A random variable X has the following distribution
`{:(X=x_(1):,-2,-1,0,1,2,3),(P(X=x_(1)):,0.1,k,0.2,2k,0.3,k):}`
The value of k and its mean and variance are

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the value of k

We know that the sum of all probabilities must equal 1. The given probabilities are:

- \( P(X = -2) = 0.1 \)
- \( P(X = -1) = k \)
- \( P(X = 0) = 0.2 \)
- \( P(X = 1) = 2k \)
- \( P(X = 2) = 0.3 \)
- \( P(X = 3) = k \)

Setting up the equation:

\[ 0.1 + k + 0.2 + 2k + 0.3 + k = 1 \]

Combining like terms:

\[ 0.1 + 0.2 + 0.3 + (k + 2k + k) = 1 \]
\[ 0.6 + 4k = 1 \]

Now, solving for \( k \):

\[ 4k = 1 - 0.6 \]
\[ 4k = 0.4 \]
\[ k = \frac{0.4}{4} = 0.1 \]

### Step 2: Update the probability distribution

Now that we have \( k = 0.1 \), we can update the probabilities:

- \( P(X = -2) = 0.1 \)
- \( P(X = -1) = 0.1 \)
- \( P(X = 0) = 0.2 \)
- \( P(X = 1) = 2(0.1) = 0.2 \)
- \( P(X = 2) = 0.3 \)
- \( P(X = 3) = 0.1 \)

### Step 3: Calculate the mean

The mean \( \mu \) of a discrete random variable is given by:

\[ \mu = \sum (x_i \cdot P(X = x_i)) \]

Calculating the mean:

\[ \mu = (-2)(0.1) + (-1)(0.1) + (0)(0.2) + (1)(0.2) + (2)(0.3) + (3)(0.1) \]
\[ = -0.2 - 0.1 + 0 + 0.2 + 0.6 + 0.3 \]
\[ = -0.3 + 1.1 = 0.8 \]

### Step 4: Calculate the variance

The variance \( \sigma^2 \) is given by:

\[ \sigma^2 = \sum (x_i^2 \cdot P(X = x_i)) - \mu^2 \]

First, we calculate \( \sum (x_i^2 \cdot P(X = x_i)) \):

\[ = (-2)^2(0.1) + (-1)^2(0.1) + (0)^2(0.2) + (1)^2(0.2) + (2)^2(0.3) + (3)^2(0.1) \]
\[ = 4(0.1) + 1(0.1) + 0 + 1(0.2) + 4(0.3) + 9(0.1) \]
\[ = 0.4 + 0.1 + 0 + 0.2 + 1.2 + 0.9 \]
\[ = 0.4 + 0.1 + 0.2 + 1.2 + 0.9 = 2.8 \]

Now, substituting into the variance formula:

\[ \sigma^2 = 2.8 - (0.8)^2 \]
\[ = 2.8 - 0.64 = 2.16 \]

### Final Results

- The value of \( k \) is \( 0.1 \)
- The mean \( \mu \) is \( 0.8 \)
- The variance \( \sigma^2 \) is \( 2.16 \)