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To complete the provided probability distribution and solve the given questions, we will follow these steps:

### 1. Complete the Probability Distribution:
First, we need to find the missing values in the table, specifically for \( (X - \mu)^2 P(X) \). Given values are:
- \( \mu = E(X) = 3.8200 \)

The formula for each term in \( (X - \mu)^2 P(X) \) is:
\[
(X - \mu)^2 P(X)
\]
Where \( X \) is the value from the first column and \( P(X) \) is the probability from the second column.

Let's compute these values for each row:

#### For \( X = 2 \):
\[
(X - \mu)^2 = (2 - 3.82)^2 = (-1.82)^2 = 3.3124
\]
\[
(X - \mu)^2 P(X) = 3.3124 \times 0.25 = 0.8281
\]

#### For \( X = 3 \):
\[
(X - \mu)^2 = (3 - 3.82)^2 = (-0.82)^2 = 0.6724
\]
\[
(X - \mu)^2 P(X) = 0.6724 \times 0.14 = 0.0941
\]

#### For \( X = 4 \):
\[
(X - \mu)^2 = (4 - 3.82)^2 = 0.0324
\]
\[
(X - \mu)^2 P(X) = 0.0324 \times 0.2 = 0.0065
\]

#### For \( X = 5 \):
\[
(X - \mu)^2 = (5 - 3.82)^2 = 1.3924
\]
\[
(X - \mu)^2 P(X) = 1.3924 \times 0.38 = 0.5291
\]

#### For \( X = 6 \):
\[
(X - \mu)^2 = (6 - 3.82)^2 = 4.7524
\]
\[
(X - \mu)^2 P(X) = 4.7524 \times 0.01 = 0.0475
\]

#### For \( X = 7 \):
\[
(X - \mu)^2 = (7 - 3.82)^2 = 10.1824
\]
\[
(X - \mu)^2 P(X) = 10.1824 \times 0.02 = 0.2036
\]

Adding these values will give us \( \sum (X - \mu)^2 P(X) \).

### 2. Find the Variance:
The variance is given by:
\[
\sigma^2 = \sum (X - \mu)^2 P(X)
\]
Adding up the calculated values:
\[
0.8281 + 0.0941 + 0.0065 + 0.5291 + 0.0475 + 0.2036 = 1.7089
\]

### 3. Find the Standard Deviation:
The standard deviation is the square root of the variance:
\[
\sigma = \sqrt{1.7089} \approx 1.3073
\]

### Summary:
- \( \sum (X - \mu)^2 P(X) = 1.7089 \)
- Variance (\( \sigma^2 \)) = 1.7089
- Standard Deviation (\( \sigma \)) ≈ 1.3073

Would you like further details or have additional questions?

### Related Questions:
1. How do you find the mean of a probability distribution?
2. What is the significance of variance in probability theory?
3. How would the calculations change if the probability distribution was not valid (i.e., didn't sum to 1)?
4. Can variance be a negative value? Why or why not?
5. How can the variance and standard deviation help understand the spread of data?

### Tip:
Always check that the sum of probabilities in a probability distribution equals 1. This ensures that the distribution is valid.

Complete the probability distribution and calculate the expected value, variance, and standard deviation.

Learn how to complete a probability distribution table and calculate the expected value, variance, and standard deviation of a random variable. This example demonstrates key statistical concepts such as probability distributions, variance, and more.

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