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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 mean marks in accountancy is 44 and the variance of marks in statistics is `(9/16)^(th)` of the variance of marks in accountancy. Find the mean marks in statistics and the correlation coefficient between marks in the two subjects.
उत्तर
Given n = 50
Regression line of x on y is
3y – 5x + 180 = 0
⇒ 5x = `[3y + 180]/5`
∴ x = `3/5 y + 180/5`
∴ `b_(xy) = "coefficient of y" = 3/5`
Variance of marks in statistics = `9/16`
Variance of marks in accountancy.
i.e. V(x) = `9/16`V(y)
⇒ `V_(x)/V_(y) = 9/16`
Taking square roots,
`(σ_x)/(σ_y) = 3/4`
We have, `b_(xy) = r. (σ_x)/(σ_y)`
⇒ `3/5 = r xx 3/4`
⇒ `r = 4/5 = 0.8`
Given `bary` = 44
∴ Substituting y = 44 is 3y – 5x + 180 = 0
⇒ 3(44) – 5x + 18 = 0
⇒ 132 – 5x + 180 = 0
⇒ 5x = 132 + 180 = 312
⇒ x = `312/5 ` = 62.4
∴ `barx` = 62.4
∴ Mean marks in statistics `barx` = 62.4
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