Question

Mean deviation of the observations 70, 42, 63, 34, 44, 54, 55, 46, 38, 48 from median is
A. 7.8
B. 8.5
C. 7.6
D. 8.8

Answer 1

Hint: The question is related to the statistics topic. Here we have to determine the mean deviation about the median for the following observations. The observation is in the form of ungrouped data and the median for the ungrouped data is the middle term of the observation. By using the formula \[\dfrac{{\sum {\left| {{x_i} - M} \right|} }}{n}\], we determine the mean deviation about the median.

Complete step-by-step answer:
The observation is in the form of ungrouped data where we do not know the value of frequency. In the ungrouped data first we arrange the observations in the ascending order.
now consider the given observations.
70, 42, 63, 34, 44, 54, 55, 46, 38, 48
On arranging the above observations in the ascending order, we have
34, 38, 42, 44, 46, 48, 54, 55, 63, 70.
The number of observations are 10. To determine the median for the ungrouped data we have two formulas. If the number of observations is in odd number the formula is \[\dfrac{{n + 1}}{2}\,th\] observation if it is in even number we have two observations then the formula is \[\dfrac{n}{2}\,\,th\]observation and \[\dfrac{n}{2} + 1\,\,th\] observations
The number 10 is the even number, to determine the median we have to use the formula \[\dfrac{n}{2}\,th\] observation. and \[\dfrac{n}{2} + 1\,\,th\] observations
On substituting the value of n we have
 \[ \Rightarrow M = \dfrac{n}{2}\,th\,\,and\,\,\dfrac{n}{2} + 1\,\,th\] observations
 \[ \Rightarrow M = 5th\,\,and\,\,6\,th\] observations
The 5th observation is 46 and the 6th observation is 48.
Therefore the median is given by
 \[ \Rightarrow M = \dfrac{{46 + 48}}{2}\]
on simplifying this we have
 \[ \Rightarrow M = \dfrac{{94}}{2} = 47\]
Therefore the median is 47.
Now we have to determine the mean deviation about median, the formula is given as \[\dfrac{{\sum {\left| {{x_i} - M} \right|} }}{n}\], where M is the median, n is the total number of observations and \[{x_i}\] is each observation.
Therefore we have
\[M.D(M) = \dfrac{{\sum {\left| {{x_i} - M} \right|} }}{n}\]
On substituting the values we have
\[ \Rightarrow M.D(M) = \dfrac{
  |34 - 47| + |38 - 47| + |42 - 47| + |44 - 47| + |46 - 47| + |54 - 47| + |55 - 46| + \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|63 - 47| + |70 - 47| \\
 }{{10}}\]
On simplifying we have
\[ \Rightarrow M.D(M) = \dfrac{{| - 13| + | - 9| + | - 5| + | - 3| + | - 1| + |7| + |8| + |16| + |23|}}{{10}}\]
when we have a negative value in the mod, we will consider it has positive so we have
\[ \Rightarrow M.D(M) = \dfrac{{13 + 9 + 5 + 3 + 1 + 7 + 8 + 16 + 23}}{{10}}\]
On adding these numbers we have
\[ \Rightarrow M.D(M) = \dfrac{{85}}{{10}}\]
on dividing we have
\[ \Rightarrow M.D(M) = 8.5\]
Hence the median is 47 and mean deviation about median is 8.5
The option is the correct one.
So, the correct answer is “Option B”.

Note: The mean deviation is calculated for the both mean and median. If they mention mean deviation about mean there is another formula. The first thing is first we have to determine the median value correctly, if we went wrong in determining the value of the median then the whole solution will be wrong.