Statistics Harmonic Mean Formula with Example

All about Harmonic Mean with Formula And Example

 

Harmonic Mean

Today we will learn Harmonic Mean Formula With Example. It is the total number of items of a value divided by the sum of reciprocal values of variables. It is a specified average that solves problems involving variables expressed within ‘Time rates’ that vary according to time.

If x1, x2, …, xn are ‘n’ values for discrete series without frequency, then their Harmonic Mean (HM) is,

Note: For better understanding, I suggest you check my post on Arithmetic Mean Formula With Example and Geometric Mean Formula With Example.

Individual Series

\[HM=\frac{n}{\sum{\frac{1}{{{x}_{i}}}}}\]

Discrete Series

\[HM=\frac{n}{\sum{{{f}_{i}}(\frac{1}{{{x}_{i}}})}}\]

Continuous Series

\[HM=\frac{n}{\sum{{{f}_{i}}(\frac{1}{{{m}_{i}}})}}\]
mi = Mid value of each class interval.

 

Weighted Harmonic Mean

\[{{H}_{W}}=\frac{\sum{{{W}_{i}}}}{\sum{{{W}_{i}}(\frac{1}{{{x}_{i}}})}}=\frac{\sum{W{{x}_{i}}}}{\sum{{{W}_{i}}}}\]

 Example 01

The daily incomes of 5 families in a very rural village are given below. Compute HM.

Family12345
Income($)8590705060

Solution:

FamilyIncome (x)Reciprocal (1/x)
1850.0117
2900.0111
3700.0143
4500.02
5600.0167
  ∑ (1/x) = 0.0738

\[HM=\frac{n}{\sum{\frac{1}{{{x}_{i}}}}}=\frac{5}{0.0738}=67.75\]

 Example 02

A man travel by a car for 3 days he covered 480 km each day. On the first day he drives for 10 hrs at the rate of 48 KMPH, on the second day for 12 hrs at the rate of 40 KMPH, and the third day for 15 hrs at the rate of 32 KMPH. Compute HM and weighted mean and compare them.

Solution:

x1/x
480.02083
400.025
320.03125
 ∑ (1/x) = 0.07708

\[HM=\frac{n}{\sum{\frac{1}{{{x}_{i}}}}}=\frac{3}{0.07708}=38.92\]
Now, we will calculate weighted Mean,

WxWx
1048480
1240480
1532480
∑ W = 37 ∑ Wx = 1440

\[{{H}_{W}}=\frac{\sum{W{{x}_{i}}}}{\sum{{{W}_{i}}}}=\frac{1440}{37}=38.92\]

 Example 03

Find the HM of the following data.

Class0-1010-2020-3030-4040-50
Frequency5152587

Solution:

Class IntervalFrequency (f)Mid-value (m)Reciprocal (1/m)f(1/m)
0-10550.21
10-2015150.06660.999
20-3025250.041
30-408350.02850.228
40-507450.02220.1554
 ∑ f = 60  ∑ f(1/m) = 3.3824

\[HM=\frac{n}{\sum{{{f}_{i}}(\frac{1}{{{m}_{i}}})}}=\frac{60}{3.3824}=17.74\]

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