Let X be a random variable with the following probability distribution.

Value x of X

P(X=x)

2

0.15

1

0.20

0

0.10

1

0.10

2

0.45

Complete the following. (if necessary, consult a list of formulas.)

(9) Find the expectation E(X) l X. E(X)=

(b) Find the variance Var(X) of X var(x)=

Given the probability distribution of the random variable \( X \), we can calculate the expectation \( E(X) \) and the variance \( \text{Var}(X) \).

### Probability Distribution:
| Value \( x \) of \( X \) | \( P(X = x) \) |
|:------------------------:|:--------------:|
| 2                        | 0.15           |
| 1                        | 0.20           |
| 0                        | 0.10           |
| 1                        | 0.10           |
| 2                        | 0.45           |

However, notice that the values of \( X \) have repeated probabilities. To simplify, we should group the same values of \( X \) together:

- For \( X = 2 \): \( P(X = 2) = 0.15 + 0.45 = 0.60 \)
- For \( X = 1 \): \( P(X = 1) = 0.20 + 0.10 = 0.30 \)
- For \( X = 0 \): \( P(X = 0) = 0.10 \)

### Simplified Probability Distribution:
| Value \( x \) of \( X \) | \( P(X = x) \) |
|:------------------------:|:--------------:|
| 2                        | 0.60           |
| 1                        | 0.30           |
| 0                        | 0.10           |

### (a) Expectation \( E(X) \):
The expectation \( E(X) \) is calculated using the formula:
\[
E(X) = \sum_{i} x_i \cdot P(X = x_i)
\]
Substituting the values:
\[
E(X) = (2 \times 0.60) + (1 \times 0.30) + (0 \times 0.10) = 1.20 + 0.30 + 0 = 1.50
\]
So, \( E(X) = 1.50 \).

### (b) Variance \( \text{Var}(X) \):
The variance \( \text{Var}(X) \) is calculated using the formula:
\[
\text{Var}(X) = E(X^2) - [E(X)]^2
\]
First, we calculate \( E(X^2) \):
\[
E(X^2) = \sum_{i} x_i^2 \cdot P(X = x_i)
\]
Substituting the values:
\[
E(X^2) = (2^2 \times 0.60) + (1^2 \times 0.30) + (0^2 \times 0.10) = (4 \times 0.60) + (1 \times 0.30) + (0 \times 0.10) = 2.40 + 0.30 + 0 = 2.70
\]

Now, using \( E(X) = 1.50 \):
\[
\text{Var}(X) = E(X^2) - [E(X)]^2 = 2.70 - (1.50)^2 = 2.70 - 2.25 = 0.45
\]

### Final Answers:
- \( E(X) = 1.50 \)
- \( \text{Var}(X) = 0.45 \)

Would you like more details or have any questions?

### Related Questions:
1. How do you interpret the expectation \( E(X) \) in the context of a probability distribution?
2. How does the variance \( \text{Var}(X) \) reflect the spread of the distribution?
3. What is the standard deviation of \( X \) given the variance \( \text{Var}(X) \)?
4. How would the expectation change if \( P(X = 0) \) were increased to 0.20?
5. What is the cumulative distribution function (CDF) for this probability distribution?
6. How do you compute \( E(X^3) \) for this distribution?
7. What would happen to the variance if all probabilities were uniformly distributed?
8. How does variance relate to the concept of risk in probability?

### Tip:
When calculating expectation and variance, always check for any repeated values in the distribution to simplify the calculations.

Learn how to calculate the expectation and variance of a random variable with a given probability distribution. Understand the concepts of probability theory and apply formulas to find expectation and variance effectively. Suitable for college-level students and learners interested in probability theory.

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