15.6 Analysis of Frequency Distributions

In earlier sections, we have studied about some types of measures of dispersion. The mean deviation and the standard deviation have the same units in which the data are given. Whenever we want to compare the variability of two series with same mean, which are measured in different units, we do not merely calculate the measures of dispersion but we require such measures which are independent of the units. The measure of variability which is independent of units is called coefficient of variation (denoted as C.V.)
The coefficient of variation is defined as
\(
\text { C.V. }=\frac{\sigma}{\bar{x}} \times 100, \bar{x} \neq 0,
\)
where \(\sigma\) and \(\bar{x}\) are the standard deviation and mean of the data.
For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

15.6.1 Comparison of two frequency distributions with same mean Let \(\bar{x}_1\) and \(\sigma_1\) be the mean and standard deviation of the first distribution, and \(\bar{x}_2\) and \(\sigma_2\) be the mean and standard deviation of the second distribution.
Then C. V. (1st distribution) \(=\frac{\sigma_1}{\bar{x}_1} \times 100\)
and C.V. (2nd distribution) \(=\frac{\sigma_2}{\bar{x}_2} \times 100\)
Given \(\bar{x}_1=\bar{x}_2=\bar{x}\) (say)
Therefore C.V. (1st distribution) \(=\frac{\sigma_1}{\bar{x}} \times 100 \dots(1)\)
and C.V. (2nd distribution) \(=\frac{\sigma_2}{\bar{x}} \times 100 \dots(2)\)
It is clear from (1) and (2) that the two C.Vs. can be compared on the basis of values of \(\sigma_1\) and \(\sigma_2\) only.

Thus, we say that for two series with equal means, the series with greater standard deviation (or variance) is called more variable or dispersed than the other. Also, the series with lesser value of standard deviation (or variance) is said to be more consistent than the other.
Let us now take following examples:

Example 17: Two plants \(A\) and \(B\) of a factory show following results about the number of workers and the wages paid to them.
\(
\begin{array}{lcc}
& \text { A } & \text { B } \\
\text { No. of workers } & 5000 & 6000 \\
\text { Average monthly wages } & \text { Rs } 2500 & \text { Rs } 2500 \\
\begin{array}{l}
\text { Variance of distribution } \\
\text { of wages }
\end{array} & 81 & 100
\end{array}
\)
In which plant, \(A\) or \(B\) is there greater variability in individual wages?

Solution: The variance of the distribution of wages in plant \(A \left(\sigma_1^2\right)=81\)
Therefore, standard deviation of the distribution of wages in plant \(A \left(\sigma_1\right)=9\)
Also, the variance of the distribution of wages in plant \(B \left(\sigma_2{ }^2\right)=100\)
Therefore, standard deviation of the distribution of wages in plant \(B \left(\sigma_2\right)=10\)
Since the average monthly wages in both the plants is same, i.e., Rs.2500, therefore, the plant with greater standard deviation will have more variability. Thus, the plant \(B\) has greater variability in the individual wages.

Example 18: Coefficient of variation of two distributions are 60 and 70, and their standard deviations are 21 and 16 , respectively. What are their arithmetic means.

Solution: Given
C.V. (1st distribution) \(=60, \sigma_1=21\)
C.V. (2nd distribution) \(=70, \sigma_2=16\)
Let \(\bar{x}_1\) and \(\bar{x}_2\) be the means of 1st and 2nd distribution, respectively. Then
\(
\text { C.V. (1st distribution) }=\frac{c_1}{\bar{x}_1} \times 100
\)
Therefore \(60=\frac{21}{\bar{x}_1} \times 100 \text { or } \bar{x}_1=\frac{21}{60} \times 100=35\)
and C.V. (2nd distribution) \(=\frac{c_2}{\bar{x}_2} \times 100\)
i.e. \(70=\frac{16}{\bar{x}_2} \times 100 \text { or } \bar{x}_2=\frac{16}{70} \times 100=22.85\)

Example 19: The following values are calculated in respect of heights and weights of the students of a section of Class XI :
\(
\begin{array}{lcl}
& \text { Height } & \text { Weight } \\
\text { Mean } & 162.6 cm & 52.36 kg \\
\text { Variance } & 127.69 cm ^2 & 23.1361 kg ^2
\end{array}
\)
Can we say that the weights show greater variation than the heights?

Solution:

\(
\begin{aligned}
&\text { To compare the variability, we have to calculate their coefficients of variation. }\\
&\begin{array}{ll}
\text { Given } & \text { Variance of height }=127.69 cm ^2 \\
\text { Therefore } & \text { Standard deviation of height }=\sqrt{127.69} cm =11.3 cm \\
\text { Also } & \text { Variance of weight }=23.1361 kg ^2
\end{array}
\end{aligned}
\)
Therefore \(\quad\) Standard deviation of weight \(=\sqrt{23.1361} kg =4.81 kg\) Now, the coefficient of variations (C.V.) are given by
\(
\begin{aligned}
\text { (C.V.) in heights } & =\frac{\text { Standard Deviation }}{\text { Mean }} \times 100 \\
& =\frac{11.3}{162.6} \times 100=6.95
\end{aligned}
\)
and (C.V.) in weights \(=\frac{4.81}{52.36} \times 100=9.18\)
Clearly C.V. in weights is greater than the C.V. in heights Therefore, we can say that weights show more variability than heights.