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Question
The two regression lines between height (X) in inches and weight (Y) in kgs of girls are,
4y − 15x + 500 = 0
and 20x − 3y − 900 = 0
Find the mean height and weight of the group. Also, estimate the weight of a girl whose height is 70 inches.
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Solution
Given, X = Height (in inches), Y = weight (in Kg)
The equation of regression are
4y - 15x + 500 = 0
i.e., –15x + 4y = – 500 …(i)
and 20x – 3y – 900 = 0
i.e., 20x – 3y = 900 …(ii)
By 3 × (i) + 4 × (ii), we get
- 45x + 12y = - 1500
+ 80x - 12y = 3600
35x = 2100
∴ x = 60
Substituting x = 60 in (i), we get
–15(60) + 4y = –500
∴ 4y = 900 – 500
∴ y = 100
Since the point of intersection of two regression lines is `bar x, bar y`,
`bar x` = mean height of the group = 60 inches, and
`bar y` = mean weight of the group = 100 kg.
Let 4y – 15x + 500 = 0 be the regression equation of Y on X.
∴ The equation becomes 4y = 15x – 500
i.e., Y = `15/4"X" - 500/4` ...(i)
Comparing it with Y = bYX X + a, we get
∴ `"b"_"YX" = 15/4`
∴ Now, other equation 20x – 3y – 900 = 0 be the regression equation of X on Y
∴The equation becomes 20x – 3y – 900 = 0
i.e., 20x = 3y + 900
X = `3/20"Y" + 900/20`
Comparing it with X = bXY Y + a',
∴ `"b"_"XY" = 3/20`
Now, `"b"_"YX" * "b"_"XY" = 15/4 * 3/20 = 0.5625`
i.e., bXY . bYX < 1
∴ Assumption of regression equations is true.
Now, substituting x = 70 in (i) we get
y = `15/4 xx 70 - 500/4 = (1050 - 500)/4 = 550/4 = 137.5`
∴ Weight of girl having height 70 inches is 137.5 kg
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