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.

Explanation

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