Question

Calculate the number of items, when
(i) Standard deviation of series Y = 10
(ii) Coefficient of correlation = 0.6
(iii) Sum of the product of deviation of X and Y from actual means = 150
(iv) Sum of squares of deviations of X from actual means = 125

Answer 1

$$\begin{gathered} \mathrm{Given},\mathrm{\sigma y}=10 \\ \Rightarrow {\mathrm{\sigma }}_{\mathrm{y}}^2=100 \\ r=0.6 \\ \mathrm{\Sigma }xy=150 \\ \mathrm{\Sigma }{x}^2=125 \\ {\mathrm{\sigma }}_y^2=\frac{\mathrm{\Sigma }{y}^2}{\mathrm{N}} \\ 100=\frac{{\mathrm{\Sigma y}}^2}{\mathrm{N}} \\ \Rightarrow {\mathrm{\Sigma y}}^2=100\mathrm{N} \\ \mathrm{Now}, \\ r=\frac{\mathrm{\Sigma }xy}{\sqrt{\mathrm{\Sigma }{x}^2\times \mathrm{\Sigma }{y}^2}} \\ 0.6=\frac{150}{\sqrt{125}\sqrt{100\mathrm{N}}} \\ 0.6=\frac{150}{\sqrt{12500\mathrm{N}}} \end{gathered}$$

Squaring both sides, we get
$$\begin{gathered} 0.36=\frac{22500}{12500\mathrm{N}} \\ \Rightarrow \mathrm{N}=5 \end{gathered}$$

Hence, number of items = 5