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Question

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumptionNumber of consumers(in units)6585485105510512513125145201451651416518581852054

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Solution

Class IntervalFrequency(fi)Cumulative frequency (cf)6585448510559105125132212514520421451651456165185864185205468

Here, n=68 and n2=34. Therefore, median class is 125145

Here, l=125,f=20,cf=22 and h=20

Median=l+⎜ ⎜n2cff⎟ ⎟×h

=125+(342220)×20=125+12=137

Therefore, the median of the given data is 137 units.


Class IntervalFrequency (fi)Class mark (xi) ui=(xi135h) (fiui)6585475312851055952101051251311511312514520135001451651415511416518581752161852054195312fi=68fiui=7

Here, a=135 and h=20

Mean=a+hfiuifi

=135+20(768)=135+2.05=137.05

Therefore, the mean of the monthly consumption of electricity is 137.05 units.


Highest frequency is 20, therefore the modal class is 125145

Here, l=125,f1=20,f0=13,f2=14 and h=20

Mode=l+[(f1f0)(2f1f0f2)]×h

=125+[2013401314]×20

=125+(713)×20=125+10.77=135.77

Therefore, the mode of the monthly consumption of electricity is 135.77 units.



The median of the monthly consumption of electricity is 137 units

The mean of the monthly consumption of electricity is 137.05 units

The mode of the monthly consumption of electricity is 135.77 units.

The three measures are approximately the same in this case.



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