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प्रश्न
Given the following information about the production and demand of a commodity.
Obtain the two regression lines:
| Production (X) |
Demand (Y) |
|
| Mean | 85 | 90 |
| Variance | 25 | 36 |
Coefficient of correlation between X and Y is 0.6. Also estimate the demand when the production is 100 units.
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उत्तर
Given, `bar(x)` = 85, `bar(y)` = 90, `sigma_x^2` = 25, `sigma_y^2` = 36, r = 0.6
∴ `sigma_x` = 5, `sigma_y` = 6
byx = `"r" sigma_y/sigma_x = 0.6 xx 6/5` = 0.72
bxy = `"r" sigma_x/sigma_y = 0.6 xx 5/6` = 0.5
The regression equation of Y on X is given by `("Y" - bary) = "b"_(xy) ("X" - barx)`
(Y – 90) = 0.72(X – 85)
Y – 90 = 0.72X – 61.2
Y = 0.72X – 61.2 + 90
Y = 28.8 + 0.72X ......(i)
The regression equation of X on Y is given by `("X" - barx) = "b"_(xy) ("Y" - bary)`
(X – 85) = 0.5(Y – 90)
X – 85 = 0.5Y – 45
X = 0.5Y – 45 + 85
X = 40 + 05Y ......(ii)
For X = 100, from equation (i) we get
Y = 28.8 + 0.72(100)
= 28.8 + 72
= 100.8
∴ The production is 90 when demand is 100.
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| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
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| 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`
Mean of x = 25
Mean of y = 20
`sigma_x` = 4
`sigma_y` = 3
r = 0.5
byx = `square`
bxy = `square`
when x = 10,
`y - square = square (10 - square)`
∴ y = `square`
