Page 127 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Quantitative Techniques – I

Notes Solution:
Calculation of Harmonic Mean
X 10 11 12 13 14 Total
Frequency ( ) 5 8 10 9 6 38
f
1
f 0.5000 0.7273 0.8333 0.6923 0.4286 3.1815
X
38
= = 11.94
3.1815
Continuous Frequency Distribution

In case of a continuous frequency distribution, the class intervals are given. The mid-values of
the first, second ...... nth classes are denoted by X , X , ...... X . The formula for the harmonic mean
1 2 n
is same, as given in (b) above.

Example: Find the harmonic mean of the following distribution :
Class Intervals : 0 -10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
Frequency : 5 8 11 21 35 30 22 18
Solution.
Calculation of Harmonic Mean

Class Intervals 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total
f
Frequency ( ) 5 8 11 21 35 30 22 18 150
Mid -Values ( ) 5 15 25 35 45 55 65 75
X
f
1.0000 0.5333 0.4400 0.6000 0.7778 0.5455 0.3385 0.2400 4.4751
X
150
HM = = 33.52 Ans.
4.4751
6.7.2 Weighted Harmonic Mean

If X , X , ...... X are n observations with weights w ,w , ...... w respectively, their weighted
1 2 n 1 2 n
harmonic mean is defined as follows :
w i
HM = w
i
X i

Example: A train travels 50 kms at a speed of 40 kms/hour, 60 kms at a speed of 50 kms/
hour and 40 kms at a speed of 60 kms/hour. Calculate the weighted harmonic mean of the speed
of the train taking distances travelled as weights. Verify that this harmonic mean represents an
appropriate average of the speed of train.

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