A) \[0.6;\text{ }16\]
B) \[0.16;\text{ }16\]
C) \[0.3;\text{ }4\]
D) \[0.6;\text{ }4\]
Correct Answer: A
Solution :
Given that, \[\sigma _{x}^{2}=9\] The given lines of regression are \[4x-5y+33=0,\text{ }20x-9y-10=0\] ie, \[y=\frac{4}{5}x+\frac{33}{5}\]and\[x=\frac{9}{20}y+\frac{10}{20}\] \[\therefore \]Regression coefficients are \[{{b}_{yx}}=\frac{4}{5}\]and\[{{b}_{xy}}=\frac{9}{20}\] Now, \[{{b}_{yx}}=\frac{\operatorname{cov}(x,y)}{\sigma _{x}^{2}}\] \[\Rightarrow \] \[\operatorname{cov}(x,y)=\frac{4}{5}\times 9=\frac{36}{5}\] and \[{{b}_{xy}}=\frac{\operatorname{cov}(x,y)}{\sigma _{y}^{2}}\] \[\Rightarrow \] \[\sigma _{y}^{2}=\frac{36}{5}\times \frac{20}{9}=16\] Again Now\[\rho (x,y)=\frac{\operatorname{cov}(x,y)}{{{\sigma }_{x}}.{{\sigma }_{y}}}\] \[=\frac{36}{5\times 3\times 4}=0.6\] Hence coefficient of correlation = 0.6 and variance of \[y=16\]
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