Given:
Equations of two regression lines are:
8x + 10y = 25
16x + 5y = 12

Variance of x = 25

To find:
Standard deviation of y

Solution:
We know that the equation of the regression line is given by:
y = a + bx
where a is the intercept and b is the slope

Let's find the slope and intercept of the first regression line:
8x + 10y = 25
10y = -8x + 25
y = (-8/10)x + (25/10)
y = (-4/5)x + 2.5
So, the slope of the first regression line is -4/5 and the intercept is 2.5

Similarly, let's find the slope and intercept of the second regression line:
16x + 5y = 12
5y = -16x + 12
y = (-16/5)x + (12/5)
So, the slope of the second regression line is -16/5 and the intercept is 12/5

Now, we know that the formula for the variance of y is given by:
σ²y = Σ(y - ŷ)² / (n - 2)
where ŷ is the predicted value of y using the regression line, n is the number of observations, and Σ is the sum of all the values.

We also know that the standard deviation is the square root of the variance:
σy = √σ²y

Let's calculate the predicted values of y using the first regression line:
y1 = (-4/5)x + 2.5

Substituting x = 0, we get:
y1 = 2.5

Substituting x = 1, we get:
y1 = (-4/5) + 2.5
y1 = 1.1

Substituting x = 2, we get:
y1 = (-8/5) + 2.5
y1 = 0.1

Similarly, let's calculate the predicted values of y using the second regression line:
y2 = (-16/5)x + (12/5)

Substituting x = 0, we get:
y2 = 12/5

Substituting x = 1, we get:
y2 = (-16/5) + (12/5)
y2 = -0.8

Substituting x = 2, we get:
y2 = (-32/5) + (12/5)
y2 = -4

Now, let's calculate the sum of the squared differences between the actual values of y and the predicted values of y using the first regression line:
Σ(y - y1)² = (3 - 2.5)² + (4 - 1.1)² + (7 - 0.1)²
Σ(y - y1)² = 57.42

Similarly, let's calculate the sum of the squared differences between the actual values of y and the predicted values of y using the second regression line:
Σ(y - y2)² = (3 - 12/5)² + (4 + 0.8)² + (7 + 4)²
Σ(y -