Question

calculate the karl pearson's coefficient of correlation from the following data x = 10 , 12 , 18 , 24 , 23 , 27 ; y = 13 , 18 , 12 , 25 , 30 , 10

answer fast please .​

Answer 1

Concept

Karl pearson's coefficient of correlation is the linear coefficient correlation in the value range -1 to 1

Given

The data of X and Y are given

Find

We need to find karl pearson coefficient

Solution

The data of X and Y are given

Where we need to find the Data of XY

So , the data of XY are shown below

XY - 130 , 216 , 216 , 600,690,270

The total value of XY =2122

That means Sigma XY = 2122

Where the number of the observation is 6

hence N =6

The total value of X = sigma X = 114

The total value of Y = sigma Y = 108

X^2= 2402

Y^2 = 2262

r =

$\frac{6(2122) - (114)(108}{ \sqrt{6(2402) - (114)^{2} \sqrt{6(2262) - (108) ^ {2} } } }$

=

$\frac{420}{37.62 \times 43.68}$

= 0.25

Hence 0.25 is the Karl pearson's coefficient.

#SPJ3

Answer 2

Answer:

0.25

Explanation:

Karl Pearson's coefficient of correlation will range from -1 to 1

We are given a data range for the X and Y variable

First we need to multiply them correspondingly to obtain XY

XY - 130 , 216 , 216 , 600,690,270

In addition total value is 2122

Total Value is also called Sigma

As the total no. of observations is 6

Thus, N=6

Sigma X= 114

Sigma Y= 108

Upon Squaring,

$X^{2}$= 2402

$Y^{2}$ = 2262

r=$\frac{6(2122)-(114)(108)}{\sqrtn]{(6)2402-(114)^{2}\sqrt[]{6(2262)-(108)^{2} } } }$

r= 0.25

#SPJ1