Question

If a cubical die is thrown, then the mean and variance of the random variable \( X \), giving the number on the face that shows up, are respectively:

• A. \( \frac{2}{7}, \frac{12}{35} \)
• B. \( \frac{7}{2}, \frac{12}{35} \)
• C. \( \frac{1}{7}, \frac{1}{12} \)
• D. \( \frac{7}{2}, \frac{35}{12} \)

Hint:

For a fair \( n \)-sided die, the mean is \( \frac{n+1}{2} \) and the variance is \( \frac{n^2 - 1}{12} \). For a standard 6-sided die, mean = \( \frac{6+1}{2} = \frac{7}{2} \) and variance = \( \frac{6^2 - 1}{12} = \frac{35}{12} \).

Answer 1

The Correct Option is D

Step 1: Identify the possible values and their probabilities for the random variable \( X \).
When a fair cubical die is thrown, the possible outcomes for the number on the face that shows up are \( \{1, 2, 3, 4, 5, 6\} \). Since the die is fair, each outcome has an equal probability of \( \frac{1}{6} \). Thus, the probability distribution of \( X \) is: \[ P(X = k) = \frac{1}{6}, \quad \text{for } k = 1, 2, 3, 4, 5, 6 \]
Step 2: Calculate the mean (expected value) of the random variable \( X \).
The mean \( \mu \) or \( E(X) \) is given by: \[ \mu = E(X) = \sum_{k=1}^{6} k \cdot P(X = k) = \sum_{k=1}^{6} k \cdot \frac{1}{6} = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) \] The sum of the first \( n \) natural numbers is \( \frac{n(n+1)}{2} \). For \( n = 6 \), the sum is \( \frac{6(6+1)}{2} = \frac{6 \times 7}{2} = 21 \). \[ \mu = E(X) = \frac{1}{6} (21) = \frac{21}{6} = \frac{7}{2} \]
Step 3: Calculate the variance of the random variable \( X \).
The variance \( \sigma^2 \) or \( Var(X) \) is given by \( Var(X) = E(X^2) - [E(X)]^2 \). First, let's calculate \( E(X^2) \): \[ E(X^2) = \sum_{k=1}^{6} k^2 \cdot P(X = k) = \sum_{k=1}^{6} k^2 \cdot \frac{1}{6} = \frac{1}{6} (1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2) \] The sum of the squares of the first \( n \) natural numbers is \( \frac{n(n+1)(2n+1)}{6} \). For \( n = 6 \), the sum is \( \frac{6(6+1)(2 \times 6 + 1)}{6} = \frac{6 \times 7 \times 13}{6} = 7 \times 13 = 91 \). \[ E(X^2) = \frac{1}{6} (91) = \frac{91}{6} \] Now, we can calculate the variance: \[ Var(X) = E(X^2) - [E(X)]^2 = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{91}{6} - \frac{49}{4} \] To subtract these fractions, find a common denominator, which is 12: \[ Var(X) = \frac{91 \times 2}{12} - \frac{49 \times 3}{12} = \frac{182 - 147}{12} = \frac{35}{12} \]
Step 4: State the mean and variance.
The mean of the random variable \( X \) is \( \frac{7}{2} \) and the variance is \( \frac{35}{12} \). This corresponds to option (4).