X is a random variable with distribution given below
`{:(x:,0,1,2,3),(P(X=x):,k,3k,3k,k):}`
The value of k and its variance are

Answer

To solve the problem, we need to find the value of \( k \) and the variance of the random variable \( X \) with the given probability distribution.

The distribution is given as:
- \( P(X=0) = k \)
- \( P(X=1) = 3k \)
- \( P(X=2) = 3k \)
- \( P(X=3) = k \)

### Step 1: Find the value of \( k \)

We know that the sum of all probabilities must equal 1:
\[ P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1 \]
Substituting the probabilities, we have:
\[ k + 3k + 3k + k = 1 \]
This simplifies to:
\[ 8k = 1 \]
Now, solving for \( k \):
\[ k = \frac{1}{8} \]

### Step 2: Write the probabilities with the value of \( k \)

Now that we have \( k \), we can express the probabilities:
- \( P(X=0) = k = \frac{1}{8} \)
- \( P(X=1) = 3k = \frac{3}{8} \)
- \( P(X=2) = 3k = \frac{3}{8} \)
- \( P(X=3) = k = \frac{1}{8} \)

### Step 3: Calculate the mean \( \mu \)

The mean \( \mu \) of the random variable \( X \) is calculated as:
\[ \mu = E(X) = \sum (x_i \cdot P(X=x_i)) \]
Calculating this:
\[ \mu = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8} \]
\[ \mu = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2} \]

### Step 4: Calculate \( E(X^2) \)

Next, we need to calculate \( E(X^2) \):
\[ E(X^2) = \sum (x_i^2 \cdot P(X=x_i)) \]
Calculating this:
\[ E(X^2) = 0^2 \cdot \frac{1}{8} + 1^2 \cdot \frac{3}{8} + 2^2 \cdot \frac{3}{8} + 3^2 \cdot \frac{1}{8} \]
\[ E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3 \]

### Step 5: Calculate the variance \( \sigma^2 \)

The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = E(X^2) - (E(X))^2 \]
Substituting the values we calculated:
\[ \sigma^2 = 3 - \left(\frac{3}{2}\right)^2 \]
\[ \sigma^2 = 3 - \frac{9}{4} \]
To subtract, we convert 3 to a fraction:
\[ \sigma^2 = \frac{12}{4} - \frac{9}{4} = \frac{3}{4} \]

### Final Answers

Thus, the value of \( k \) is \( \frac{1}{8} \) and the variance is \( \frac{3}{4} \).

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