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प्रश्न
Given the following information about the production and demand of a commodity.
Obtain the two regression lines:
| ADVERTISEMENT (x) (₹ in lakhs) |
DEMAND (y) (₹ in lakhs) |
|
| Mean | 10 | 90 |
| Variance | 9 | 144 |
Coefficient of correlation between x and y is 0.8.
What should be the advertising budget if the company wants to attain the sales target of ₹ 150 lakhs?
उत्तर
Given, `bar(x)` = 10, `bar(y)` = 90, `sigma_x^2` = 9, `sigma_y^2` = 144, r = 0.8
∴ `sigma_x` = 3, `sigma_y` = 12
byx = `"r" sigma_y/sigma_x = 0.8 xx 12/3` = 0.8 × 4 = 3.2
bxy = `"r" sigma_x/sigma_y = 0.8 xx 3/12` = 0.8 × 0.25 = 0.2
The regression equation of Y on X is
`("Y" - bary) = "b"_(yx) ("X" - barx)`
∴ (Y – 90) = 3.2 (X – 10)
∴ Y – 90 = 3.2 X – 32
∴ Y = 3.2 X – 32 + 90
∴ Y = 3.2 X + 58 ......(i)
The regression equation of X on Y is
`("X" - barx) = "b"_(xy) ("Y" - bary)`
∴ (X – 10) = 0.2 (Y – 90)
∴ X – 10 = 0.2 Y – 18
∴ X = 0.2 Y – 18 + 10
∴ X = 0.2 Y – 8 ......(ii)
When the company wants to attain the sales target of ₹ 150 lakhs,
Put Y = 150 lakh in equation (ii)
∴ X = 0.2 × 150 – 8 = 30 – 8 = 22
∴ The advertising budget should be ₹ 22 lakhs if the company wants to attain the sales target of ₹ 150 lakhs.
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| 1 | 5 | – 2 | – 4 | 8 | 4 | 16 |
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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 = 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`
The regression equation of y on x is 2x – 5y + 60 = 0
Mean of x = 18
`2 square - 5 bary + 60` = 0
∴ `bary = square`
`sigma_x : sigma_y` = 3 : 2
∴ byx = `square/square`
∴ byx = `square/square`
∴ r = `square`
