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.

Explanation

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