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Question

Mathematics Question on Median of Grouped Data

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 consumers**
65 - 854
85 - 1055
105 - 12513
125 - 14520
145 - 16514
165 - 1858
185 - 2054
Answer

The cumulative frequencies with their respective class intervals are as follows.

Monthly consumption

(in units)** Number of consumers**Cumulative frequency
60 - 8544
85 - 10554 + 5 = 9
105 - 125139 + 13 = 22
125 - 1452022 + 20 = 42
145 - 1651442 + 14 = 56
165 - 185856 + 8 = 64
185 - 205464 + 4 = 68
Total(n)68

From the table, we obtain
n = 68
Cumulative frequency just greater n2(i.e.,682=34)\frac{n}2 ( i.e., \frac{68}2 = 34) than is 42, belonging to class interval 125 - 145.
Median class = 125 - 145
Lower limit (ll) of median class = 125
Frequency (ff) of median class = 20
Cumulative frequency (cfcf) of median class = 22
Class size (hh) = 20

Median = l+(n2cff×h)l + (\frac{\frac{n}2 - cf}f \times h)

Median = 125+(342220×20)125 + (\frac{34 - 22}{20} \times 20)

Median = 125 +12
Median = 137


To find the class mark (xi) for each interval, the following relation is used.

Class mark (xi)(x_i) = Upper limit + Lower limit2\frac {\text{Upper \,limit + Lower \,limit}}{2}

Taking 11.5 as assumed mean (a), did_i, uiu_i, and fiuif_iu_i are calculated according to step deviation method as follows.

Monthly consumption

(in units)** Number of consumers**** xi\bf{x_i} **di=xi11.5\bf{d_i = x_i -11.5}ui=di3\bf{u_i = \frac{d_i}{3}}fiui\bf{f_iu_i}
60 - 85475-60-3-12
85 - 105595-40-2-10
105 - 12513115-20-1-13
125 - 14520135000
145 - 1651415520114
165 - 185817540216
185 - 205419560312
Total687

From the table, it can be observed that

fi=68\sum f_i = 68
fiui=7\sum f_iu_i = 7

Mean, x=a+(fiuifi)×h\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i})\times h

x\overset{-}{x} = 135+(768)×20135 + (\frac{7 }{68})\times 20

x\overset{-}{x} = 135 + 14068\frac{140}{68}
Mean, x\overset{-}{x} = 137.058


The data in the given table can be written as

Monthly consumption

(in units)** Number of consumers**
65 - 854
85 - 1055
105 - 12513
125 - 14520
145 - 16514
165 - 1858
185 - 2054

From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.

Therefore, Modal class = 125 − 145
Lower limit (ll) of modal class = 125
Frequency (f1f_1) of modal class = 40
Frequency (f0f_0) of class preceding the modal class = 13
Frequency (f2f_2) of class succeeding the modal class = 14
Class size (hh) = 20

Mode = ll + (f1f02f1f0f2)×h(\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h

Mode = 125+(20132(20)1314)×20125 + (\frac{20 - 13 }{ 2(20) - 13 - 14}) \times 20

Mode =125+[713]×20125+ [\frac{7}{13}] \times 20

Mode = 125+(14013)125 +( \frac{ 140}{ 13})

Mode = 135.76

Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively. The three measures are approximately the same in this case.