From the data given below, find the number of items (N), r=0.5,`sumxy=120`, Standard Deviation of `Y(sigma_(y))=8,sumx^(2)=90` where , x and y are deviations from arithmetic mean.

From the data given below, find the number of items (N), r=0.5,`sumxy=

1.	From the data given below, find the number of items (N), r=0.5,`sumxy=120`, Standard Deviation of `Y(sigma_(y))=8,sumx^(2)=90` where , x and y are deviations from arithmetic mean.
Answer» Given `: r=0.5,sumxy=120,sumx^(2)=90, sigma_(y)=8` Now, ` sigma_(y)=sqrt((sumy^(2))/(N))` when `y=Y-bar(Y)` [formula of Standard Deviation ] `8=sqrt((sumy^(2))/(N))` Squaring both sides, we get `64=(sumy^(2))/(N) implies sumy^(2)=64N` Now, `r=(sumxy)/(sqrt(sumx^(2)xxsumy^(2)))` `implies 0.5=(120)/(sqrt(90xx64N))` Squaring both sides `0.25=((120)^(2))/(90xx64N)` `implies0.25=(14,400)/(5,760 N)` `implies (0.25)(5,760)N=14,400` `implies(1,440)N=14,400` `:. N=(14,400)/(1,440)=10` Number of Items=10