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 x_{1}, x_{2}, …, x_{n} 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
\[HM=\frac{n}{\sum{{{f}_{i}}(\frac{1}{{{m}_{i}}})}}\]
m_{i} = 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.
Family  1  2  3  4  5 
Income($)  85  90  70  50  60 
Solution:
Family 
Income (x) 
Reciprocal (1/x) 
1  85 
0.0117 
2 
90  0.0111 
3  70 
0.0143 
4 
50  0.02 
5  60 
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 
x  Wx 
10  48 
480 
12 
40  480 
15  32 
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.
Class  010  1020  2030  3040  4050 
Frequency  5  15  25  8  7 
Solution:
Class Interval 
Frequency (f)  Midvalue (m)  Reciprocal (1/m)  f(1/m) 
010  5  5  0.2 
1 
1020 
15  15  0.0666  0.999 
2030  25  25  0.04 
1 
3040 
8  35  0.0285  0.228 
4050  7  45  0.0222 
0.1554 
∑ f = 60 
∑ f(1/m) = 3.3824 
\[HM=\frac{n}{\sum{{{f}_{i}}(\frac{1}{{{m}_{i}}})}}=\frac{60}{3.3824}=17.74\]