122,56,103,56. We count the number of data values present in the set and find it to be 10. So the number of data values is even and we have
\[n=10\]
So we need to arrange the data in ascending order. We have 22,45,45,56,56,56,56,103,122,123. We can denote them as
\[{{x}_{1}}=22,{{x}_{2}}=45,...,{{x}_{10}}=123\]
We have to find the two middle values as the number of data values is even. The position of the middle values in the arranged data in ascending order is
\[\begin{align}
  & {{\left( \dfrac{n}{2} \right)}^{\text{th}}}={{\left( \dfrac{10}{2} \right)}^{\text{th}}}={{5}^{\text{th}}} \\
 & {{\left( \dfrac{n}{2}+1 \right)}^{\text{th}}}={{\left( \dfrac{10}{2}+1 \right)}^{\text{th}}}={{\left( 5+1 \right)}^{\text{th}}}={{6}^{\text{th}}} \\
\end{align}\]
The data values at the ${{5}^{\text{th}}}$ and ${{6}^{\text{th}}}$positions are
\[{{x}_{5}}=56,{{x}_{6}}=56\]
The median is the average of the middle values. We take the average of data values at the ${{5}^{\text{th}}}$ and ${{6}^{\text{th}}}$positions and have the median as
\[m=\dfrac{{{x}_{5}}+{{x}_{6}}}{2}=\dfrac{56+56}{2}=\dfrac{112}{2}=56\]
So the correct option is C.\[\]

Note: We need to be careful of the confusion between median, mean, and mode. Mean is the sum of values divided by the number of values and mode is the highest value. If we divide the whole data into four parts having equal numbers of data values, the median is called the second quartile.