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~consumption~(in~units)Number~of~consumers65−85485−1055105−12513125−14520145−16514165−1858185−2054
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~consumption~(in~units)Number~of~consumers65−85485−1055105−12513125−14520145−16514165−1858185−2054
We may find the class marks by using the relation:
Class~mark=upper~class~limit~+~lower~class~limit2
Taking 135$as assumed mean a, we may find di,ui,fiui, according to step deviation method as following:
Monthly ConsuptionNumber of Consumers (fi)Class Mark(xi)di=xi−135ui=di20fiui65−85475−60−3−1285105595−40−2−10105−12513115−20−1−1312514520135000145−1651415520114165−185817540216185205419560312Total687
From the table, we may observe that:
∑fiui=7∑fi=68, class size h = 20, Mean ¯x=a+(∑fiui∑fi)×h=135+768×20=135+14068=137.058
Now, from table, it is clear that maximum class frequency is 20 belonging to class interval 125−145.
Modal class = 125−145
Lower limit l of modal class = 125
Class size h = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14
Mode=l+(f1−f02f1−f0−f2)×h=125+[20−132(20)−13−14]×20=125+713×20=125+14013=135.76
We know that:
3 median = mode + 2 mean
= 135.76+2(137.058)
= 135.76+274.116
= 409.876
Median = 136.625
So, median, mode, mean of given data is 136.625,135.76,137.05 respectively.
We may find the class marks by using the relation:
Class~mark=upper~class~limit~+~lower~class~limit2
Taking 135$as assumed mean a, we may find di,ui,fiui, according to step deviation method as following:
Monthly ConsuptionNumber of Consumers (fi)Class Mark(xi)di=xi−135ui=di20fiui65−85475−60−3−1285105595−40−2−10105−12513115−20−1−1312514520135000145−1651415520114165−185817540216185205419560312Total687
From the table, we may observe that:
∑fiui=7∑fi=68, class size h = 20, Mean ¯x=a+(∑fiui∑fi)×h=135+768×20=135+14068=137.058
Now, from table, it is clear that maximum class frequency is 20 belonging to class interval 125−145.
Modal class = 125−145
Lower limit l of modal class = 125
Class size h = 20
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding the modal class = 14
Mode=l+(f1−f02f1−f0−f2)×h=125+[20−132(20)−13−14]×20=125+713×20=125+14013=135.76
We know that:
3 median = mode + 2 mean
= 135.76+2(137.058)
= 135.76+274.116
= 409.876
Median = 136.625
So, median, mode, mean of given data is 136.625,135.76,137.05 respectively.
