If `4x - 5y + 33 = 0` and `20 x - 9y - 107 = 0` are two lines of regression, find the standard deviation of Y, the variance of x is 9.

Options

A. 4
B. 16
C. 0.4
D. 1.6

Answer

To solve the problem, we need to find the standard deviation of Y given the lines of regression and the variance of X. Let's break down the solution step by step.

### Step 1: Identify the equations of the lines of regression
We have two lines of regression:
1. \( 4x - 5y + 33 = 0 \)
2. \( 20x - 9y - 107 = 0 \)

### Step 2: Rearranging the equations
We can rearrange these equations to express \( y \) in terms of \( x \) and vice versa.

For the first equation:
\[ 4x - 5y + 33 = 0 \implies 5y = 4x + 33 \implies y = \frac{4}{5}x + \frac{33}{5} \]

For the second equation:
\[ 20x - 9y - 107 = 0 \implies 9y = 20x - 107 \implies y = \frac{20}{9}x - \frac{107}{9} \]

### Step 3: Identify the regression coefficients
From the equations, we can identify the regression coefficients:
- \( b_{yx} = \frac{4}{5} \) (slope of Y on X)
- \( b_{xy} = \frac{20}{9} \) (slope of X on Y)

### Step 4: Calculate the correlation coefficient \( r \)
The relationship between the regression coefficients and the correlation coefficient \( r \) is given by:
\[ r^2 = b_{yx} \cdot b_{xy} \]
Substituting the values:
\[ r^2 = \left(\frac{4}{5}\right) \cdot \left(\frac{20}{9}\right) = \frac{80}{45} = \frac{16}{9} \]
Taking the square root gives:
\[ r = \sqrt{\frac{16}{9}} = \frac{4}{3} \quad \text{(We will take the positive root since correlation is positive)} \]

### Step 5: Use the variance of X to find the standard deviation of Y
We know the variance of X is given as \( \sigma_x^2 = 9 \), which means:
\[ \sigma_x = 3 \]

Using the formula relating the standard deviations and the correlation coefficient:
\[ b_{yx} = r \cdot \frac{\sigma_y}{\sigma_x} \]
Substituting the known values:
\[ \frac{4}{5} = \frac{4}{3} \cdot \frac{\sigma_y}{3} \]

### Step 6: Solve for \( \sigma_y \)
Rearranging the equation:
\[ \frac{4}{5} = \frac{4 \sigma_y}{9} \implies 4 \sigma_y = \frac{4 \cdot 9}{5} \implies \sigma_y = \frac{36}{20} = \frac{9}{5} \]

### Step 7: Final result
Thus, the standard deviation of Y is:
\[ \sigma_y = \frac{9}{5} \approx 1.8 \]

### Conclusion
The standard deviation of Y is \( \sigma_y = 1.8 \).