Question

Calculate mean deviation from median for the following:
Marks (less than) 80 70 60 50 40 30 20 10
No. of students 100 90 80 60 32 20 13 5

Answer 1

Hint: This type of problem is based on the concept of median and mean deviation. First, we have to convert the marks obtained into class intervals that are from 0-10, 10-20, etc. consider the no. of students to be cumulative frequency. We have to find the frequency of each class interval by subtracting the cumulative frequency of upper class interval by lower class interval. We find that the median class is 40-50. To find the median M, we have to use the formula \[M=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h\]. Here, cf is the cumulative frequency of interval 40-50, N is the total number of class intervals, f is the frequency of the interval 40-50 and h is 10. Then, we have to find the mean deviation from median using the formula \[\dfrac{\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|}}{\sum{{{f}_{i}}}}\]. Substitute the values and do necessary calculations to find the required answer.

Complete step-by-step answer:
According to the question, we have to find the mean deviation from the median of the given data.
Let us tabulate the given data.

Marks (less than)	80	70	60	50	40	30	20	10
No. of students	100	90	80	60	32	20	13	5

Let us now convert the given marks into class intervals and find the frequency of all the intervals. Here, the cumulative frequency is no. of students.
To find the frequency of each interval, we have to subtract the cumulative frequency of the upper interval by the lower interval.
Let us tabulate and find the frequencies.
We have to first consider the lowest class interval.

Class Intervals	Cumulative frequency (cf)	Frequency (f)
0-10	5	5
10-20	13	13-5=8
20-30	20	20-13=7
30-40	32	32-20=12
40-50	60	60-32=28
50-60	80	80-60=20
60-70	90	90-80=10
70-80	100	100-90=10

Now, we have found the frequency of each interval.
We have to now find the median of the above data.
We know that the total number of students in the class is 100.
Let N=100.
To find the median interval, we have to find \[\dfrac{N}{2}\].
On substituting the value of N, we get
\[\dfrac{N}{2}=\dfrac{100}{2}\]
\[\Rightarrow \dfrac{N}{2}=50\]
In the class interval 40-50, we find that 50 is the upper limit.
Therefore, the median interval is 40-50.
We know that the formula to find the median in the median interval is
\[M=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h\]
Where l is lower interval of the median interval, N is the total number of students, cf is the cumulative frequency of the median interval, f is the frequency of the median interval and h is the difference of the intervals.
We find that l=40, N=100, cf=32, f=28 and h=10.
On substituting these values, we get
\[M=40+\dfrac{\dfrac{100}{2}-32}{28}\times 10\]
We know that \[\dfrac{100}{2}=50\].
\[\Rightarrow M=40+\dfrac{50-32}{28}\times 10\]
On further simplification, we get
\[M=40+\dfrac{18}{28}\times 10\]
We know that \[\dfrac{18}{28}=0.643\] approximately.
\[\Rightarrow M=40+0.643\times 10\]
\[\Rightarrow M=40+6.43\]
\[\therefore M=46.43\]
Therefore, the median is 46.43.
Now, we have to find \[{{x}_{i}}\] which is equal to the sum of upper interval and lower interval divided by 2.
Then, we have to find the modulus of \[{{x}_{i}}\] subtracted by M, that is \[\left| {{x}_{i}}-M \right|\].
And we should multiply the respective frequencies with \[\left| {{x}_{i}}-M \right|\] to find \[{{f}_{i}}\left| {{x}_{i}}-M \right|\].

Class interval	Cumulative frequency (cf)	Frequency (\[{{f}_{i}}\])	\[{{x}_{i}}\]	\[\left| {{x}_{i}}-M \right|\]	\[{{f}_{i}}\left| {{x}_{i}}-M \right|\]
0-10	5	5	\[\dfrac{10+0}{2}=5\]	\[\left| 5-46.43 \right|=41.43\]	5(41.43)=207.15
10-20	13	8	\[\dfrac{10+20}{2}=15\]	\[\left| 15-46.43 \right|=31.43\]	8(31.43)=251.44
20-30	20	7	\[\dfrac{30+20}{2}=35\]	\[\left| 25-46.43 \right|=21.43\]	7(21.43)=150.01
30-40	32	12	\[\dfrac{30+40}{2}=35\]	\[\left| 35-46.43 \right|=11.43\]	12(11.43)=137.16
40-50	60	28	\[\dfrac{50+40}{2}=45\]	\[\left| 45-46.43 \right|=1.43\]	28(1.43)=40.04
50-60	80	20	\[\dfrac{50+60}{2}=55\]	\[\left| 55-46.43 \right|=8.57\]	20(8.57)=171.4
60-70	90	10	\[\dfrac{70+60}{2}=65\]	\[\left| 65-46.43 \right|=18.57\]	10(18.57)=185.7
70-80	100	10	\[\dfrac{70+80}{2}=75\]	\[\left| 75-46.43 \right|=28.57\]	10(28.57)=285.7
		\[\sum{{{f}_{i}}=100}\]			\[\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|=1428.6}\]

We know that the formula to find the Mean deviation from median is
\[\dfrac{\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|}}{\sum{{{f}_{i}}}}\]
We knave found that \[\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|=1428.6}\] and \[\sum{{{f}_{i}}=100}\].
On substituting the values, we get
\[\dfrac{\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|}}{\sum{{{f}_{i}}}}=\dfrac{1428.6}{100}\]
On further simplification, we get
\[\dfrac{\sum{{{f}_{i}}\left| {{x}_{i}}-M \right|}}{\sum{{{f}_{i}}}}=14.286\]
Therefore, the mean deviation from median in the given data is 14.286.

Note: We should be very careful with the calculations. Do not get confused by the formula for mean deviation about mean and mean deviation about median. We have to first find the median and then solve the rest. Avoid calculation mistakes based on sign conventions. It is advisable to plot the graph of the given function to get more clarification for the answer.