Question

A random variable \( X \) has p.m.f. \( P(X = x) = \frac{{}^{4}C_x}{2^4}, \quad x = 0, 1, 2, 3, 4 \), and \( \mu \) and \( \sigma^2 \) are the mean and variance respectively of the random variable \( X \), then: 

• A. \( \mu = 2, \sigma^2 = 4 \)
• B. \( \mu = 2, \sigma^2 = 1 \)
• C. \( \mu = 3, \sigma^2 = 4 \)
• D. \( \mu = 2, \sigma^2 = 5 \)

Hint:

For discrete random variables, calculate the mean and variance using the definitions of expectation: \( E[X] \) for the mean and \( E[X^2] - \mu^2 \) for the variance.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The p.m.f. given \( P(X=x) = \binom{4}{x} (1/2)^x (1/2)^{4-x} \) describes a Binomial Distribution \( X \sim \text{B}(n, p) \).
Step 2: Key Formula or Approach:
For \( \text{B}(n, p) \):
Mean \( \mu = np \).
Variance \( \sigma^2 = npq \).
Step 3: Detailed Explanation:
Comparing \( \text{P}(X = x) = \frac{\binom{4}{x}}{16} \) with the standard form:
\( n = 4 \).
\( p = 1/2 \), \( q = 1 - 1/2 = 1/2 \).
Mean \( \mu = 4 \times (1/2) = 2 \).
Variance \( \sigma^2 = 4 \times (1/2) \times (1/2) = 1 \).
Step 4: Final Answer:
The values are \( \mu = 2 \) and \( \sigma^2 = 1 \).