Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution 50.


x: 10



30



50



70



90



y: 17



f1



32



f2



19



X



y



yx



10



17



170



30



f1



30f1



50



32



1600



70



f2



70f2



90



19



1710




N = 120



yx = 30f1 + 70f2 + 3480



Given, mean = 50


= 50


= 50


30f1 + 70f2 + 3480 = 50 * 120


30f1 + 70f2 + 3480 = 6000 (i)


Also, y = 120


17 + f1 + 32 + f2 + 19 = 120


f1 + f2 = 52


f1 = 52 – f2


Substituting value of f1 in (i), we get


30 (52 – f2) + 70f2 + 3480 = 6000


40f2 = 960


f2 = 24


Hence, f1 = 52 – 24 = 28


Therefore, f1 = 28 and f2 = 24


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