For a bivariate date, you are given following information:
` sum x = 125 , sum x^(2) = 1650 , sum y = 100 , sum y^(2) = 1500 , sum xy= 50 , n= 25`
Find line of regression.

Answer

To find the line of regression for the given bivariate data, we will use the formulas for the coefficients \( a \) and \( b \) in the regression equation \( y = a + bx \).

### Step 1: Identify the given values
We are given:
- \( \Sigma x = 125 \)
- \( \Sigma x^2 = 1650 \)
- \( \Sigma y = 100 \)
- \( \Sigma y^2 = 1500 \)
- \( \Sigma xy = 50 \)
- \( n = 25 \)

### Step 2: Calculate the slope \( b \)
The formula for \( b \) is:
\[ b = \frac{n \Sigma xy - \Sigma x \Sigma y}{n \Sigma x^2 - (\Sigma x)^2} \]

Substituting the given values:
\[ b = \frac{25 \cdot 50 - 125 \cdot 100}{25 \cdot 1650 - (125)^2} \]

Calculating the numerator:
\[ 25 \cdot 50 = 1250 \]
\[ 125 \cdot 100 = 12500 \]
Thus,
\[ n \Sigma xy - \Sigma x \Sigma y = 1250 - 12500 = -11250 \]

Calculating the denominator:
\[ 25 \cdot 1650 = 41250 \]
\[ (125)^2 = 15625 \]
Thus,
\[ n \Sigma x^2 - (\Sigma x)^2 = 41250 - 15625 = 25625 \]

Now substituting back into the formula for \( b \):
\[ b = \frac{-11250}{25625} \approx -0.439 \]

### Step 3: Calculate the intercept \( a \)
The formula for \( a \) is:
\[ a = \frac{\Sigma y \Sigma x^2 - \Sigma x \Sigma xy}{n \Sigma x^2 - (\Sigma x)^2} \]

Substituting the values:
\[ a = \frac{100 \cdot 1650 - 125 \cdot 50}{25 \cdot 1650 - (125)^2} \]

Calculating the numerator:
\[ 100 \cdot 1650 = 165000 \]
\[ 125 \cdot 50 = 6250 \]
Thus,
\[ \Sigma y \Sigma x^2 - \Sigma x \Sigma xy = 165000 - 6250 = 158750 \]

Now substituting back into the formula for \( a \):
\[ a = \frac{158750}{25625} \approx 6.194 \]

### Step 4: Write the regression equation
Now that we have \( a \) and \( b \), we can write the equation of the regression line:
\[ y = a + bx \]
Substituting the values of \( a \) and \( b \):
\[ y = 6.194 - 0.439x \]

### Final Answer
The equation of the regression line is:
\[ y = 6.194 - 0.439x \]