If the two lines of regression are `4x-5y+33=0and20x-9y-107=0`, find the variance of y when `sigma_(x)=3`.

Answer

To find the variance of \( y \) given the two lines of regression and \( \sigma_x = 3 \), we can follow these steps:

### Step 1: Write the equations of the lines of regression
The two lines of regression are given as:
1. \( 4x - 5y + 33 = 0 \)
2. \( 20x - 9y - 107 = 0 \)

### Step 2: Rearrange the first line to find \( b_{xy} \)
From the first line, we can express \( y \) in terms of \( x \):
\[ 4x - 5y + 33 = 0 \implies 5y = 4x + 33 \implies y = \frac{4}{5}x + \frac{33}{5} \]
Thus, the slope \( b_{xy} = \frac{4}{5} \).

### Step 3: Rearrange the second line to find \( b_{yx} \)
From the second line, we can express \( y \) in terms of \( x \):
\[ 20x - 9y - 107 = 0 \implies 9y = 20x - 107 \implies y = \frac{20}{9}x - \frac{107}{9} \]
Thus, the slope \( b_{yx} = \frac{20}{9} \).

### Step 4: Calculate the correlation coefficient \( r \)
The correlation coefficient \( r \) can be calculated using the formula:
\[ r = \sqrt{b_{xy} \cdot b_{yx}} \]
Substituting the values:
\[ r = \sqrt{\left(\frac{4}{5}\right) \cdot \left(\frac{20}{9}\right)} = \sqrt{\frac{80}{45}} = \sqrt{\frac{16}{9}} = \frac{4}{3} \]

### Step 5: Use the formula for \( b_{xy} \)
We know that:
\[ b_{xy} = r \cdot \frac{\sigma_x}{\sigma_y} \]
Substituting the known values:
\[ \frac{4}{5} = \frac{4}{3} \cdot \frac{3}{\sigma_y} \]

### Step 6: Solve for \( \sigma_y \)
Cross-multiplying gives:
\[ 4 \cdot \sigma_y = 5 \cdot 3 \implies 4 \cdot \sigma_y = 15 \implies \sigma_y = \frac{15}{4} \]

### Step 7: Calculate the variance of \( y \)
The variance \( \sigma_y^2 \) is given by:
\[ \sigma_y^2 = \left(\frac{15}{4}\right)^2 = \frac{225}{16} \]

### Final Answer
Thus, the variance of \( y \) is:
\[ \boxed{\frac{225}{16}} \]