NCERT Solutions
Class 10 Maths
Statistics

Ex.13.3 Q.1
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.
To find the class mark (xi) for each interval, the following relation is used.
Class mark xi = (Upper limit + Lower limit) ÷ 2
Taking 135 as assured mean (a), di, ui, and fiui can be calculated according to step deviation method as follows:
From the table, it can be observed that
Σfi = 68
Σfiui = 7
Now, Mean = a + (Σfiui ÷ Σfi ) h
= 135 + (7 ÷ 68) × 20
= 135 + (140 ÷ 68)
= 135 + 2.058
= 137.058
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 (l) of modal class = 125
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 13
Frequency (f2) of class succeeding modal class = 14
Class size = 20
Mode = l + {(f1 – f0) ÷ (2f1 – f0 – f2)} × h
= 125 + {(20 - 13) ÷ (2 × 20 – 13 – 14)} × 20
= 125 + {7 ÷ (40 - 27)} × 20
= 125 + (140 ÷ 13)
= 125 + 10.76
= 135.76
To find the median of the given data, cumulative frequency is calculated as follows:
From the table, we obtain n = 68
Cumulative frequency (cf) just greater than is (i.e.,
= 34) is 42, belonging to interval 125 − 145.
Therefore, median class = 125 − 145
Lower limit (l) of median class = 125
Class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Median = l + {(n2 – cf) ÷ f} × h
= 125 + {(34 - 22) ÷ 20} × 20
= 125 + 12
= 137
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.