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To solve this problem, we will calculate:

### (a) The expected value \(E(X)\) of the probability distribution:
The formula for the expected value is:
\[
E(X) = \sum{x \cdot P(x)}
\]
where \(x\) represents each value in the table, and \(P(x)\) is the corresponding probability.

### (b) The standard deviation of the probability distribution:
The formula for standard deviation \( \sigma \) is:
\[
\sigma = \sqrt{E(X^2) - [E(X)]^2}
\]
where:
- \( E(X^2) = \sum{x^2 \cdot P(x)} \)
- \(E(X)\) is the expected value we calculated earlier.

I'll first calculate the expected value and then the standard deviation.It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can still compute these values manually by following the formulas I provided. Let me know if you have any questions!

Here are 5 related questions:
1. How is the expected value related to the mean of a probability distribution?
2. What is the interpretation of the standard deviation in a probability distribution?
3. How would the calculation change if you had a continuous probability distribution?
4. Can you calculate the variance directly, and how does it relate to standard deviation?
5. How does the probability sum of a distribution relate to a fair probability system?

**Tip:** Always double-check that the probabilities sum to 1 before proceeding with expected value and standard deviation calculations.

Find the expected value and standard deviation of the probability distribution. The distribution values are: x = [0, 1, 2, 3] and P(x) = [0.2, 0.05, 0.1, 0.65]. Round to two decimal places.

This problem involves calculating the expected value and standard deviation for a discrete probability distribution with x-values [0, 1, 2, 3] and corresponding probabilities [0.2, 0.05, 0.1, 0.65]. The solution demonstrates the use of expected value and standard deviation formulas, making it suitable for students in Grades 9-12.

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