Page 124 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Unit 6: Measures of Central Tendency

period and P the population after n years, using Equation (2), we can write the expression for Notes
n
1
P n n
the average rate of change of population per annum as r = – 1.
P 0
Similarly, Equation (4), given above, can be used to find the average rate of growth of population
when its rates of growth in various years are given.
Remarks: The formulae of price and population changes, considered above, can also be extended
to various other situations like growth of money, capital, output, etc.

Example: The population of a country increased from 2,00,000 to 2,40,000 within a period
of 10 years. Find the average rate of growth of population per year.
Solution:
Let r be the average rate of growth of population per year for the period of 10 years. Let P be
0
initial and P be the final population for this period.
10
We are given P = 2,00,000 and P = 2,40,000.
0 10
1 1
P 10 10 2,40,000 10
r = 1 = 1
P 0 2,00,000
1
24 10 1
Now antilog log24 log20
20 10
1
antilog 1.3802 1.3010 antilog 0.0079 =1.018
10
Thus, r = 1.018 - 1 = 0.018.
Hence, the percentage rate of growth = 0.018 × 100 = 1.8% p. a.

6.6.5 Suitability of Geometric Mean for Averaging Ratios

It will be shown here that the geometric mean is more appropriate than arithmetic mean while
averaging ratios.

Let there be two values of each of the variables x and y, as given below:
x y
Ratio Ratio
x y y x
40 60 2/3 3/ 2
20 80 1/ 4 4

2 1 3 4
Now AM of (x/y) ratios = 3 4 11 and the AM of (y/x) ratios = 2 11 . We note that their
2 24 2 4
product is not equal to unity.
1
However, the product of their respective geometric means, i.e., and 6 , is equal to unity.
6
Since it is desirable that a method of average should be independent of the way in which a ratio
is expressed, it seems reasonable to regard geometric mean as more appropriate than arithmetic
mean while averaging ratios.

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