Let a random variable `X` have a binomial distribution with mean `8` and variance `4`. If `P(X lt= 2)=(k)/(2^(16))`, then `k` is equal to :

Options

A. `17`
B. `121`
C. `1`
D. `137`

Answer

To solve the problem, we need to analyze the given information about the random variable \(X\) which follows a binomial distribution. The mean and variance are provided, and we need to find the value of \(k\) in the expression \(P(X \leq 2) = \frac{k}{2^{16}}\).

### Step-by-Step Solution:

1. **Identify Mean and Variance**:
The mean \(\mu\) of a binomial distribution is given by:
\[ \mu = N \cdot P \]
where \(N\) is the number of trials and \(P\) is the probability of success.

The variance \(\sigma^2\) is given by:
\[ \sigma^2 = N \cdot P \cdot (1 - P) = N \cdot P \cdot Q \]
where \(Q = 1 - P\).

From the problem, we know:
\[ \mu = 8 \quad \text{and} \quad \sigma^2 = 4 \]

2. **Set Up Equations**:
From the mean:
\[ N \cdot P = 8 \quad \text{(1)} \]
From the variance:
\[ N \cdot P \cdot Q = 4 \quad \text{(2)} \]

3. **Express Q in terms of P**:
Since \(Q = 1 - P\), we can substitute \(Q\) in equation (2):
\[ N \cdot P \cdot (1 - P) = 4 \]

4. **Substitute \(N\) from Equation (1)**:
From equation (1), we can express \(N\) as:
\[ N = \frac{8}{P} \]
Substituting this into equation (2):
\[ \frac{8}{P} \cdot P \cdot (1 - P) = 4 \]
Simplifying gives:
\[ 8(1 - P) = 4 \]
\[ 1 - P = \frac{1}{2} \quad \Rightarrow \quad P = \frac{1}{2} \]

5. **Find N**:
Substitute \(P\) back into equation (1):
\[ N \cdot \frac{1}{2} = 8 \quad \Rightarrow \quad N = 16 \]

6. **Identify the Distribution**:
Thus, \(X\) follows a binomial distribution \(X \sim \text{Binomial}(16, \frac{1}{2})\).

7. **Calculate \(P(X \leq 2)\)**:
We need to calculate:
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
Using the binomial probability formula:
\[ P(X = k) = \binom{N}{k} P^k Q^{N-k} \]
where \(P = \frac{1}{2}\) and \(Q = \frac{1}{2}\).

- For \(X = 0\):
\[ P(X = 0) = \binom{16}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^{16} = 1 \cdot 1 \cdot \frac{1}{2^{16}} = \frac{1}{2^{16}} \]

- For \(X = 1\):
\[ P(X = 1) = \binom{16}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^{15} = 16 \cdot \frac{1}{2} \cdot \frac{1}{2^{15}} = \frac{16}{2^{16}} \]

- For \(X = 2\):
\[ P(X = 2) = \binom{16}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^{14} = 120 \cdot \frac{1}{4} \cdot \frac{1}{2^{14}} = \frac{120}{2^{16}} \]

8. **Combine the Probabilities**:
Now, combine the probabilities:
\[ P(X \leq 2) = \frac{1}{2^{16}} + \frac{16}{2^{16}} + \frac{120}{2^{16}} = \frac{1 + 16 + 120}{2^{16}} = \frac{137}{2^{16}} \]

9. **Find \(k\)**:
From the problem statement:
\[ P(X \leq 2) = \frac{k}{2^{16}} \quad \Rightarrow \quad k = 137 \]

### Final Answer:
Thus, the value of \(k\) is \( \boxed{137} \).