1.

Find the median of the following frequency distibution :

Answer»

Solution :First of all we will PREPARE an EXCLUSIVE series from the GIVEN (inclusive) series. To make it we subtract and add the same number, i.e.,`(70-68)/(2)=1` in each class. So, new table is :

Here, `(N)/(2)=(98)/(2)=49`
So, cumulative frequency just greater than or equal to 49 is 70 and its corresponding class is the medianl class.
`:. """Median"=l_(1)+((N)/(2)-C)/(f)xxi=89+(49-40)/(30)xx10+3=92`


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