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Worked Examples · Example 2

Q.For a bivariate distribution, r=0.6r = 0.6, σx=5\sigma_x = 5, σy=4\sigma_y = 4, Xˉ=25\bar X = 25 and Yˉ=30\bar Y = 30. Find the two regression equations, and estimate YY when X=30X = 30.

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Step 1 — Regression coefficients from r,σx,σyr,\sigma_x,\sigma_y.

byx=r⋅σyσx=0.6×45=0.6×0.8=0.48b_{yx} = r\cdot\frac{\sigma_y}{\sigma_x} = 0.6 \times \frac{4}{5} = 0.6 \times 0.8 = 0.48

bxy=r⋅σxσy=0.6×54=0.6×1.25=0.75b_{xy} = r\cdot\frac{\sigma_x}{\sigma_y} = 0.6 \times \frac{5}{4} = 0.6 \times 1.25 = 0.75

Step 2 — Independent check. byx⋅bxy=0.48×0.75=0.36b_{yx}\cdot b_{xy} = 0.48\times 0.75 = 0.36, and 0.36=0.6=r\sqrt{0.36}=0.6=r ✓ — confirms both coefficients.

Step 3 — Regression equations.

YY on XX: Y−30=0.48(X−25)=0.48X−12⇒Y=0.48X+18Y - 30 = 0.48(X-25) = 0.48X - 12 \Rightarrow Y = 0.48X + 18

XX on YY: X−25=0.75(Y−30)=0.75Y−22.5⇒X=0.75Y+2.5X - 25 = 0.75(Y-30) = 0.75Y - 22.5 \Rightarrow X = 0.75Y + 2.5

Step 4 — Prediction. At X=30X=30: Y=0.48(30)+18=14.4+18=32.4Y = 0.48(30)+18 = 14.4+18 = 32.4.

✓Final answer

Y=0.48X+18Y = 0.48X+18, X=0.75Y+2.5X=0.75Y+2.5; estimated YY at X=30X=30 is 32.4\mathbf{32.4}.

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