(a) Scatter Plot of the data is-
(b)
E ( X Y ) = ∑ X Y n = 30055 5 = 6011 E ( X ) = ∑ X n = 10015 5 = 2003 E ( Y ) = ∑ Y n = 15 5 = 3 σ X = ( x − x ˉ ) 2 n = 10 5 = 2 σ Y = ( y − y ˉ ) 2 n = 10 5 = 2 E(XY)=\dfrac{\sum XY}{n}=\dfrac{30055}{5}=6011
\\[9pt]
E(X)=\dfrac{\sum X}{n}=\dfrac{10015}{5}=2003
\\[9pt]
E(Y)=\dfrac{\sum Y}{n}=\dfrac{15}{5}=3
\\[9pt]
\sigma_X=\sqrt{\dfrac{ (x-\bar{x})^2}{n}}=\sqrt{\dfrac{10}{5}}=\sqrt{2}
\\[9pt]
\sigma_Y=\sqrt{\dfrac{(y-\bar{y})^2}{n}}=\sqrt{\dfrac{10}{5}}=\sqrt{2} E ( X Y ) = n ∑ X Y = 5 30055 = 6011 E ( X ) = n ∑ X = 5 10015 = 2003 E ( Y ) = n ∑ Y = 5 15 = 3 σ X = n ( x − x ˉ ) 2 = 5 10 = 2 σ Y = n ( y − y ˉ ) 2 = 5 10 = 2
Pearson correlation coefficient r = C o v ( X , Y ) σ X , σ Y r=\dfrac{Cov(X,Y)}{\sigma_X,\sigma_Y} r = σ X , σ Y C o v ( X , Y )
= E ( X Y ) − E ( X ) E ( Y ) σ X . σ Y = 6011 − 2003 ( 3 ) 2 . 2 = 2 2 = 1 =\dfrac{E(XY)-E(X)E(Y)}{\sigma_X.\sigma_Y}
\\[9pt]
=\dfrac{6011-2003(3)}{\sqrt{2}.\sqrt{2}}\\[9pt]
=\dfrac{2}{2}=1 = σ X . σ Y E ( X Y ) − E ( X ) E ( Y ) = 2 . 2 6011 − 2003 ( 3 ) = 2 2 = 1
(c) spearmen rank correlation coefficient-
= 1 − 6 d 2 n 3 − n =1-\dfrac{6d^2}{n^3-n} = 1 − n 3 − n 6 d 2
where, d= Difference between x and Y
= 1 − 6 × 20000000 ( 5 ) 3 − 5 =1-\dfrac{6\times 20000000}{(5)^3-5} = 1 − ( 5 ) 3 − 5 6 × 20000000
= 1 − 1200000000 120 = 1 − 1 0 6 =1-\dfrac{1200000000}{120}=1-10^6 = 1 − 120 1200000000 = 1 − 1 0 6
Comments