Page 128 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Unit 6: Measures of Central Tendency

Solution: Notes

w
i 150 150
HM = w i 50 60 40 1.25 1.20 0.67 .... (1)
X 40 50 60
i
= 48.13 kms/hour

Total distance travelled
Verification: Average speed
Total time taken
We note that the numerator of Equation (1) gives the total distance travelled by train. Further, its
50
denominator represents total time taken by the train in travelling 150 kms, since is time
40
60 40
taken by the train in travelling 50 kms at a speed of 40 kms/hour. Similarly and are time
50 60
taken by the train in travelling 60 kms and 40 kms at the speeds of 50 kms./hour and 60 kms/
hour respectively. Hence, weighted harmonic mean is most appropriate average in this case.

Example: Ram goes from his house to office on a cycle at a speed of 12 kms/hour and
returns at a speed of 14 kms/hour. Find his average speed.
Solution:
Since the distances of travel at various speeds are equal, the average speed of Ram will be given
by the simple harmonic mean of the given speeds.
2 2
Average speed 12.92 kms/hour
1 1 0.1547
12 14

6.7.3 Merits and Demerits of Harmonic Mean

Merits

1. It is a rigidly defined average.
2. It is based on all the observations.

3. It gives less weight to large items and vice-versa.
4. It is capable of further mathematical treatment.
5. It is suitable in computing average rate under certain conditions.

Demerits

1. It is not easy to compute and is difficult to understand.
2. It may not be an actual item of the given observations.

3. It cannot be calculated if one or more observations are equal to zero.
4. It may not be representative of the data if small observations are given correspondingly
small weights.

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