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Question: If \(Var(x) = 8.25\), \(Var(y) = 33.96\) and \(Cov(x,y) = 10.2\) then the correlation coefficient is...

If Var(x)=8.25Var(x) = 8.25, Var(y)=33.96Var(y) = 33.96 and Cov(x,y)=10.2Cov(x,y) = 10.2 then the correlation coefficient is
A)0.89A)0.89
B)−0.98B) - 0.98
C)0.61C)0.61
D)−0.16D) - 0.16

Explanation

Solution

First, we will need to know about the concept of the correlation coefficient.
The coefficient of the correlation is used to measure the relationship extent between 22 separate intervals or variables.
Denoted by the symbol rr. Where r is the value of positive or negative. Thus, this will further be generalized into the form of Pearson’s correlation coefficient.
We will simply use the formula of standard deviation and variance to get the correlation coefficient of the problem.
Formula used:
r=Cov(x,y)σXσYr = \dfrac{{Cov(x,y)}}{{{\sigma _X}{\sigma _Y}}}, where σX{\sigma _X}is the standard deviation of X and σY{\sigma _Y} is the standard deviation of Y.

Complete step-by-step solution:
Since from the given that we have, Var(x)=8.25Var(x) = 8.25, Var(y)=33.96Var(y) = 33.96 and Cov(x,y)=10.2Cov(x,y) = 10.2
Now let us find the standard deviation for the X, which is σX=VarX=8.25{\sigma _X} = \sqrt {VarX} = \sqrt {8.25}
Similarly, we can also able to find the standard deviation for Y, which is σY=VarY=33.96{\sigma _Y} = \sqrt {VarY} = \sqrt {33.96}
Also, from the given that, we know Cov(x,y)=10.2Cov(x,y) = 10.2
Now substituting all the know values into the given correlation formula we get r=Cov(x,y)σXσY⇒10.28.2533.96r = \dfrac{{Cov(x,y)}}{{{\sigma _X}{\sigma _Y}}} \Rightarrow \dfrac{{10.2}}{{\sqrt {8.25} \sqrt {33.96} }}
Since 8.25=2.872\sqrt {8.25} = 2.872 and 33.96=5.827\sqrt {33.96} = 5.827 , the values of the root
Thus, we get r=Cov(x,y)σXσY=10.28.2533.96=10.22.872×5.827r = \dfrac{{Cov(x,y)}}{{{\sigma _X}{\sigma _Y}}} = \dfrac{{10.2}}{{\sqrt {8.25} \sqrt {33.96} }} = \dfrac{{10.2}}{{2.872 \times 5.827}}
With the help of multiplication operation, we get r=Cov(x,y)σXσY=10.22.872×5.827=10.216.735r = \dfrac{{Cov(x,y)}}{{{\sigma _X}{\sigma _Y}}} = \dfrac{{10.2}}{{2.872 \times 5.827}} = \dfrac{{10.2}}{{16.735}}
With the help of division operation, we get r=Cov(x,y)σXσY=10.216.735=0.609r = \dfrac{{Cov(x,y)}}{{{\sigma _X}{\sigma _Y}}} = \dfrac{{10.2}}{{16.735}} = 0.609 which is approximately 0.610.61
Hence, the option C)0.61C)0.61 is correct.

Note: The standard formula for the correlation coefficient:
Let us consider two different variables x and y that are related commonly, to find the extent of the link between the given numbers x and y, we will choose Pearson's coefficient r method.
In that process, the formula given is used to identify the extent or range of the two variables' equality.
Which is r=n∑xy−∑x∑y[n∑(y)2−(∑x)2][n∑(y)2−(∑y)2]r = \dfrac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {[n{{\sum {(y)} }^2} - (\sum {x{)^2}} ][n{{\sum {(y)} }^2} - (\sum {y{)^2}} ]} }}
In this formula r=n∑xy−∑x∑y[n∑(y)2−(∑x)2][n∑(y)2−(∑y)2]r = \dfrac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {[n{{\sum {(y)} }^2} - (\sum {x{)^2}} ][n{{\sum {(y)} }^2} - (\sum {y{)^2}} ]} }}
∑x\sum x denotes the number of first variable values.
∑y\sum y denotes the count of the second variable values.
∑x2{\sum x ^2} denotes the addition of a square for the first value.
  ∑y2\;{\sum y ^2} denotes the sum of the second values. And n denotes the total count data quantity.