If r=0.25,sumxy=45,sigma_(y)=3,sumx^(2)=50, where x and y denote deviation from their respective means, find the number of items.

If r=0.25,sumxy=45,sigma_(y)=3,sumx^(2)=50, where x and y denote devia

1.	If r=0.25,sumxy=45,sigma_(y)=3,sumx^(2)=50, where x and y denote deviation from their respective means, find the number of items.
Answer» Solution :Given `: r=0.25,sumxy=45, sigma_(y)=3,sumx^(2)=50` Now, ` sigma_(y)=sqrt((sumy^(2))/(N))` when `y=Y-bar(Y)` [formula of STANDARD Deviation ] `3=sqrt((sumy^(2))/(N))` Squaring both sides, we GET `9=(sumy^(2))/(N) IMPLIES sumy^(2)=9N` Now, `r=(sumxy)/(sqrt(sumx^(2)xxsumy^(2)))` `implies 0.25=(45)/(sqrt(50xx9N))` Squaring both sides `0.0625=((45)^(2))/(50xx9N)` `0.0625=(2,025)/(50xx9N)` `implies 0.0625=(2,025)/(450N)` `implies (0.0625)(450)N)=2,025` `implies (28.125)N=2,025` `:.N=(2,025)/(28.125)=72` Number of Items =72