To find the variance of the discrete random variable $X$, we will follow these steps:

1. List of values and probabilities:

   • $X = -13, -10, -9, -5, 2, 10$
   • Corresponding probabilities: $P(X = -13) = \frac{1}{49}, P(X = -10) = \frac{13}{49}, P(X = -9) = \frac{9}{49}, P(X = -5) = \frac{10}{49}, P(X = 2) = \frac{11}{49}, P(X = 10) = \frac{5}{49}$

2. Compute the expected value $\mathbb{E}[X]$: $\mathbb{E}[X] = \sum_i x_i P(x_i) = (-13)\frac{1}{49} + (-10)\frac{13}{49} + (-9)\frac{9}{49} + (-5)\frac{10}{49} + 2\frac{11}{49} + 10\frac{5}{49}$

3. Compute the expected value of $X^2$ (i.e., $\mathbb{E}[X^2]$): $\mathbb{E}[X^2] = \sum_i x_i^2 P(x_i) = (-13)^2\frac{1}{49} + (-10)^2\frac{13}{49} + (-9)^2\frac{9}{49} + (-5)^2\frac{10}{49} + 2^2\frac{11}{49} + 10^2\frac{5}{49}$

4. Find the variance $\text{Var}(X)$: [ \text{Var}(X) = \mathbb{E}[X^2] - (\mathbb{E}[X])^2 ]

Let me calculate each step.Here are the results for the discrete random variable $X$:

1. Expected value $\mathbb{E}[X]$: $\mathbb{E}[X] = -4.1224$

2. Expected value of $X^2$, i.e., $\mathbb{E}[X^2]$: $\mathbb{E}[X^2] = 61.0612$

3. Variance of $X$, i.e., $\text{Var}(X)$: $\text{Var}(X) = 44.0666$

If you have any questions or would like more details, feel free to ask!

Here are 5 related questions:

1. How would the variance change if we adjusted the probabilities?
2. What is the significance of a negative expected value?
3. How is variance affected if the data points are further spread out?
4. How does the expected value of $X^2$ contribute to calculating variance?
5. Can you derive the standard deviation from the variance?

Tip: The standard deviation is the square root of the variance, giving a measure of spread in the same units as the data itself.