Question

The mean deviation about median of variates 13, 14, 15 …99, 100 is
A.1936
B.21.5
C.23.5
D.22

Answer 1

Hint: As we know, to find the mean deviation about median, we need to first calculate median. It is known that the median of ungrouped data is the middle term of the sorted data given in the question. Also, the median of the ungrouped data is different for even and odd numbers of terms. Firstly, we need to observe the data whether it has an even or odd number of terms then we will find the middle term of the sorted data. For an odd number of terms, the middle term can simply be calculated by $\dfrac{{n + 1}}{2}$ and this term will be the median. For even number of terms, there will be two middle terms which is calculated by $\dfrac{n}{2},\dfrac{n}{2} + 1$ and then these two terms can be added and divided by 2 to get the median of data. Then, we can calculate the mean deviation about the median by $\dfrac{{\sum {\left| {{x_i} - M} \right|} }}{n}$ . Now, put the value of median in the formula and mean deviation about median is obtained.

Complete step-by-step answer:
The ungrouped data is given in the question.
The given data is 13, 14, 15 …99, 100.
It is sorted. So, we can directly calculate the median.
To calculate the median of the ungrouped data, we have to observe the number of terms given.
There are (100-12) = 88 terms which is even.
So, median can be calculated by finding middle terms.
The middle terms of the given data are:
$\dfrac{n}{2},\dfrac{n}{2} + 1$
Here, we have 88 numbers of terms which means $n = 88$.
So, middle terms will be $\left( {\dfrac{{88}}{2},\dfrac{{88}}{2} + 1} \right)$ which are $44,45$ .
44th and 45th terms are 56 and 57.
Therefore, median of the given data is:
$
  M = \dfrac{{56 + 57}}{2} \\
\Rightarrow M = 56.5 \\
 $
Now, we have to calculate mean deviation about the median.
Mean deviation about median is given by $\dfrac{{\sum {\left| {{x_i} - M} \right|} }}{n}$
On putting values we get,
Mean deviation = \[\dfrac{{\sum\nolimits_{i = 1}^{88} {\left| {{x_i} - M} \right|} }}{n} = \dfrac{{\sum\nolimits_{i = 1}^{88} {\left| {{x_i} - M} \right|} }}{{88}}\]
Here, first and last terms of AP are given.
Therefore, using sum formula of AP, we get:
Mean deviation $ = \dfrac{{22\left( {43.5 + 0.5} \right) + 22\left( {0.5 + 43.5} \right)}}{{88}} = 22$
Therefore, the mean deviation about median is 22.
Hence, median is 56.5 and mean deviation about median is 22.
Hence, option D is correct.

Note: As we know, high attention is needed to do these types of questions because we have to take a lot of terms in calculation and we cannot make mistakes in the number of terms. The calculation of median and mean deviation about median is simple but it should also be rechecked twice to avoid any mistakes in these types of questions.