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 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

We will find mean by step deviation method.

By formula,

Class mark = $\dfrac{\text{Lower limit + Upper limit}}{2}$

Here, h (class size) = 20.

By formula,

Mean = $a + \dfrac{Σf$iui}{Σf_i} \times h

Substituting values we get :

$\text{Mean } = 135 + \dfrac{7}{68} \times 20 \\[1em] = 135 + \dfrac{35}{17} \\[1em] = 135 + 2.05 = 137.05$

Cumulative frequency distribution table is as follows :

Here, n = 68, which is even

$\dfrac{n}{2} = \dfrac{68}{2}$ = 34.

Cumulative frequency just greater than $\dfrac{n}{2}$ is 42, belonging to class-interval 125 - 145.

∴ Median class = 125 - 145

⇒ Lower limit of median class (l) = 125

⇒ Frequency of median class (f) = 20

⇒ Cumulative frequency of class preceding median class (cf) = 22

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\text{Median} = 125 + \dfrac{34 - 22}{20} \times 20 \\[1em] = 125 + \dfrac{12}{20} \times 20 \\[1em] = 125 + 12 \\[1em] = 137.$

By formula,

Mode = l + $\Big(\dfrac{f$1 - f0}{2f1 - f0 - f_2}\Big) \times h

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

From table,

Class 125 - 145 has the highest frequency.

∴ It is the modal class.

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

Substituting values we get :

$\text{Mode} = 125 + \Big(\dfrac{20 - 13}{2 \times 20 - 13 - 14}\Big) \times 20 \\[1em] = 125 + \dfrac{7}{40 - 27} \times 20 \\[1em] = 125 + \dfrac{140}{13} \\[1em] = 125 + 10.76 \\[1em] = 135.76$

Hence, mean = 137.05, median = 137 and mode = 135.76