Harmonic Mean
Harmonic mean is another measure of central tendency and is also based on mathematics like arithmetic mean and geometric mean. Like arithmetic mean and geometric mean, harmonic mean is also useful for quantitative data. Harmonic mean is defined as:
Harmonic mean is the quotient of the “number of the given values” and the“sum of the reciprocals of the given values”.
Harmonic mean in mathematical terms is defined as follows:
For Ungrouped Data

For Grouped Data

$$H.M{\text{ of }}X = \overline X = \frac{n}{{\sum \left( {\frac{1}{x}} \right)}}$$

$$H.M{\text{ of }}X = \overline X = \frac{{\sum f}}{{\sum \left( {\frac{f}{x}} \right)}}$$

Example:
Calculate the harmonic mean of the numbers 13.5, 14.5, 14.8, 15.2 and 16.1
Solution:
The harmonic mean is calculated as below:
$$x$$

$$\frac{1}{x}$$

$$13.2$$

$$0.0758$$

$$14.2$$

$$0.0704$$

$$14.8$$

$$0.0676$$

$$15.2$$

$$0.0658$$

$$16.1$$

$$0.0621$$

Total

$$\sum \left( {\frac{1}{x}} \right) = 0.3417$$

$$H.M{\text{ of }}X = \overline X = \frac{n}{{\sum \left( {\frac{1}{x}} \right)}}$$
\[H.M{\text{ of }}X = \overline X = \frac{5}{{0.3417}} = 14.63\]
Example:
Given the following frequency distribution of first year students of a particular college, calculate the harmonic mean.
Age (Years)

$$13$$

$$14$$

$$15$$

$$16$$

$$17$$

Number of Students

$$2$$

$$5$$

$$13$$

$$7$$

$$3$$

Solution:
The given distribution is grouped data and the variable involved is the ages of first year students, while the number of students represents frequencies.
Ages (Years)
$$x$$ 
Number of Students
$$f$$ 
$$\frac{1}{x}$$

$$13$$

$$2$$

$$0.1538$$

$$14$$

$$5$$

$$0.3571$$

$$15$$

$$13$$

$$0.8667$$

$$16$$

$$7$$

$$0.4375$$

$$17$$

$$3$$

$$0.1765$$

Total

$$\sum f = 30$$

$$\sum \left( {\frac{f}{x}} \right) = 1.9916$$

Now we will find the harmonic mean as
$$\overline X = \frac{{\sum f}}{{\sum \left( {\frac{f}{x}} \right)}} = \frac{{30}}{{1.9916}} = 15.0631 \approx 15$$ years.
Example:
Calculate the harmonic mean for the data given below:
Marks

$$30 – 39$$

$$40 – 49$$

$$50 – 59$$

$$60 – 69$$

$$70 – 79$$

$$80 – 89$$

$$90 – 99$$

$$f$$

$$2$$

$$3$$

$$11$$

$$20$$

$$32$$

$$25$$

$$7$$

Solution:
The necessary calculations are given below:
Marks

$$x$$

$$f$$

$$\frac{f}{x}$$

$$30 – 39$$

$$34.5$$

$$2$$

$$0.0580$$

$$40 – 49$$

$$44.5$$

$$3$$

$$0.0674$$

$$50 – 59$$

$$54.5$$

$$11$$

$$0.2018$$

$$60 – 69$$

$$64.5$$

$$20$$

$$0.3101$$

$$70 – 79$$

$$74.5$$

$$32$$

$$0.4295$$

$$80 – 89$$

$$84.5$$

$$25$$

$$0.2959$$

$$90 – 99$$

$$94.5$$

$$7$$

$$0.0741$$

Total

$$\sum f = 100$$

$$\sum \left( {\frac{f}{x}} \right) = 1.4368$$

Now we will find the harmonic mean as
\[\overline X = \frac{{\sum f}}{{\sum \left( {\frac{f}{x}} \right)}} = \frac{{100}}{{1.4368}} = 69.60\]
Rakesh Sudan
December 28 @ 11:20 pm
Why we not use GM or HM while we convert equation of motion, v=u +at into distance equation, S=ut +1/2 at^2 ?
Sinoya gift
March 18 @ 1:29 am
This is very helpful.
But am still struggling how some other books r calculating this harmonic mean.
How can we find a data set of 22, 25, 15, 10.
Since they r finding 15.87 by multiplying 4 by 825 and divide it by 208 #4×825/208.
Where those figures are they taking?
Please help with some extra examples if necessary