i) Since both the lines of regression pass through the point (\overline{X},\overline{Y}) , we have

8\overline{X}-10\overline{Y}+ 66 = 0 and 40\overline{X}-18\overline{Y} = 214

Solving, we get \overline{X} = 13, \overline{Y} = 17

ii) Let 8X - 10 Y + 66 = 0 and 40X - 18Y = 214 be the lines of regression of Y on X and X on Y respectively. These equations can be put in the form:

Y = \frac{8}{10}X + \frac{66}{10} and X = \frac{18}{40}Y + \frac{214}{40}

So,

b_{YX} = Regression coefficient of Y on X = \frac{8}{10} = \frac{4}{5}

and

b_{XY} = Regression coefficient of X on Y = \frac{18}{40} = \frac{9}{20}

Hence r^{2} = b_{YX}\cdot b_{XY} = \frac{4}{5}\cdot \frac{9}{20} = \frac{9}{25}

But since both the regression coefficients are positive we take r = +0.6

iii) b_{YX} = r \cdot \frac{\sigma_y}{\sigma_x}

\Rightarrow \frac{4}{5} = \frac{3}{5} \times \frac{\sigma_{y}}{3} (since \sigma_{x}^{2} = 9 (given)

So \sigma_{Y} = 4

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Answer in Statistics and Probability for Anurodh #55140

Sep 29, 2015 **...** In a **partially destroyed laboratory**, **record** of an **analysis** of **correlation** of **data**, only the following results are legible: Variance of X = 9

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