Question

find the correlation coefficient from the following information of rainfall (x) (in cm) and yield (y) (tons per hectare) for the last 10 years of a district, n=10, cov (x,y)=30, S.D of X=5 and variance of Y=144​

Answer 1

Given,

number of years=10, Covariance of (x,y)=30,Standard variance of x= 5 and

variance of y =144.

To find,

The correlation coefficient of rainfall (x) (in cm) and yield (y) (tons per hectare) for the last 10 years of a district.

Solution,

We need to apply some simple statistical formulas,

 n = 10,   Cov (x , y) = 30,   S.D. of x = 5, and   Variance of y= 144(given)

In the first step we have to find, the standard variance of y

⇒the variance of y =144

⇒s.d of y=√144

⇒12.

After finding the s.d of y, we have to now apply the formula of correlation of coefficient of (x,y).

Now, r= $\frac{(cov x.y)}{s.d_{x}*s.d_{y}}$

⇒r=$\frac{30}{12*5}$

⇒r=0.5

∴The correlation coefficient from the following information of rainfall (x) (in cm) and yield (y) (tons per hectare) for the last 10 years of a district is 0.5.

 

Answer 2

Answer:

Given,

number of years=10, Covariance of (x,y)=30,Standard variance of x= 5 and

variance of y =144.

To find,

The correlation coefficient of rainfall (x) (in cm) and yield (y) (tons per hectare) for the last 10 years of a district.

Solution,

We need to apply some simple statistical formulas,

n = 10,   Cov (x , y) = 30,   S.D. of x = 5, and   Variance of y= 144(given)

In the first step we have to find, the standard variance of y

⇒the variance of y =144

⇒s.d of y=√144

⇒12.

After finding the s.d of y, we have to now apply the formula of correlation of coefficient of (x,y).

Now, r=  

⇒r=

⇒r=0.5

∴The correlation coefficient from the following information of rainfall (x) (in cm) and yield (y) (tons per hectare) for the last 10 years of a district is 0.5.