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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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