For the given data, the calculation corresponding to all value of pairs $(x,y)$is following ${\sum {(x - \overline x )} ^2} = 36,{\sum {(y - \overline y )} ^2} = 25,\sum {(x - \overline x )\sum {(y - \overline y )} } = 20$ Then the Karl Pearson’s correlation coefficient is
$A)0.2$
$B)0.5$
$C)0.66$
$D)0.33$
Answer
574.2k+ views
Hint: 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 $2$ separate intervals or variables.
Denoted by the symbol $r$. Where r is the value of positive or negative. Thus, this will further be generalized into the form of Pearson’s correlation coefficient. The formula for Pearson’s correlation is given below.
Formula used:
\[r = \dfrac{{\sum {(x - \overline x )\sum {(y - \overline y )} } }}{{\sqrt {{{\sum {(x - \overline x )} }^2}} \sqrt {{{\sum {(y - \overline y )} }^2}} }}\] is the Pearson’s correlation coefficient for the particularly given value.
Complete step-by-step solution:
Since from the given that we have, ${\sum {(x - \overline x )} ^2} = 36,{\sum {(y - \overline y )} ^2} = 25,\sum {(x - \overline x )\sum {(y - \overline y )} } = 20$where these are the calculation corresponding all value of pairs $(x,y)$
Let us find the square root of the first two terms, which are ${\sum {(x - \overline x )} ^2} = 36 \Rightarrow \sqrt {\sum {(x - \overline x )}^2 } = 6, {\sum {(y - \overline y )} ^2} = 25 \Rightarrow \sqrt {\sum {(y - \overline y )}^2 } = 5$ where the square root of $\sqrt {36} = 6,$ and the square root of $\sqrt {25} = 5$
Now substitute the values into the given formula, we get \[r = \dfrac{{\sum {(x - \overline x )\sum {(y - \overline y )} } }}{{\sqrt {{{\sum {(x - \overline x )} }^2}} \sqrt {{{\sum {(y - \overline y )} }^2}} }} = \dfrac{{20}}{{6 \times 5}}\]
Further solving we get, \[r = \dfrac{{20}}{{6 \times 5}} = \dfrac{{20}}{{30}} = \dfrac{2}{3} = 0.66\]
Hence, the option $C)0.66$ is correct.
Additional information:
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 $r = \dfrac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {[n{{\sum {(y)} }^2} - (\sum {x{)^2}} ][n{{\sum {(y)} }^2} - (\sum {y{)^2}} ]} }}$.
Note: In this formula $r = \dfrac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {[n{{\sum {(y)} }^2} - (\sum {x{)^2}} ][n{{\sum {(y)} }^2} - (\sum {y{)^2}} ]} }}$
$\sum x $denotes the number of first variable values.
\[\sum y \] denotes the count of the second variable values.
${\sum x ^2}$ denotes the addition of a square for the first value.
\[\;{\sum y ^2}\] denotes the sum of the second values. And n denotes the total count data quantity.
The coefficient of the correlation is used to measure the relationship extent between $2$ separate intervals or variables.
Denoted by the symbol $r$. Where r is the value of positive or negative. Thus, this will further be generalized into the form of Pearson’s correlation coefficient. The formula for Pearson’s correlation is given below.
Formula used:
\[r = \dfrac{{\sum {(x - \overline x )\sum {(y - \overline y )} } }}{{\sqrt {{{\sum {(x - \overline x )} }^2}} \sqrt {{{\sum {(y - \overline y )} }^2}} }}\] is the Pearson’s correlation coefficient for the particularly given value.
Complete step-by-step solution:
Since from the given that we have, ${\sum {(x - \overline x )} ^2} = 36,{\sum {(y - \overline y )} ^2} = 25,\sum {(x - \overline x )\sum {(y - \overline y )} } = 20$where these are the calculation corresponding all value of pairs $(x,y)$
Let us find the square root of the first two terms, which are ${\sum {(x - \overline x )} ^2} = 36 \Rightarrow \sqrt {\sum {(x - \overline x )}^2 } = 6, {\sum {(y - \overline y )} ^2} = 25 \Rightarrow \sqrt {\sum {(y - \overline y )}^2 } = 5$ where the square root of $\sqrt {36} = 6,$ and the square root of $\sqrt {25} = 5$
Now substitute the values into the given formula, we get \[r = \dfrac{{\sum {(x - \overline x )\sum {(y - \overline y )} } }}{{\sqrt {{{\sum {(x - \overline x )} }^2}} \sqrt {{{\sum {(y - \overline y )} }^2}} }} = \dfrac{{20}}{{6 \times 5}}\]
Further solving we get, \[r = \dfrac{{20}}{{6 \times 5}} = \dfrac{{20}}{{30}} = \dfrac{2}{3} = 0.66\]
Hence, the option $C)0.66$ is correct.
Additional information:
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 $r = \dfrac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {[n{{\sum {(y)} }^2} - (\sum {x{)^2}} ][n{{\sum {(y)} }^2} - (\sum {y{)^2}} ]} }}$.
Note: In this formula $r = \dfrac{{n\sum {xy} - \sum x \sum y }}{{\sqrt {[n{{\sum {(y)} }^2} - (\sum {x{)^2}} ][n{{\sum {(y)} }^2} - (\sum {y{)^2}} ]} }}$
$\sum x $denotes the number of first variable values.
\[\sum y \] denotes the count of the second variable values.
${\sum x ^2}$ denotes the addition of a square for the first value.
\[\;{\sum y ^2}\] denotes the sum of the second values. And n denotes the total count data quantity.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

