In a binomial distribution `n=12, p=1//3`. Then mean and variance are

Options

A. a) `4 , (8)/(3)`
B. b) `2 , (6)/(3)`
C. c) `2, (5)/(2)`
D. d) `4 , (2)/(3)`

Answer

To solve the problem of finding the mean and variance of a binomial distribution with parameters \( n = 12 \) and \( p = \frac{1}{3} \), we can follow these steps:

### Step 1: Identify the parameters
We have:
- \( n = 12 \) (the number of trials)
- \( p = \frac{1}{3} \) (the probability of success)

### Step 2: Calculate the mean
The mean \( \mu \) of a binomial distribution is given by the formula:
\[ \mu = n \cdot p \]
Substituting the values:
\[ \mu = 12 \cdot \frac{1}{3} \]
Calculating this gives:
\[ \mu = 12 \cdot \frac{1}{3} = 4 \]

### Step 3: Calculate the variance
The variance \( \sigma^2 \) of a binomial distribution is given by the formula:
\[ \sigma^2 = n \cdot p \cdot q \]
where \( q = 1 - p \). First, we need to calculate \( q \):
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]
Now we can substitute \( n \), \( p \), and \( q \) into the variance formula:
\[ \sigma^2 = 12 \cdot \frac{1}{3} \cdot \frac{2}{3} \]
Calculating this step-by-step:
1. Calculate \( n \cdot p \):
\[ 12 \cdot \frac{1}{3} = 4 \]
2. Now calculate \( 4 \cdot \frac{2}{3} \):
\[ 4 \cdot \frac{2}{3} = \frac{8}{3} \]

### Final Results
- Mean \( \mu = 4 \)
- Variance \( \sigma^2 = \frac{8}{3} \)

Thus, the mean is 4 and the variance is \( \frac{8}{3} \).