If `r=0.997,sumxy=46,bar(X)=4,bar(Y)=8,sumx^(2)=28`, what will be the value of `sumy^(2)`?

If `r=0.997,sumxy=46,bar(X)=4,bar(Y)=8,sumx^(2)=28`, what will be the

1.	If `r=0.997,sumxy=46,bar(X)=4,bar(Y)=8,sumx^(2)=28`, what will be the value of `sumy^(2)`?
Answer» `r=(sumxy)/(sqrt(sumx^(2)xxsumy^(2)))` Given : `r=0.997, sum xyl =46 , bar (X)=4, bar (Y)=8, sum x^(2)=28` Substituting the values, we get `0.997=(46)/(sqrt(28xx sumy^(2)))` Squaring both side , we get `0.994=(2,116)/(28xxsumy^(2))` `implies 27.832xx sumy^(2)=2,116` `sumy^(2)=76.03` Value of `sumy^(2)=76.03`