Question

Consider a population list x, with a CV= 10%. A second population list, y, has a CV= 20%. Assume that the average of list y is 10 times the average of list x. What is the variance of list y if the variance of list x is 100?

Answer 1

Answer:

$s^2_Y = 200^2= 40000$

Explanation:

Data given

From the info given we have the following conditions:

$\bar Y = 10 \bar X$

$s^2_x = 100$

$s_x = 10$

$CV_x = 10\% = 0.1$

$CV_y = 20\% = 0.2$

And we want to find $s^2_y =?$

Solution to the problem

From the definition of variation coefficient we know this:

$CV_x = 0.1=\frac{S_x}{\bar X}$

From this condition we can solve for $\bar X$ and we got:

$\bar X = \frac{10}{0.1}=100$

Now we can find $\bar Y$ like this:

$\bar Y = 10*100 = 1000$

Now from the definition for coefficient of variation for Y we have:

$CV_y = 0.2=\frac{S_y}{\bar Y}$

We can solve for $S_y$ and we got:

$S_y = 0.2*1000= 200$

And finally the variance would be:

$s^2_Y = 200^2= 40000$