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 consumption (in units)Number of consumers (fi)xi class markdi=xi−135ui=di20fiui65−85475−60−3−1285−105595−40−2−10105−12513115−20−1−13125−14520135000145−1651415520114165−185817540216185−205419560312Total68 7
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=137.06
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=135.77
We know that:
3 median = mode + 2 mean
= 135.77 + 2 (137.06)
= 135.77 + 274.12 = 409.89
Median = 409.9 / 3 = 136.63
So, median, mode, mean of given data is 136.63, 135.77, 137.06 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 consumption (in units)Number of consumers (fi)xi class markdi=xi−135ui=di20fiui65−85475−60−3−1285−105595−40−2−10105−12513115−20−1−13125−14520135000145−1651415520114165−185817540216185−205419560312Total68 7
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=137.06
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=135.77
We know that:
3 median = mode + 2 mean
= 135.77 + 2 (137.06)
= 135.77 + 274.12 = 409.89
Median = 409.9 / 3 = 136.63
So, median, mode, mean of given data is 136.63, 135.77, 137.06 respectively.
