Question

The following table gives weekly consumption of electricity of 56 families : weekly consumption 0-10=16 10-20=12 20-30=18 30-40=6 40-50=4 find median

Answer 1

Answer:

the median weekly consumption would be 20 units

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Answer 2

Given is the frequency distribution of electricity consumption, Find their median.

Explanation:

• For the given distribution we have total frequencies and class size as, $N=56\ \ \ \ \ C=10$
• The table for the frequency distribution and the cumulative frequencies is given below: $Electricity \ units\ \ \ \ \ \ \ \ \ \ \ No.\ of \ families\ \ \ \ \ \ \ \ \ \ \ \ cf\\0-10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 16\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 16\\10-20\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 28 \\20-30\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 18\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 46$$30-40\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 52\\40-50\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 56$  
• Here we get the median class as, $\frac{N}{2}=\frac{56}{2}=28\ \ \ \ \ \ \ \ \ \ \ ->median\ class : (10-20)$  
• Here, the lower boundary of the median class $l=10$ , cf of class before the median class $F=16$, frequency of the median class $f_m=12$.
• Hence we know that median is given by, $M_d=l+(\frac{\frac{N}{2}-F }{f_m} )C$                 $->M_d=10+(\frac{28-16}{12})10\\ ->M_d=20$        ------>Answer