Question

2) Karl Pearson's coefficient of correlation between two variable x and y is 0.52. Their covariance
is + 7.8. If the variance of x is 16. Find the standard deviation of y.​

Answer 1

Answer:

Using fact coefficient of correlation rxy=σxσyCov(xy)=3×48=32.

Answer 2

Answer:

The standard deviation of $y$ calculated is $3.75$.

Step-by-step explanation:

Given data,

Karl Pearson's coefficient of correlation between two variables $x$ and $y$ = $0.52$

The covariance between the two variables $x$ and $y$ = $+ 7.8$

The variance of $x$, v($x$) = $16$

The standard deviation of $y$ =?

Now, as we know,

• Correlation coefficient = $\frac{cov x,y}{(std. x)\times ( std. y)}$    -------equation (1)

Here,

• std.$x$ = standard deviation of $x$
• std.$y$ = standard deviation of $y$

Now, as given, the variance of $x$ = $16$

Therefore,

• std.$x$ = $\sqrt{v(x)}$ = $\sqrt{16}$ = $4$

Now, after putting the given values and the calculated values in equation (1), we get:

• Correlation coefficient = $\frac{cov x,y}{(std. x)\times ( std. y)}$
• $0.52= \frac{7.8}{4\times std.y}$
• $std.y = \frac{7.8}{4\times 0.52}$ = $3.75$

Hence, the standard deviation of $y$ calculated is $3.75$.