To find the mean and variance of a probability distribution, we use the following formulas:

1. Mean (Expected Value), $E(X)$: $E(X) = \sum x_i \cdot P(x_i)$ where $x_i$ is the value of the random variable and $P(x_i)$ is the probability associated with $x_i$.

2. Variance, $\text{Var}(X)$: $\text{Var}(X) = E(X^2) - (E(X))^2$ where $E(X^2) = \sum x_i^2 \cdot P(x_i)$.

Given Data

• Probabilities: $P(0) = 0.4$, $P(1) = 0.3$, $P(2) = 0.2$, $P(3) = 0.1$
• Values of the random variable $X$: $0, 1, 2, 3$

Step 1: Calculate the Mean (Expected Value)

$E(X) = 0 \cdot 0.4 + 1 \cdot 0.3 + 2 \cdot 0.2 + 3 \cdot 0.1$

Step 2: Calculate $E(X^2)$

$E(X^2) = 0^2 \cdot 0.4 + 1^2 \cdot 0.3 + 2^2 \cdot 0.2 + 3^2 \cdot 0.1$

Step 3: Calculate the Variance

$\text{Var}(X) = E(X^2) - (E(X))^2$

Let's compute these values.The mean (expected value) and variance of the probability distribution are both 1.0.

Summary:

• Mean (Expected Value), $E(X)$: 1.0
• Variance, $\text{Var}(X)$: 1.0

Would you like more details or have any questions?

Related Questions:

1. What is the standard deviation of this distribution?
2. How would the mean and variance change if the probabilities were altered?
3. Can we determine the skewness of this distribution?
4. What is the cumulative distribution function (CDF) for this probability distribution?
5. How do we interpret these statistics in real-world terms for power failures?

Tip:

Understanding variance helps in assessing the spread or variability of a distribution, which is crucial for risk assessment and decision-making in real-world scenarios.