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Exercises · Q8

Q.The two regression lines for a bivariate distribution are 3X+2Y=263X+2Y=26 and 6X+Y=316X+Y=31. Find

(i) the means of XX and YY, and
(ii) the coefficient of correlation between XX and YY.
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Step 1 — Solve for the means. Both regression lines pass through (Xˉ,Yˉ)(\bar X,\bar Y), so solve the two equations simultaneously:

3X+2Y=26...(i)6X+Y=31...(ii)3X+2Y=26 \quad \text{...(i)} \qquad 6X+Y=31 \quad \text{...(ii)}

From (ii): Y=31−6XY=31-6X. Substituting into (i): 3X+2(31−6X)=26⇒3X+62−12X=26⇒−9X=−36⇒X=43X+2(31-6X)=26 \Rightarrow 3X+62-12X=26 \Rightarrow -9X=-36 \Rightarrow X=4.

Then Y=31−6(4)=31−24=7Y=31-6(4)=31-24=7. So Xˉ=4\bar X=4, Yˉ=7\bar Y=7.

Step 2 — Identify which line is which. Test both assignments, since the problem does not label them, and only one assignment is mathematically valid (the product of coefficients cannot exceed 1).

Assignment A: treat (i) as YY on XX: 2Y=26−3X⇒Y=13−1.5X⇒byx=−1.52Y=26-3X \Rightarrow Y=13-1.5X \Rightarrow b_{yx}=-1.5. Treat (ii) as XX on YY: 6X=31−Y⇒X=316−16Y⇒bxy=−16≈−0.1676X=31-Y \Rightarrow X=\frac{31}{6}-\frac{1}{6}Y \Rightarrow b_{xy}=-\frac{1}{6}\approx-0.167.

Check: byx⋅bxy=(−1.5)(−0.167)=0.25≤1b_{yx}\cdot b_{xy}=(-1.5)(-0.167)=0.25 \le 1 ✓ valid. …

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