A cubical die is thrown. Find the Mean and variance of X, giving the number on the face that shows up.

Answer

To find the mean and variance of the random variable \( X \), which represents the number on the face of a thrown cubical die, we can follow these steps:

### Step 1: Identify the Random Variable and its Probability Distribution
The random variable \( X \) can take values from 1 to 6, corresponding to the faces of the die. The probability \( P(X = x) \) for each face is equal since the die is fair.

\[ P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = \frac{1}{6} \]

### Step 2: Calculate the Mean (Expected Value)
The mean \( \mu \) (or expected value \( E(X) \)) is calculated using the formula:

\[ E(X) = \sum_{i=1}^{n} x_i P(X = x_i) \]

Substituting the values:

\[ E(X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \]

\[ E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} \]

### Step 3: Calculate the Sum of the First 6 Natural Numbers
The sum of the first \( n \) natural numbers is given by the formula:

\[ \text{Sum} = \frac{n(n + 1)}{2} \]

For \( n = 6 \):

\[ \text{Sum} = \frac{6 \cdot (6 + 1)}{2} = \frac{6 \cdot 7}{2} = 21 \]

### Step 4: Substitute Back to Find the Mean
Now substituting back into the mean formula:

\[ E(X) = \frac{21}{6} = \frac{7}{2} \]

### Step 5: Calculate the Variance
The variance \( \sigma^2 \) is calculated using the formula:

\[ \sigma^2 = E(X^2) - (E(X))^2 \]

#### Step 5.1: Calculate \( E(X^2) \)
First, we need to find \( E(X^2) \):

\[ E(X^2) = \sum_{i=1}^{n} x_i^2 P(X = x_i) \]

Calculating \( E(X^2) \):

\[ E(X^2) = 1^2 \cdot \frac{1}{6} + 2^2 \cdot \frac{1}{6} + 3^2 \cdot \frac{1}{6} + 4^2 \cdot \frac{1}{6} + 5^2 \cdot \frac{1}{6} + 6^2 \cdot \frac{1}{6} \]

\[ E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} \]

#### Step 5.2: Calculate the Sum of Squares
Calculating the sum:

\[ 1 + 4 + 9 + 16 + 25 + 36 = 91 \]

So,

\[ E(X^2) = \frac{91}{6} \]

#### Step 5.3: Substitute to Find Variance
Now, substituting into the variance formula:

\[ \sigma^2 = E(X^2) - (E(X))^2 = \frac{91}{6} - \left(\frac{7}{2}\right)^2 \]

Calculating \( (E(X))^2 \):

\[ \left(\frac{7}{2}\right)^2 = \frac{49}{4} \]

Convert \( \frac{49}{4} \) to have a common denominator with \( \frac{91}{6} \):

\[ \frac{49}{4} = \frac{49 \cdot 3}{4 \cdot 3} = \frac{147}{12} \]

Convert \( \frac{91}{6} \):

\[ \frac{91}{6} = \frac{91 \cdot 2}{6 \cdot 2} = \frac{182}{12} \]

Now substituting back:

\[ \sigma^2 = \frac{182}{12} - \frac{147}{12} = \frac{35}{12} \]

### Final Results
- Mean \( E(X) = \frac{7}{2} \)
- Variance \( \sigma^2 = \frac{35}{12} \)