Question

If ${G_1},{G_2}$ are geometric means of two series and G is the GM of the ratios of the corresponding observations then G is equal to
A)\[\dfrac{{{G_1}}}{{{G_2}}}\]
 B) \[\log {G_1} - \log {G_2}\]
C) \[\dfrac{{\log {G_1}}}{{\log {G_2}}}\]
D) \[\log \left( {{G_1}{G_2}} \right)\]

Answer 1

Hint: The above question is based on the concept of geometric mean. The main approach towards solving this problem is to find out individual geometric means of ${G_1},{G_2}$ which is the root of the product of all the values and then taking the ratio of both the geometric means of the series.

Complete step-by-step answer:
Geometric mean is the average value which states the central tendency of a set of numbers by finding the product of values. It is basically multiplying all the numbers in the given data set and taking the \[{n^{th}}\] root of the product. For example ,consider the given series 4,10,16,24.Therefore by taking the fourth root of the product of all the numbers we get the geometric mean as 11.13.
So now in the above given question,it said that there are two series whose geometric mean is ${G_1},{G_2}$.So we need to find G which is the geometric mean of the ratios corresponding observations.
Let \[{x_1},{x_2}....{x_n}\] be n observation for variate x and \[{y_1},{y_2}....{y_n}\]be n observation for variate y .Then according to the definition,
\[{G_1} = {\left( {{x_1},{x_2}....{x_n}} \right)^{\dfrac{1}{n}}}\] and \[{G_1} = {\left( {{y_1},{y_2}....{y_n}} \right)^{\dfrac{1}{n}}}\]
Therefore, by taking the ratio we get,
\[G = \dfrac{{{x_1}}}{{{y_1}}} \times \dfrac{{{x_2}}}{{{y_2}}}......\dfrac{{{x_n}}}{{{y_n}}} = \dfrac{{{G_1}}}{{{G_2}}}\]

So, the correct answer is “Option A”.

Note: An important thing to note is that all the numbers cannot be added to find the mean because it will form an arithmetic mean .Therefore the product of all the numbers with n root is said to be the geometric mean.