Question

If the mean and variance of a binomial distribution are 18 and 12 respectively, then \( n \) is equal to .......

• A. 36
• B. 54
• C. 16
• D. 27

Hint:

For binomial distributions, use the relationships between mean, variance, and \( p \) to solve for unknowns.

Answer 1

The Correct Option is C

Step 1: Recall the formulas.
For a binomial distribution, the mean \( \mu \) and variance \( \sigma^2 \) are given by: \[ \mu = n \cdot p \text{and} \sigma^2 = n \cdot p \cdot (1 - p) \]

Step 2: Set up the system of equations.
We are given: \[ \mu = 18, \sigma^2 = 12 \] From the first equation: \[ n \cdot p = 18 $\Rightarrow$ p = \frac{18}{n} \] Substitute this into the variance equation: \[ 12 = n \cdot \frac{18}{n} \cdot \left( 1 - \frac{18}{n} \right) \]

Step 3: Simplify and solve.
\[ 12 = 18 \left( 1 - \frac{18}{n} \right) \] \[ 12 = 18 - \frac{324}{n} \] \[ \frac{324}{n} = 6 $\Rightarrow$ n = 54 \]

Step 4: Conclude.
Thus, \( n = 54 \), so the correct answer is option (ii).

Final Answer: \[ \boxed{54} \]