If two variables x and y are related by 2x 3y – 7 = 0 and the mean and mean deviation about mean of x are 1 and 0.3 respectively, then the co-effectively, then the coefficient of mean deviation of y about mean is?

Question

If two variables x and y are related by 2x   3y – 7 = 0 and the mean and mean deviation about mean of x are 1 and 0.3 respectively, then the co-effectively, then the coefficient of mean deviation of y about mean is?

Answer

Given:

Two variables x and y are related by the equation 2x + 3y – 7 = 0. The mean and mean deviation about the mean of x are 1 and 0.3 respectively.

To Find:

The coefficient of mean deviation of y about the mean.

Solution:

Step 1: Find the mean of y.

The equation 2x + 3y – 7 = 0 can be written as 3y = 7 - 2x. Hence, y = (7 - 2x)/3. Now, we can find the mean of y as follows:

$\begin{align*} \bar{y} &= \frac{1}{n}\sum_{i=1}^{n}y_i\ &= \frac{1}{n}\sum_{i=1}^{n}\frac{7-2x_i}{3}\ &= \frac{7}{3}-\frac{2}{3}\bar{x} \end{align*}$

Substituting the given value of mean of x as 1, we get:

$\bar{y}=\frac{7}{3}-\frac{2}{3}(1)=\frac{5}{3}$

Hence, the mean of y is 5/3.

Step 2: Find the mean deviation of y about the mean.

The mean deviation of y about the mean is given by the formula:

$\frac{1}{n}\sum_{i=1}^{n}|y_i-\bar{y}|$

Substituting the given value of mean of y as 5/3, we get:

$\begin{align*} ext{Mean Deviation of y about mean} &= \frac{1}{n}\sum_{i=1}^{n}\left|\frac{7-2x_i}{3}-\frac{5}{3} ight|\ &= \frac{1}{n}\sum_{i=1}^{n}\left|\frac{2}{3}-\frac{2}{3}x_i ight|\ &= \frac{2}{3}\left(\frac{1}{n}\sum_{i=1}^{n}|1-x_i| ight)\ &= \frac{2}{3} ext{MD}_x \end{align*}$

Where MDx is the mean deviation about the mean of x, given as 0.3.

Hence, the mean deviation of y about the mean is (2/3) x 0.3 = 0.2.

Step 3: Find the coefficient of mean deviation of y about the mean.

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