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# 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

**Solution:**

We will find the mean by step-deviation method

Mean, (x) = a + (Σfᵢuᵢ/Σfᵢ) × h

Modal Class is the class with the highest frequency

Mode = l + [(f₁ - f₀) / (2f₁ - f₀ - f₂)] × h

Here,

Class size, h

Lower limit of modal class, l

Frequency of modal class, f₁

Frequency of class preceding modal class, f₀

Frequency of class succeeding the modal class, f₂

Median Class is the class having Cumulative frequency(cf) just greater than n/2

Median = l + [(n/2 - cf)/f] × h

Class size, h

Number of observations, n

Lower limit of median class, l

Frequency of median class, f

Cumulative frequency of class preceding median class, cf

Class mark, xᵢ = (Upper class limit + Lower class limit)/2

Class size, h = 20

Taking assumed mean, a = 135

From the table, we obtain, Σfᵢ = 68 and Σfᵢuᵢ = 7

Mean, (x) = a + (Σfᵢuᵢ/Σfᵢ) × h

= 135 + (7/68) × 20

= 135 + 140/68

= 135 + 2.05

= 137.05

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

Class size, h = 20

Modal class = 125 − 145

Lower limit of modal class, l = 125

Frequency of modal class, f₁ = 20

Frequency of class preceding modal class, f₀ = 13

Frequency of class succeeding the modal class, f₂ = 14

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × h

= 125 + [(20 - 13)/(2 × 20 - 13 - 14)] × 20

= 125 + [7/(40 - 27)] × 20

= 125 + (7/13) × 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 ⇒ n/2 = 34

Cumulative frequency(cf) just greater than n/2 is 42, belonging to class-interval 125 − 145. Therefore, median class = 125 - 145

Therefore, median class = 125 - 145

Class size, h = 20

Lower limit of median class, l = 125

Frequency of median class, f = 20

Cumulative frequency of class preceding median class, cf = 22

Median = l + [(n/2 - 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.

So, mean > median > mode

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 14

**Video Solution:**

## 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.

NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.3 Question 1

**Summary:**

The median, mean and mode based on the frequency distribution data of the monthly consumption of electricity of 68 consumers of a locality are 137,135.76,and 137.05 respectively. So, mean > median > mode

**☛ Related Questions:**

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