The Binomial distribution whose mean is 3 and whose standard deviation is 3/2 is

Options

A. `(1//2+1//2)^(12)`
B. `(1//4+3//4)^(12)`
C. `(3//4+1//4)^(12)`
D. none

Answer

To solve the problem of finding the Binomial distribution with a mean of 3 and a standard deviation of \( \frac{3}{2} \), we can follow these steps:

### Step 1: Understand the Mean and Standard Deviation of a Binomial Distribution
The mean (\( \mu \)) and standard deviation (\( \sigma \)) of a binomial distribution can be expressed as:
- Mean: \( \mu = n \cdot p \)
- Standard Deviation: \( \sigma = \sqrt{n \cdot p \cdot q} \)

Where:
- \( n \) = number of trials
- \( p \) = probability of success
- \( q \) = probability of failure (where \( q = 1 - p \))

### Step 2: Set Up the Equations
From the problem, we know:
- \( \mu = 3 \)
- \( \sigma = \frac{3}{2} \)

Using the mean equation:
\[ n \cdot p = 3 \quad \text{(1)} \]

Using the standard deviation equation:
\[ \sqrt{n \cdot p \cdot q} = \frac{3}{2} \quad \text{(2)} \]

### Step 3: Substitute \( q \) in Terms of \( p \)
Since \( q = 1 - p \), we can substitute \( q \) in equation (2):
\[ \sqrt{n \cdot p \cdot (1 - p)} = \frac{3}{2} \]

### Step 4: Square Both Sides of the Equation
Squaring both sides of equation (2):
\[ n \cdot p \cdot (1 - p) = \left(\frac{3}{2}\right)^2 \]
\[ n \cdot p \cdot (1 - p) = \frac{9}{4} \quad \text{(3)} \]

### Step 5: Substitute \( n \) from Equation (1) into Equation (3)
From equation (1), we can express \( n \) as:
\[ n = \frac{3}{p} \]

Substituting this into equation (3):
\[ \frac{3}{p} \cdot p \cdot (1 - p) = \frac{9}{4} \]
\[ 3(1 - p) = \frac{9}{4} \]

### Step 6: Solve for \( p \)
Now, we can solve for \( p \):
\[ 3 - 3p = \frac{9}{4} \]
\[ -3p = \frac{9}{4} - 3 \]
\[ -3p = \frac{9}{4} - \frac{12}{4} \]
\[ -3p = -\frac{3}{4} \]
\[ p = \frac{1}{4} \]

### Step 7: Calculate \( q \)
Now that we have \( p \), we can find \( q \):
\[ q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 8: Calculate \( n \)
Using equation (1) to find \( n \):
\[ n \cdot p = 3 \]
\[ n \cdot \frac{1}{4} = 3 \]
\[ n = 3 \cdot 4 = 12 \]

### Step 9: Write the Binomial Distribution
Now we can write the binomial distribution:
\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]
Substituting the values of \( n \), \( p \), and \( q \):
\[ P(X = k) = \binom{12}{k} \left(\frac{1}{4}\right)^k \left(\frac{3}{4}\right)^{12-k} \]

### Final Answer
Thus, the binomial distribution is:
\[ P(X = k) = \binom{12}{k} \left(\frac{1}{4}\right)^k \left(\frac{3}{4}\right)^{12-k} \]