Certainly, let’s solve this step by step.

The table provides the values of a random variable $X$ along with their associated probabilities $P(x)$. The goal is to find the expected value $E(-4X + \Delta)$, where $\Delta$ is a constant.

Step 1: Calculate $E(X)$

The expectation $E(X)$ is calculated as follows: $E(X) = \sum_{i} x_i P(x_i)$ where $x_i$ are the values that $X$ can take, and $P(x_i)$ are the probabilities.

From the table:

• $x = 0$ with $P(x) = 0.25$
• $x = 1$ with $P(x) = 0.21$
• $x = 2$ with $P(x) = 0.15$
• $x = 3$ with $P(x) = 0.14$

Calculating $E(X)$: $E(X) = (0 \times 0.25) + (1 \times 0.21) + (2 \times 0.15) + (3 \times 0.14)$ $E(X) = 0 + 0.21 + 0.30 + 0.42 = 0.93$

Step 2: Calculate $E(-4X + \Delta)$

Using the linearity of expectation, we can separate the expectation: $E(-4X + \Delta) = -4E(X) + E(\Delta)$ Since $\Delta$ is a constant, $E(\Delta) = \Delta$.

Thus, $E(-4X + \Delta) = -4 \cdot 0.93 + \Delta$ $E(-4X + \Delta) = -3.72 + \Delta$

Final Answer

The expected value $E(-4X + \Delta) = -3.72 + \Delta$.

Would you like further clarification on any step or additional details?

Here are some related questions that might interest you:

1. What if the constant $\Delta$ was specified? How would that affect the expectation?
2. How would you calculate $E(X^2)$ using the same table of values?
3. Can you determine the variance $\text{Var}(X)$ with the given information?
4. How would $E(aX + b)$ change for arbitrary constants $a$ and $b$?
5. What if the probabilities were adjusted? How would that impact $E(X)$?

Tip: Always remember that constants factor out in expectations, simplifying calculations greatly!