A fair die is tossed. Let X denote twice the number appearing. Find the probability distribution, mean and variance of X.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the Sample Space
When a fair die is tossed, the possible outcomes (sample space) are:
\[ S = \{1, 2, 3, 4, 5, 6\} \]

### Step 2: Define the Random Variable X
The random variable \( X \) is defined as twice the number appearing on the die. Therefore, the possible values of \( X \) are:
- If the die shows 1, \( X = 2 \times 1 = 2 \)
- If the die shows 2, \( X = 2 \times 2 = 4 \)
- If the die shows 3, \( X = 2 \times 3 = 6 \)
- If the die shows 4, \( X = 2 \times 4 = 8 \)
- If the die shows 5, \( X = 2 \times 5 = 10 \)
- If the die shows 6, \( X = 2 \times 6 = 12 \)

Thus, the possible values of \( X \) are:
\[ X = \{2, 4, 6, 8, 10, 12\} \]

### Step 3: Calculate the Probability Distribution
Since the die is fair, each outcome has an equal probability of occurring. The probability for each value of \( X \) is:
- \( P(X = 2) = \frac{1}{6} \)
- \( P(X = 4) = \frac{1}{6} \)
- \( P(X = 6) = \frac{1}{6} \)
- \( P(X = 8) = \frac{1}{6} \)
- \( P(X = 10) = \frac{1}{6} \)
- \( P(X = 12) = \frac{1}{6} \)

Thus, the probability distribution of \( X \) is:
\[ \begin{align*} X & : 2 \quad 4 \quad 6 \quad 8 \quad 10 \quad 12 \\ P(X) & : \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \end{align*} \]

### Step 4: Calculate the Mean (Expected Value) of X
The mean (expected value) of \( X \) is calculated as:
\[ E(X) = \sum (X_i \cdot P(X_i)) = 2 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} + 8 \cdot \frac{1}{6} + 10 \cdot \frac{1}{6} + 12 \cdot \frac{1}{6} \]
\[ E(X) = \frac{1}{6} (2 + 4 + 6 + 8 + 10 + 12) \]
Calculating the sum inside the parentheses:
\[ 2 + 4 = 6, \quad 6 + 6 = 12, \quad 12 + 8 = 20, \quad 20 + 10 = 30, \quad 30 + 12 = 42 \]
Thus,
\[ E(X) = \frac{42}{6} = 7 \]

### Step 5: Calculate the Variance of X
To find the variance, we first need to calculate \( E(X^2) \):
\[ E(X^2) = \sum (X_i^2 \cdot P(X_i)) = 2^2 \cdot \frac{1}{6} + 4^2 \cdot \frac{1}{6} + 6^2 \cdot \frac{1}{6} + 8^2 \cdot \frac{1}{6} + 10^2 \cdot \frac{1}{6} + 12^2 \cdot \frac{1}{6} \]
Calculating \( X_i^2 \):
\[ E(X^2) = \frac{1}{6} (4 + 16 + 36 + 64 + 100 + 144) \]
Calculating the sum:
\[ 4 + 16 = 20, \quad 20 + 36 = 56, \quad 56 + 64 = 120, \quad 120 + 100 = 220, \quad 220 + 144 = 364 \]
Thus,
\[ E(X^2) = \frac{364}{6} \approx 60.67 \]

Now, we can find the variance:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 = 60.67 - 7^2 = 60.67 - 49 = 11.67 \]

### Final Results
- Probability Distribution of \( X \):
\[ \begin{align*} X & : 2 \quad 4 \quad 6 \quad 8 \quad 10 \quad 12 \\ P(X) & : \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \quad \frac{1}{6} \end{align*} \]
- Mean \( E(X) = 7 \)
- Variance \( \text{Var}(X) = 11.67 \)