Regression equations are 3x 5y 13 and 2x y 7 what will be the estimate value of x when y 10

(i) Given equations of regression are

3x + 2y - 26 = 0

i.e., 3x + 2y = 26       ….(i)

and 6x + y - 31 = 0

i.e., 6x + y = 31       ….(ii)

By (i) - 2 × (ii), we get

3x + 2y = 26

12x + 2y = 62

-      -       -    
- 9x = - 36

∴ x = `(-36)/-9 = 4`

Substituting x = 4 in (ii), we get

6 × 4 + y = 31

∴ 24 + y = 31

∴ y = 31 - 24

∴ y = 7

Since the point of intersection of two regression lines is `(bar x, bar y)`, 

`bar x` = mean of X = 4, and

`bar y` = mean of Y = 7

(ii) Let 3x + 2y 26 = 0 be the regression equation of Y on X.

∴ The equation becomes 2Y = 3X + 26

i.e., Y = `(-3)/2"X" + 26/2`

Comparing it with Y = bYX X + a, we get

`"b"_"YX" = (- 3)/2`

Now, the other equation 6x + y - 31 = 0 is the regression equation of X on Y.

∴ The equation becomes 6X = - Y + 31

i.e., X = `(-1)/6 "Y" + 31/6`

Comparing it with X = bXY Y+ a', we get

`"b"_"XY" = (-1)/6`

∴ r = `+-sqrt("b"_"XY" * "b"_"YX")`

`= +- sqrt((- 1)/6 xx (- 3)/2) = +- sqrt(1/4) = +- 1/2 = +- 0.5`

Since the values of bXY and bYX are negative,

r is also negative.

∴ r = - 0.5

(iii) The regression equation of Y on X is

Y = `(- 3)/2 "X" + 26/2`

For X = 2, we get

Y = `(- 3)/2 xx 2 + 26/2 = - 3 + 13 = 10`

(iv) Given, Var (Y) = 36, i.e., `sigma_"Y"^2` = 36

∴ `sigma_"Y" = 6`

Since `"b"_"XY" = "r" xx sigma_"X"/sigma_"Y"`

`(- 1)/6 = - 0.5 xx sigma_"X"/6`

∴ `sigma_"X" = (-6)/(- 6 xx 0.5) = 2`

∴ `sigma_"X"^2` = Var(X) = 4

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