Let  be a random variable with the following probability distribution.
Value x of X    P(X=x)
0	                      0.10
10	                 0.25
20	                 0.05
30	                 0.15
40	                 0.30
50	                 0.15
Complete the following. (If necessary, consult a list of formulas.)

(a) Find the expectation E (X) of X.
 
(b) Find the variance Var (X) of X.


Let's calculate the expectation (mean) \( E(X) \) and the variance \( \text{Var}(X) \) of the random variable \( X \) based on the given probability distribution.

### (a) **Find the expectation \( E(X) \) of \( X \):**

The expectation \( E(X) \) is given by the formula:

\[
E(X) = \sum_x x \cdot P(X = x)
\]

Using the given probability distribution:

\[
E(X) = (0 \cdot 0.10) + (10 \cdot 0.25) + (20 \cdot 0.05) + (30 \cdot 0.15) + (40 \cdot 0.30) + (50 \cdot 0.15)
\]

Let's compute this:

\[
E(X) = (0) + (2.5) + (1.0) + (4.5) + (12.0) + (7.5) = 27.5
\]

So, the expectation \( E(X) = 27.5 \).

---

### (b) **Find the variance \( \text{Var}(X) \) of \( X \):**

The variance \( \text{Var}(X) \) is calculated using the formula:

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

First, we need to calculate \( E(X^2) \):

\[
E(X^2) = \sum_x x^2 \cdot P(X = x)
\]

So:

\[
E(X^2) = (0^2 \cdot 0.10) + (10^2 \cdot 0.25) + (20^2 \cdot 0.05) + (30^2 \cdot 0.15) + (40^2 \cdot 0.30) + (50^2 \cdot 0.15)
\]

Let's compute this:

\[
E(X^2) = (0) + (100 \cdot 0.25) + (400 \cdot 0.05) + (900 \cdot 0.15) + (1600 \cdot 0.30) + (2500 \cdot 0.15)
\]

\[
E(X^2) = (0) + (25) + (20) + (135) + (480) + (375) = 1035
\]

Now, using the variance formula:

\[
\text{Var}(X) = E(X^2) - [E(X)]^2 = 1035 - (27.5)^2
\]

\[
\text{Var}(X) = 1035 - 756.25 = 278.75
\]

Thus, the variance \( \text{Var}(X) = 278.75 \).

---

### Summary:
- (a) \( E(X) = 27.5 \)
- (b) \( \text{Var}(X) = 278.75 \)

---

Would you like more details or have any questions?

Here are 5 related questions you can explore:
1. How is the expectation affected if all probabilities are doubled?
2. What is the standard deviation \( \sigma(X) \) based on this variance?
3. How would you compute the cumulative distribution function (CDF) for this data?
4. What is the skewness of this distribution?
5. How does the variance change if the value of \( X = 50 \) is reduced to \( X = 45 \)?

**Tip**: Variance is always a non-negative value because it represents the spread of the data around the mean.

Let X be a random variable with the following probability distribution. Value x of X: 0 (P(X=0) = 0.10), 10 (P(X=10) = 0.25), 20 (P(X=20) = 0.05), 30 (P(X=30) = 0.15), 40 (P(X=40) = 0.30), 50 (P(X=50) = 0.15). Complete the following: (a) Find the expectation E(X) of X. (b) Find the variance Var(X) of X.

Learn how to calculate the expectation (mean) and variance of a discrete random variable using a given probability distribution. This example includes step-by-step computations for E(X) = 27.5 and Var(X) = 278.75. Suitable for students in Grades 10-12 studying probability and statistics.

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