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प्रश्न
For 50 students of a class, the regression equation of marks in statistics (X) on the marks in accountancy (Y) is 3y − 5x + 180 = 0. The variance of marks in statistics is `(9/16)^"th"` of the variance of marks in accountancy. Find the correlation coefficient between marks in two subjects.
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उत्तर
Given, n = 50, X = marks in Statistics,
Y = marks in Accountancy,
Regression equation of X on Y is
3y – 5x + 180 = 0,
`bar y = 44, sigma_X^2 = 9/16 sigma_Y^2`
Now, 3y – 5x + 180 = 0 is the regression equation of X on Y.
∴ The equation becomes 5X = 3Y + 180
i.e., X = `3/5` Y + `180/5`
Comparing it with X = bXY Y + a', we get
`b_(XY) = 3/5, a = 180/5` = 36
a = `barx - b_(XY) bary`
∴ 36 = `bar x - 3/5 xx 44`
∴ 36 = `bar x` – 26.4
∴ `bar x` = 36 + 26.4 = 62.4
Also, `sigma_X^2 = 9/16 sigma_Y^2`
∴ `sigma_X^2/sigma_Y^2 = 9/16`
∴ `sigma_X/sigma_Y = 3/4`
`b_(XY) = r xx sigma_X/sigma_Y`
∴ `3/5 = r xx 3/4`
∴ `3/5 xx 4/3` = r
∴ r = `4/5` = 0.8
∴ Mean marks in statistics `(barx)` are 62.4 and correlation coefficient (r) between marks in the two subjects is 0.8.
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संबंधित प्रश्न
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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 |
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