If X is a binomial variate with mean 6 and variance 2, then the value of `P(5leXle7)` is

Options

A. `(4762)/(6561)`
B. `(4672)/(6561)`
C. `(5264)/(6561)`
D. `(5462)/(6651)`

Answer

To solve the problem, we need to find the probability \( P(5 \leq X \leq 7) \) where \( X \) is a binomial random variable with a mean of 6 and a variance of 2.

### Step-by-Step Solution

1. **Identify the parameters of the binomial distribution**:
- The mean \( \mu \) of a binomial distribution is given by \( \mu = n \cdot p \).
- The variance \( \sigma^2 \) is given by \( \sigma^2 = n \cdot p \cdot q \) where \( q = 1 - p \).
- From the problem, we have:
\[ \mu = 6 \quad \text{and} \quad \sigma^2 = 2 \]

2. **Set up the equations**:
- From the mean:
\[ n \cdot p = 6 \quad \text{(1)} \]
- From the variance:
\[ n \cdot p \cdot q = 2 \quad \text{(2)} \]

3. **Express \( q \) in terms of \( p \)**:
- Since \( q = 1 - p \), substitute \( q \) into equation (2):
\[ n \cdot p \cdot (1 - p) = 2 \]

4. **Substituting \( n \cdot p \) from equation (1)** into the modified equation (2):
- Substitute \( n \cdot p = 6 \):
\[ 6 \cdot (1 - p) = 2 \]
- Solving for \( p \):
\[ 6 - 6p = 2 \implies 6p = 4 \implies p = \frac{2}{3} \]

5. **Finding \( n \)**:
- Substitute \( p \) back into equation (1):
\[ n \cdot \frac{2}{3} = 6 \implies n = 6 \cdot \frac{3}{2} = 9 \]

6. **Calculate \( q \)**:
- Since \( q = 1 - p \):
\[ q = 1 - \frac{2}{3} = \frac{1}{3} \]

7. **Now, we have \( n = 9 \), \( p = \frac{2}{3} \), and \( q = \frac{1}{3} \)**. We need to find \( P(5 \leq X \leq 7) \):
\[ P(5 \leq X \leq 7) = P(X = 5) + P(X = 6) + P(X = 7) \]

8. **Using the binomial probability formula**:
- The probability mass function for a binomial distribution is given by:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
- Calculate \( P(X = 5) \):
\[ P(X = 5) = \binom{9}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^{4} \]
- Calculate \( P(X = 6) \):
\[ P(X = 6) = \binom{9}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^{3} \]
- Calculate \( P(X = 7) \):
\[ P(X = 7) = \binom{9}{7} \left(\frac{2}{3}\right)^7 \left(\frac{1}{3}\right)^{2} \]

9. **Calculating each term**:
- \( P(X = 5) = \binom{9}{5} \cdot \left(\frac{2}{3}\right)^5 \cdot \left(\frac{1}{3}\right)^{4} = 126 \cdot \frac{32}{243} \cdot \frac{1}{81} = \frac{4032}{19683} \)
- \( P(X = 6) = \binom{9}{6} \cdot \left(\frac{2}{3}\right)^6 \cdot \left(\frac{1}{3}\right)^{3} = 84 \cdot \frac{64}{729} \cdot \frac{1}{27} = \frac{5376}{19683} \)
- \( P(X = 7) = \binom{9}{7} \cdot \left(\frac{2}{3}\right)^7 \cdot \left(\frac{1}{3}\right)^{2} = 36 \cdot \frac{128}{2187} \cdot \frac{1}{9} = \frac{4608}{19683} \)

10. **Adding the probabilities**:
\[ P(5 \leq X \leq 7) = \frac{4032 + 5376 + 4608}{19683} = \frac{14016}{19683} \]

### Final Answer
Thus, the value of \( P(5 \leq X \leq 7) \) is \( \frac{14016}{19683} \).