Find the mean, median and mode of the following data:


Class



0 - 10



10 - 20



20 - 30



30 - 40



40 - 50



50 - 60



60 - 70



Frequency



6



8



10



15



5



4



2



For equal class intervals, we will solve by finding mid points of these classes using direct method.


CLASS



MID - POINT(xi)



FREQUENCY(fi)



fixi



0 - 10



5



6



30



10 - 20



15



8



120



20 - 30



25



10



250



30 - 40



35



15



525



40 - 50



45



5



225



50 - 60



55



4



220



60 - 70



65



2



130



TOTAL




50



1500




We have got


Σfi = 50 and Σfixi = 1500


mean is given by





Thus, mean is 30.


To find median, Assume


Σfi = N = Sum of frequencies,


h = length of median class,


l = lower boundary of the median class,


f = frequency of median class


and Cf = cumulative frequency


Lets form a table.


CLASS



FREQUENCY(fi)



Cf



0 - 10



6



6



10 - 20



8



6 + 8 = 14



20 - 30



10



14 + 10 = 24



30 - 40



15



24 + 15 = 39



40 - 50



5



39 + 5 = 44



50 - 60



4



44 + 4 = 48



60 - 70



2



48 + 2 = 50



TOTAL



50




We have got


So, N = 50


N/2 = 50/2 = 25


The cumulative frequency just greater than (N/2 = ) 25 is 39, so the corresponding median class is 30 - 40 and accordingly we get Cf = 24(cumulative frequency before the median class).


Now, since median class is 30 - 40.


l = 30, h = 10, f = 15, N/2 = 25 and Cf = 24


Median is given by,




= 30 + 0.67


= 30.67


Thus, median is 30.67.


Since, we have got mean = 30 and median = 30.67


Applying the empirical formula,


Mode = 3(Median) – 2(Mean)


Mode = 3(30.67) – 2(30)


Mode = 92.01 – 60 = 32.01


Mean = 30, Median = 30.67 and Mode = 32.01


19
1