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Set up and solve for the expected value of the given scenario.

Multiply each outcome x; by the probability of each outcome P(x;) and add them together.

- (1) = X1P (x1) + X2P (x2) + X3P (X3) + x1 P(x1) + x5 P(x5)

E（X） =（1）（0.24） + （2）（0.31） + （3）（0.01） + （4）（0.15） + （5）（0.29）

E(X) = 0.24 + 0.62 + 0.03 + 0.6 + 1. 45

Find that the expected value of this scenario is 2.94.

To calculate the expected value \( E(X) \) of the scenario, the formula is:

\[
E(X) = \sum_{i=1}^n x_i P(x_i)
\]

Using the given values:

\[
E(X) = (1 \times 0.24) + (2 \times 0.31) + (3 \times 0.01) + (4 \times 0.15) + (5 \times 0.29)
\]

Let's calculate each term:

- \( 1 \times 0.24 = 0.24 \)
- \( 2 \times 0.31 = 0.62 \)
- \( 3 \times 0.01 = 0.03 \)
- \( 4 \times 0.15 = 0.60 \)
- \( 5 \times 0.29 = 1.45 \)

Now, add these results together:

\[
E(X) = 0.24 + 0.62 + 0.03 + 0.60 + 1.45 = 2.94
\]

Thus, the expected value \( E(X) \) is **2.94**.

Would you like more details on this calculation or have any questions?

### Relative Questions:
1. What is the significance of the expected value in probability?
2. How can the expected value be applied in real-world scenarios?
3. What happens to the expected value if one of the probabilities changes?
4. Can the expected value be negative, and in what scenarios might this occur?
5. How would you calculate the variance based on the given data?

**Tip:** When calculating expected value, ensure that the sum of the probabilities \( P(x_i) \) adds up to 1 for a valid probability distribution.

Calculate the expected value of the given scenario using the probabilities and outcomes in the table.

Calculate the expected value of a scenario using probabilities and outcomes. This example covers probability theory and the expected value formula, demonstrating the step-by-step calculation for E(X). Suitable for students in Grades 9-12.

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