**– Average Midterm Score = 70**

**– Standard Deviation of Midterm Score = 10**

**– Average Final Score = 70**

**– Standard Deviation of Final Score = 20**

**– Correlation Coefficient of Final Score = 0.60**

The **aim of this question** is to use the **linear regression model** to find the **dependence** of one variable on the other and then apply this model for **prediction**.

The **linear regression model** relating a variable x with a variable y can be** defined by the following formula:**

\[ y \ = \ m x \ + \ c \]

The **slope and intercept** used in above model can be calculated using the following formula:

\[ \text{ Slope } = \ m \ = r \ \dfrac{ \sigma_{ y } }{ \sigma_{ x } } \]

\[ \text{ y-intercept } = \ c \ = \ \mu_{ y} \ – \ m \mu_{ x } \]

## Expert Answer

Let’s call the **midterm score** $ x $, which is the** independent variable,** while the **final score** $ y $ is the **dependent variable**. In this case, the **given data** may be represented as follows:

\[ \text{ Average Midterm Score } = \ \mu_{ x } \ = \Â 70 \]

\[ \text{ Standard Deviation of Midterm Score } = \ \sigma_{ x } \ = \ 10 \]

\[ \text{ Average Final Score } = \ \mu_{ y } \ = \ 70 \]

\[ \text{ Standard Deviation of Final Score } = \ \sigma_{ y } \ = \ 20 \]

\[ \text{ Correlation Coefficient of Final Score } = \ r \ = \ 0.60 \]

For the case of **linear regression**, the **slope of the equation** can be calculated using the following formula:

\[ \text{ Slope } = \ m \ = r \ \dfrac{ \sigma_{ y } }{ \sigma_{ x } } \]

**Substituting values in the above equation:**

\[ m \ = 0.6 \ \dfrac{ 20 }{ 10 } \]

\[ m \ = 0.6 \times 2 \]

\[ m \ = 1.2 \]

For the case of **linear regression**, the **y-intercept of the equation** can be calculated using the following formula:

\[ \text{ y-intercept } = \ c \ = \ \mu_{ y} \ – \ m \mu_{ x } \]

**Substituting values in the above equation:**

\[ \text{ y-intercept } = \ c \ = \ 55 \ – \ ( 1.2 ) ( 70 ) \]

\[ \text{ y-intercept } = \ c \ = \ 55 \ – \ 84 \]

\[ \text{ y-intercept } = \ c \ = \ -29 \]

**So the final equation of linear regression is:**

\[ y \ = \ m x \ + \ c \]

**Substituting values in the above equation:**

\[ y \ = \ 1.2 x \ – \ 29 \]

Which is the **required result.**

## Numerical Result

\[ y \ = \ 1.2 x \ – \ 29 \]

## Example

Using the **above regression equation**, find the final **score of a student** that scored **50 marks in the mid term**.

**Given:**

\[ x \ = \ 50 \]

**Recall the linear regression equation:**

\[ y \ = \ 1.2 x \ – \ 29 \]

**Substituting the value of $ x $:**

\[ y \ = \ 1.2 ( 50 ) \ – \ 29 \]

\[ y \ = \ 60 \ – \ 29 \]

\[ y \ = \ 31 \]

Which is the **required result.**