Understanding the Problem: Variance Calculation

The question asks us to find the variance of a discrete random variable X, given its probability distribution law. The distribution provides the possible values of X and their corresponding probabilities.

Distribution Law of Random Variable X

The given probability distribution is:

First, let's verify if the probabilities sum to 1:

$$ \sum P(X=x) = 0.3 + 0.4 + 0.3 = 1.0 $$

The probabilities sum to 1, so this is a valid probability distribution.

Calculating the Expected Value (Mean) E(X)

The expected value, or mean ($\mu$), of a discrete random variable is calculated as the sum of each value multiplied by its probability:

$$ E(X) = \mu = \sum_{i} x_i P(X=x_i) $$

Using the given distribution:

$$ E(X) = (1 \times 0.3) + (2 \times 0.4) + (3 \times 0.3) $$

$$ E(X) = 0.3 + 0.8 + 0.9 $$

$$ E(X) = 2.0 $$

So, the mean of the distribution is 2.0.

Calculating the Expected Value of X squared E(X²)

To find the variance, we also need the expected value of X squared. This is calculated by summing the square of each value multiplied by its probability:

$$ E(X^2) = \sum_{i} x_i^2 P(X=x_i) $$

Using the given distribution:

$$ E(X^2) = (1^2 \times 0.3) + (2^2 \times 0.4) + (3^2 \times 0.3) $$

$$ E(X^2) = (1 \times 0.3) + (4 \times 0.4) + (9 \times 0.3) $$

$$ E(X^2) = 0.3 + 1.6 + 2.7 $$

$$ E(X^2) = 4.6 $$

So, the expected value of X squared is 4.6.

Calculating the Variance Var(X)

The variance ($Var(X)$) is calculated using the formula:

$$ Var(X) = E(X^2) - [E(X)]^2 $$

Substituting the values we calculated:

$$ Var(X) = 4.6 - (2.0)^2 $$

$$ Var(X) = 4.6 - 4.0 $$

$$ Var(X) = 0.6 $$

Conclusion

The variance of the given distribution is 0.6. Comparing this result with the options provided, option 2 matches our calculated value.