Question

If \( \mu \) and \( \sigma^2 \) are mean and variance of a random variable \( X \) whose p.m.f. is given by \[ P(X = x) = \binom{6}{x} \left( \frac{1}{4} \right)^x \left( \frac{3}{4} \right)^{6 - x}, \quad x = 0, 1, 2, 3, 4, 5, 6, \text{ then the value of } 2\mu + 12\sigma^2 = \]

• A. 4
• B. 8
• C. 20
• D. 16

Hint:

For a binomial distribution, the mean is \( \mu = n \cdot p \) and the variance is \( \sigma^2 = n \cdot p \cdot (1 - p) \).

Answer 1

The Correct Option is C

Step 1: Find the mean \( \mu \).
The mean of a binomial distribution is \( \mu = n \cdot p \), where \( n = 6 \) and \( p = \frac{1}{4} \). Thus, \[ \mu = 6 \times \frac{1}{4} = \frac{6}{4} = 1.5 \]
Step 2: Find the variance \( \sigma^2 \).
The variance of a binomial distribution is \( \sigma^2 = n \cdot p \cdot (1 - p) \). Thus, \[ \sigma^2 = 6 \times \frac{1}{4} \times \frac{3}{4} = \frac{18}{16} = 1.125 \]
Step 3: Calculate \( 2\mu + 12\sigma^2 \).
Now, calculate \( 2\mu + 12\sigma^2 \): \[ 2\mu + 12\sigma^2 = 2 \times 1.5 + 12 \times 1.125 = 3 + 13.5 = 20 \]
Step 4: Conclusion.
Thus, the value of \( 2\mu + 12\sigma^2 \) is 20, corresponding to option (C).