Question

If the coefficient of correlation between $x$ and $y$ is $0.6$ then, covariance is $16$. If the standard deviation of $x$ is $4$ , then the standard deviation of $y$ is.
$\eqalign{
  & 1)5 \cr
  & 2)10 \cr
  & 3)\dfrac{{20}}{3} \cr} $
4) None of these

Answer 1

Hint: The Pearsons’s coefficient of correlation is given by, covariance of the two variables, divided by the product of their standard deviations. There are four parameters in the formula equation. In the above question, three of them are given and we are supposed to find the fourth one. Therefore, we use the formula and substitute all the given values. Then we can rearrange accordingly and find the unknown.
The Pearson’s coefficient of Correlation is as follows:
${\rho _{xy}} = \dfrac{{\operatorname{cov} (x,y)}}{{{\sigma _x}{\sigma _y}}}$
Where,
${\rho _{xy}}$ is the coefficient of correlation
$\operatorname{cov} (x,y)$ is the covariance
${\sigma _x}{\sigma _y}$ is the standard deviation of $x$ and $y$

Complete step-by-step answer:
Given:
Coefficient of correlation, ${\rho _{xy}} = 0.6$
Covariance, $\operatorname{cov} (x,y) = 16$
Standard deviation of $x,{\sigma _x} = 4$
We need to find the standard deviation of $y,{\sigma _y}$
Now, let us substitute the given parameters in the above-mentioned formula.
We get,
$0.6 = \dfrac{{16}}{{4{\sigma _y}}}$
Let us rearrange so that we get ${\sigma _y}$ in the RHS.
$\eqalign{
  & {\sigma _y} = \dfrac{{16}}{{4 \times 0.6}} \cr
  & {\sigma _y} = \dfrac{{20}}{3} \cr} $
Therefore, the final answer is $\dfrac{{20}}{3}$
Hence, option (3) is the correct answer.
So, the correct answer is “Option 3”.

Note: Note down the given parameters first. There are only four and three are already given, so make necessary substitutions. Learn the formula for the coefficient of correlation. When parameters are taken from LHS to RHS, the operation changes from division to multiplication and vice versa. The final answer will always be a positive number because there are no negative signs involved in the formula or in the given values.