The probability distribution of a random variable X is given by.
`{:(X=x:,0,1,2,3,4),(P(X=x):,0.4,0.3,0.1,0.1,0.1):}`
The variance of X, is

Options

A. 1.76
B. 2.45
C. 3.2
D. 4.8

Answer

To find the variance of the random variable \(X\) given its probability distribution, we can follow these steps:

### Step 1: List the values and probabilities
The values of \(X\) and their corresponding probabilities are:
- \(X = 0\), \(P(X=0) = 0.4\)
- \(X = 1\), \(P(X=1) = 0.3\)
- \(X = 2\), \(P(X=2) = 0.1\)
- \(X = 3\), \(P(X=3) = 0.1\)
- \(X = 4\), \(P(X=4) = 0.1\)

### Step 2: Create a table for calculations
We will create a table to calculate \(X^2 \cdot P(X)\) and \(X \cdot P(X)\):

| \(X\) | \(P(X)\) | \(X^2\) | \(X^2 \cdot P(X)\) | \(X \cdot P(X)\) |
|-------|----------|---------|---------------------|-------------------|
| 0 | 0.4 | 0 | \(0^2 \cdot 0.4 = 0\) | \(0 \cdot 0.4 = 0\) |
| 1 | 0.3 | 1 | \(1^2 \cdot 0.3 = 0.3\) | \(1 \cdot 0.3 = 0.3\) |
| 2 | 0.1 | 4 | \(2^2 \cdot 0.1 = 0.4\) | \(2 \cdot 0.1 = 0.2\) |
| 3 | 0.1 | 9 | \(3^2 \cdot 0.1 = 0.9\) | \(3 \cdot 0.1 = 0.3\) |
| 4 | 0.1 | 16 | \(4^2 \cdot 0.1 = 1.6\) | \(4 \cdot 0.1 = 0.4\) |

### Step 3: Calculate the summations
Now, we will calculate the summations needed for the variance formula.

1. **Calculate \( \sum (X^2 \cdot P(X)) \)**:
\[ \sum (X^2 \cdot P(X)) = 0 + 0.3 + 0.4 + 0.9 + 1.6 = 3.2 \]

2. **Calculate \( \sum (X \cdot P(X)) \)**:
\[ \sum (X \cdot P(X)) = 0 + 0.3 + 0.2 + 0.3 + 0.4 = 1.2 \]

### Step 4: Calculate the variance
The formula for variance \( \sigma^2 \) is:
\[ \sigma^2 = \sum (X^2 \cdot P(X)) - \left( \sum (X \cdot P(X)) \right)^2 \]
Substituting the values we found:
\[ \sigma^2 = 3.2 - (1.2)^2 \]
Calculating \( (1.2)^2 \):
\[ (1.2)^2 = 1.44 \]
Now substituting back:
\[ \sigma^2 = 3.2 - 1.44 = 1.76 \]

### Final Answer
The variance of \(X\) is \( \sigma^2 = 1.76 \).

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