Mathematics Analysis of Frequency Distributions and Comparison of two frequency distributions with same mean

### Topics covered

star Analysis of Frequency Distributions
star Comparison of two frequency distributions with same mean

### Analysis of Frequency Distributions

color{red} ✍️ In earlier sections, we have studied about some types of measures of dispersion.

color{red} ✍️ 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.

color{red} ✍️ The measure of variability which is independent of units is color(blue)("called coefficient of variation") (denoted as color(red)(C.V.))

color(green)("The coefficient of variation is defined as")

color{purple}(C.V. = sigma/(barx) xx 100 , \ \ barx ≠ 0 ,)

where color{purple}(σ) and color(blue)(barx) are color{purple}("the standard deviation") and color{blue}("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.

### Comparison of two frequency distributions with same mean

Let color{purple}(barx_1) and color{purple}(σ_1) be the mean and standard deviation of the first distribution, and color{purple}(barx_2) and color{purple}(σ_2) be the mean and standard deviation of the second distribution.

Then color{purple}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C.V. ("1st distribution") = (sigma_1)/(barx_1) xx100)

and color{purple}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C.V. ("2nd distribution") = (sigma_1)/(barx_1) xx100)

Given color{purple}(barx_1 = barx_2 = barx) (say)

Therefore color{purple}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C.V. ("1st distribution") = (sigma_1)/(barx_1) xx100) ...........................................(1)

and color{purple}(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C.V. ("2nd distribution") = (sigma_1)/(barx_1) xx100) ...........................................(2)

It is clear from (1) and (2) that the two C.Vs. can be compared on the basis of values of color{purple}(σ_1) and color{purple}(σ_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.
Q 3106378278

Two plants A and B of a factory show following results about the number
of workers and the wages paid to them.

In which plant, A or B is there greater variability in individual wages?

Solution:

The variance of the distribution of wages in plant A ( σ_(1)^2 ) = 81

Therefore, standard deviation of the distribution of wages in plant A (σ_1 ) = 9

Also, the variance of the distribution of wages in plant B ( σ_(2)^2 ) = 100

Therefore, standard deviation of the distribution of wages in plant B (σ_2 ) = 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.
Q 3116478370

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, σ_1 = 21

C.V. (2nd distribution) = 70, 2 σ_2 = 16

Let bar (x_1)  and bar (x_2) be the means of 1st and 2nd distribution, respectively. Then

C.V. (1st distribution) = σ_1/x_1 xx 100

Therefore 160 = 21/(bar (x_1) ) xx 100 or bar (x_1) = 21/60 xx 100 = 35

and C.V. (2nd distribution)  = σ_2/(bar (x_2)) xx 100

i.e., 70 = 16/(bar x_2) xx 100  or bar (x_2) = 16/70 xx 100 = 22.85
Q 3116478379

The following values are calculated in respect of heights and weights of
the students of a section of Class XI :

Can we say that the weights show greater variation than the heights?

Solution:

To compare the variability, we have to calculate their coefficients of variation.

Given Variance of height = 127.69 cm^2
Therefore Standard deviation of height = sqrt (127.69) cm = 11.3 cm

Also Variance of weight = 23.1361 kg^2

Therefore Standard deviation of weight = sqrt (23.1361 ) kg =4.81 kg

Now, the coefficient of variations (C.V.) are given by

(C.V.) in heights = (text (Standard Deviation) )/( text (Mean)) xx 100

= 11.3/162.6 xx 100 = 6.95

and (C.V.) in weights = 4.81/52.36 xx 100 = 9.18

Clearly C.V. in weights is greater than the C.V. in heights