- JEE Questions
- If The Lines Of Regression Of Y On X And That Of X On Y Are Y Kx 4 And X 4y 5
1) k ≤ 0
2) k ≥ 0
3) 0 ≤ k ≤ 1 / 4
4) 0 ≤ k ≤ 1
Solution: (3) 0 ≤ k ≤ 1 / 4
m1m2 ≤ 1
k . 4 ≤ 1
k ≤ 1 / 4
Also, m1 and m2 must be of the same sign.
k ≥ 0
Hence 0 ≤ k ≤ 1 / 4
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if x is equal to 4 Y + fine and where is Dukh Express Pora the lines of regression or nationwide and Y on X respectively where is positive constant prove that it can take one line of regression of Y on X is given by Y is equal to X + 10 the line of origin of x on Y given by x is equal to four ok sunao general equation of a lower end of line of regression of Y on X is given by Y minus Y is equal to X
2 x minus 2 equations that Y is equal to b y x y x x we can say that the value of Y is equal to the equation of line of regression of X and Y is given by x minus 6 is equal to b h u
x x minus y again comparing the equation we can say that the value of x is equal to four ok sunao relation which is the product of V X Y and Z to put in the values we can say that ok is equal to cannot take need one cause I always like between -1 and 1
question we can say that focus on to be less than equal to 1
Given equations of regression lines are
x - 4y = 5 …(i)
16y - x = 64
i.e., - x + 16y = 64 …(ii)
Adding (i) and (ii), we get
x - 4y = 5
- x + 16y = 64
12y = 69
∴ y = `69/12 = 5.75`
Substituting y = 5.75 in (i), we get
x - 4(5.75) = 5
∴ x - 23 = 5
∴ x = 5 + 23 = 28
Since the point of intersection of two regression lines is `(bar x, bar y)`,
∴ `bar x = 28 and bar y = 5.75`
Let, x - 4y = 5 be the regression equation of X on Y
∴ The equation becomes X = 4Y + 5
Comparing it with X = bXY Y + a', we get
bXY = 4
Now, the other equation i.e. 16y - x = 64 is regression equation of Y on X
∴ The equation becomes 16Y = X + 64
i.e., Y = `1/16 "X" + 64/16`
Comparing it with Y = bYX X + a, we get
`"b"_"YX" = 1/16`
r = `+-sqrt("b"_"XY" * "b"_"YX")`
`= +- sqrt(4 xx 1/16) = +- sqrt(1/4) = +- 1/2 = +- 0.5`
Since bXY and bYX both are positive,
r is also positive.
∴ r = 0.5