The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is Rs. 18, find the missing frequency f.


Daily pocket allowance (in Rs.)



11 - 13



13 - 15



15 - 17



17 - 19



19 - 21



21 - 23



23 - 25



Number of children



7



6



9



13



F



5



4


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


DAILY POCKET ALLOWANCE (Rs.)



MID - POINT(xi)



NUMBER OF CHILDREN (fi)



fixi



11 - 13



12



7



84



13 - 15



14



6



84



15 - 17



16



9



144



17 - 19



18



13



234



19 - 21



20



f



20f



21 - 23



22



5



110



23 - 25



24



4



96



TOTAL




44 + f



752 + 20f



We have got


Σfi = 44 + f and Σfixi = 752 + 20f


∵ mean is given by



⇒ (∵ given: mean of pocket allowance is 18)


⇒ 792 + 18f = 752 + 20f


⇒ 20f – 18f = 792 – 752


⇒ 2f = 40


⇒ f = 20


Thus, f is 20.


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