Find the missing frequency in the following distribution if N is 60 and median is 40.
Marks0−1010−3030−6060−8080−90Frequency5F1F282
MarksFrequencyCumulative frequency0−105510−30F15+F130−60F25+F1+F260−80813+F1+F280−90215+F1+F2N=60
∵N=60∵15+F1+F2=60F1+F2=45Median Size=N2=602=30th item.∵M=40∴Median Class is 30−60.M=L1+N2−C.ff×c40=30+602−(5+F1)F2×3010=(25−F1)F2×3010F2=750−30F130F1+10F2=7503F1+F2=75....eq(1)F1+F2=45......eq(2)Solving eq (1) and eq (2)2F1=30F1=15F2=30