If yi=axi+b for each i=1,2,3,........n, then

Options

A. `overset(-)(y)=aoverset(-)(x)`
B. `overset(-)(y)=a overset(-)(x)+b`
C. `sigma y=sigma x`
D. `sigma y=|a| sigma x`

Answer

To solve the problem where \( y_i = ax_i + b \) for each \( i = 1, 2, 3, \ldots, n \), we need to derive the relationships for the mean and standard deviation of the \( y \) values in terms of the \( x \) values.

### Step 1: Calculate the Mean of \( y \)

1. **Write the expression for the mean \( \bar{y} \)**:
\[ \bar{y} = \frac{y_1 + y_2 + \ldots + y_n}{n} = \frac{(ax_1 + b) + (ax_2 + b) + \ldots + (ax_n + b)}{n} \]

2. **Simplify the expression**:
\[ \bar{y} = \frac{a(x_1 + x_2 + \ldots + x_n) + nb}{n} \]
\[ \bar{y} = a \left(\frac{x_1 + x_2 + \ldots + x_n}{n}\right) + b \]
\[ \bar{y} = a \bar{x} + b \]

### Step 2: Calculate the Standard Deviation of \( y \)

1. **Write the expression for the standard deviation \( \sigma_y \)**:
\[ \sigma_y = \sqrt{\frac{(y_1 - \bar{y})^2 + (y_2 - \bar{y})^2 + \ldots + (y_n - \bar{y})^2}{n}} \]

2. **Substituting \( y_i \) and \( \bar{y} \)**:
\[ \sigma_y = \sqrt{\frac{((ax_1 + b) - (a \bar{x} + b))^2 + ((ax_2 + b) - (a \bar{x} + b))^2 + \ldots + ((ax_n + b) - (a \bar{x} + b))^2}{n}} \]

3. **Simplifying the expression**:
\[ \sigma_y = \sqrt{\frac{(ax_1 - a \bar{x})^2 + (ax_2 - a \bar{x})^2 + \ldots + (ax_n - a \bar{x})^2}{n}} \]
\[ = \sqrt{\frac{a^2((x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \ldots + (x_n - \bar{x})^2)}{n}} \]
\[ = |a| \sqrt{\frac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \ldots + (x_n - \bar{x})^2}{n}} \]
\[ = |a| \sigma_x \]

### Conclusion

From the calculations, we have derived two important relationships:

1. The mean of \( y \):
\[ \bar{y} = a \bar{x} + b \]

2. The standard deviation of \( y \):
\[ \sigma_y = |a| \sigma_x \]