To calculate the expected value $E(X)$ of a probability distribution, you use the following formula:

$E(X) = \sum [x \cdot p(x)]$

In this case, we have the following values for $x$ and $p(x)$:

$E(X) = (1 \cdot 0.15) + (2 \cdot 0.35) + (3 \cdot 0.40) + (4 \cdot 0.10)$

Now, let's compute each term:

$E(X) = (1 \cdot 0.15) + (2 \cdot 0.35) + (3 \cdot 0.40) + (4 \cdot 0.10)$ $E(X) = 0.15 + 0.70 + 1.20 + 0.40$ $E(X) = 2.45$

Thus, the expected value $E(X)$ is $2.45$.

Would you like to see more details or clarifications on this?

Here are 5 related questions:

1. How is variance calculated for a probability distribution?
2. Can expected value be a value that is not in the original data set?
3. What is the meaning of expected value in a real-world context?
4. How does the expected value change if the probabilities are adjusted?
5. What is the role of expected value in decision theory?

Tip: The expected value is essentially a weighted average, with the probabilities acting as weights for the different outcomes.