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प्रश्न
The equations of two regression lines are x − 4y = 5 and 16y − x = 64. Find means of X and Y. Also, find correlation coefficient between X and Y.
उत्तर
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
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Sales (₹ in lakh) (Y) | |
| Arithmetic Mean | 10 | 90 |
| Standard Mean | 3 | 12 |
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| x | y | `x - barx` | `y - bary` | `(x - barx)(y - bary)` | `(x - barx)^2` | `(y - bary)^2` |
| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
| 2 | 7 | – 1 | – 2 | `square` | 1 | 4 |
| 3 | 9 | 0 | 0 | 0 | 0 | 0 |
| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
| 5 | 13 | 2 | 4 | 8 | 1 | 16 |
| Total = 15 | Total = 45 | Total = 0 | Total = 0 | Total = `square` | Total = 10 | Total = 40 |
Mean of x = `barx = square`
Mean of y = `bary = square`
bxy = `square/square`
byx = `square/square`
Regression equation of x on y is `(x - barx) = "b"_(xy) (y - bary)`
∴ Regression equation x on y is `square`
Regression equation of y on x is `(y - bary) = "b"_(yx) (x - barx)`
∴ Regression equation of y on x is `square`
Mean of x = 53
Mean of y = 28
Regression coefficient of y on x = – 1.2
Regression coefficient of x on y = – 0.3
a. r = `square`
b. When x = 50,
`y - square = square (50 - square)`
∴ y = `square`
c. When y = 25,
`x - square = square (25 - square)`
∴ x = `square`
bxy . byx = ______.
If byx > 1 then bxy is _______.
