Question: Find the mean deviation about the mean for the following data:

Mark30-4040-5050-6060-7070-8080-9090-100
Number Of Students371215832


Solution:

Arranging data in a table,


MarksFrequency
(f)
Mid
Value(x)
fx|x-x̄|f|x-x̄|
30-403351052781
40-5074531517113
50-601255660784
60-701565975345
70-8087560013104
80-903852552369
90-1002951903366
N= 50\sumfx
= 3100

\sumf|x-x̄|
= 568
Now,
Mean () =\dfrac{\sum fx}{N}

= \dfrac{3100}{50}

= 62

And,

Mean Deviation from Mean (M.D.) = \dfrac{\sum f|x-x̄|}{N}

= \dfrac{568}{50}

= 11.36

Hence, the required mean deviation about the mean of the given data is 11..36

Related Notes and Solutions:

Here is the website link to the notes of Statistics of Class 10.

#SciPiPupil