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प्रश्न
In a partially destroyed record, the following data are available: variance of X = 25, Regression equation of Y on X is 5y − x = 22 and regression equation of X on Y is 64x − 45y = 22 Find
- Mean values of X and Y
- Standard deviation of Y
- Coefficient of correlation between X and Y.
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उत्तर
Given, `sigma_"X"^2 = 25`
∴ `sigma_"X"` = 5
Regression equation of Y on X is
5y – x = 22
Regression equation of X on Y is
64x - 45y = 22
(i) Consider, the two regression equation
- x + 5y = 22 ....(i)
64x - 45y = 22 ....(ii)
By (i) × + (ii), we get
- 9x + 45y = 198
+ 64x - 45y = 22
55x = 220
∴ x = 4
Substituting x = 4 in (i), we get
- 4 + 5y = 22
∴ 5y = 22 + 4
∴ y = `26/5 = 5.2`
Since the point of intersection of two regression lines is `(bar x, bar y)`,
`bar x` = mean value of X = 4 and
`bar y` = mean value of Y = 5.2
(ii) To find standard deviation of Y we should first find the coefficient of correlation between X and Y.
Regression equation of Y on X is
5y - x = 22
i.e., 5Y = X + 22
i.e., Y = `"X"/5 + 22/5`
Comparing it with Y = bYX X + a, we get
`"b"_"YX" = 1/5`
Now, regression equation of X on Y is
64x - 45y = 22
i.e., 64X - 45Y = 22
i.e., 64X = 45Y + 22
i.e., X = `"45Y"/64 + 22/64`
Comparing it with X = bXY Y + a', we get
`"b"_"XY" = 45/64`
r = `+-sqrt("b"_"XY" * "b"_"YX")`
`= +- sqrt((1/5)(45/64)) = +- sqrt(9/64) = +- 3/8`
Since bYX and bXY are positive,
r is also positive.
∴ r = `3/8= 0.375`
∴ `sigma_"Y"`= Standard deviation of Y = 0.375
(iii) The correlation coefficient of X and Y =
Now, `"b"_"YX" = ("r". sigma_"Y")/sigma_"X"`
∴ `1/5 = 3/8 xx sigma_"Y"/5`
∴ `sigma_"Y" = 8/3`
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