Find the mean, variance, and standard deviation of the random variable having the probability distribution given in the following table. (Round your answers to four decimal places.)
Random variable, x	-9	-8	-7	-6	-5
P(X = x)	0.13	0.18	0.36	0.14	0.19

We are given the following probability distribution of the random variable \( X \):

\[
\begin{array}{c|ccccc}
x & -9 & -8 & -7 & -6 & -5 \\
\hline
P(X = x) & 0.13 & 0.18 & 0.36 & 0.14 & 0.19 \\
\end{array}
\]

To find the mean \( \mu \), variance \( \sigma^2 \), and standard deviation \( \sigma \), follow these steps:

### 1. Mean (\( \mu \)):
The mean of a discrete random variable is calculated as:
\[
\mu = \sum (x \cdot P(X = x))
\]
Substitute the given values:

\[
\mu = (-9 \cdot 0.13) + (-8 \cdot 0.18) + (-7 \cdot 0.36) + (-6 \cdot 0.14) + (-5 \cdot 0.19)
\]

Calculating each term:
\[
(-9 \cdot 0.13) = -1.17
\]
\[
(-8 \cdot 0.18) = -1.44
\]
\[
(-7 \cdot 0.36) = -2.52
\]
\[
(-6 \cdot 0.14) = -0.84
\]
\[
(-5 \cdot 0.19) = -0.95
\]

Now summing them up:
\[
\mu = -1.17 - 1.44 - 2.52 - 0.84 - 0.95 = -6.92
\]

### 2. Variance (\( \sigma^2 \)):
The variance is given by:
\[
\sigma^2 = \sum ((x - \mu)^2 \cdot P(X = x))
\]
Since \( \mu = -6.92 \), calculate each term \( (x - \mu)^2 \cdot P(X = x) \):

\[
(-9 - (-6.92))^2 \cdot 0.13 = ( -2.08 )^2 \cdot 0.13 = 4.3264 \cdot 0.13 = 0.5624
\]
\[
(-8 - (-6.92))^2 \cdot 0.18 = ( -1.08 )^2 \cdot 0.18 = 1.1664 \cdot 0.18 = 0.2099
\]
\[
(-7 - (-6.92))^2 \cdot 0.36 = ( -0.08 )^2 \cdot 0.36 = 0.0064 \cdot 0.36 = 0.0023
\]
\[
(-6 - (-6.92))^2 \cdot 0.14 = ( 0.92 )^2 \cdot 0.14 = 0.8464 \cdot 0.14 = 0.1185
\]
\[
(-5 - (-6.92))^2 \cdot 0.19 = ( 1.92 )^2 \cdot 0.19 = 3.6864 \cdot 0.19 = 0.7004
\]

Now sum these values:
\[
\sigma^2 = 0.5624 + 0.2099 + 0.0023 + 0.1185 + 0.7004 = 1.5935
\]

### 3. Standard Deviation (\( \sigma \)):
The standard deviation is the square root of the variance:
\[
\sigma = \sqrt{1.5935} = 1.2623
\]

### Final Answers:
- **Mean** (\( \mu \)): \( -6.9200 \)
- **Variance** (\( \sigma^2 \)): \( 1.5935 \)
- **Standard Deviation** (\( \sigma \)): \( 1.2623 \)

Would you like more details or have any questions?

Here are 5 related questions:
1. How is the expected value related to the mean in probability distributions?
2. What is the difference between population variance and sample variance?
3. How does the shape of the probability distribution affect variance?
4. What is the significance of standard deviation in real-world data?
5. How can we compute the skewness of a probability distribution?

**Tip**: When calculating variance, always ensure to first find the mean accurately before applying the formula for variance.

Find the mean, variance, and standard deviation of the random variable having the probability distribution given in the following table. (Round your answers to four decimal places.) Random variable, x: -9, -8, -7, -6, -5 P(X = x): 0.13, 0.18, 0.36, 0.14, 0.19

Learn how to find the mean, variance, and standard deviation of a random variable with the given probability distribution. This problem covers key statistical concepts, including the expected value and variance formulas. Suitable for students in Grades 10-12.

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