Harmonic Mean


Statistics / Monday, August 6th, 2018

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 of values of variables. It is specified average which solves problems involving variables expressed in 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 to 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:

Family

Income (x)

Reciprocal (1/x)

185

0.0117

2

900.0111
370

0.0143

4

500.02
560

0.0167

∑ (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:

x

1/x
48

0.02083

40

0.025
32

0.03125

∑ (1/x) = 0.07708

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

W

xWx
1048

480

12

40480
1532

480

∑ 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 Interval

Frequency (f)Mid-value (m)Reciprocal (1/m)f(1/m)
0-10550.2

1

10-20

15150.06660.999
20-3025250.04

1

30-40

8350.02850.228
40-507450.0222

0.1554

∑ 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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