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To find the **mean** and **standard deviation** of the given probability distribution, we can use the following formulas:

1. **Mean** (\(\mu\)):  
   \[
   \mu = \sum (x \cdot P(x))
   \]
   where \(x\) is each value and \(P(x)\) is the corresponding probability.

2. **Variance** (\(\sigma^2\)):  
   \[
   \sigma^2 = \sum (x^2 \cdot P(x)) - \mu^2
   \]
   where \(\mu\) is the mean.

3. **Standard Deviation** (\(\sigma\)):  
   \[
   \sigma = \sqrt{\sigma^2}
   \]

### Given:
\[
\begin{array}{c|c}
x & P(x) \\
\hline
1 & 0.2 \\
5 & 0.15 \\
7 & 0.15 \\
11 & 0.5 \\
\end{array}
\]

Let's calculate the mean and standard deviation. I'll compute these now.The **mean** of the given probability distribution is \(7.50\) (rounded to two decimal places).

The **standard deviation** is approximately \(3.94\).

Would you like further details on the steps? Here are a few related questions:

1. How does a probability distribution affect the mean?
2. What is the difference between variance and standard deviation?
3. How would the standard deviation change if the probabilities were different?
4. What happens to the mean if the largest value (\(x = 11\)) had a lower probability?
5. Can the standard deviation ever be larger than the mean?

**Tip**: For a probability distribution, the mean gives the "expected" value, while the standard deviation measures how much variation or dispersion exists from the mean.

Find the mean and standard deviation of the given probability distribution with x values: 1, 5, 7, and 11, and their respective probabilities: 0.2, 0.15, 0.15, and 0.5.

Learn how to calculate the mean and standard deviation for a probability distribution with x values: 1, 5, 7, and 11, and probabilities 0.2, 0.15, 0.15, and 0.5. This problem demonstrates key probability concepts including expected values and dispersion.

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