To find the variance of a discrete probability distribution, we use the formula:

$\text{Variance} = \sigma^2 = \sum (x_i - \mu)^2 P(X = x_i)$

Where:

• $x_i$ is each value of $x$,

• $P(X = x_i)$ is the probability of $x_i$,

• $\mu$ is the expected value (mean) of the distribution, which can be calculated using the formula:

  $\mu = \sum x_i P(X = x_i)$

Step 1: Calculate the expected value (mean) $\mu$.

$\mu = (1)(0.3) + (2)(0.1) + (3)(0.2) + (4)(0.1) + (5)(0.3)$

$\mu = 0.3 + 0.2 + 0.6 + 0.4 + 1.5 = 3.0$

Step 2: Use the variance formula.

We will now calculate $(x_i - \mu)^2$ for each $x_i$ and multiply it by the corresponding probability $P(X = x_i)$:

$(x_1 - \mu)^2 = (1 - 3.0)^2 = 4$ $(x_2 - \mu)^2 = (2 - 3.0)^2 = 1$ $(x_3 - \mu)^2 = (3 - 3.0)^2 = 0$ $(x_4 - \mu)^2 = (4 - 3.0)^2 = 1$ $(x_5 - \mu)^2 = (5 - 3.0)^2 = 4$

Now, multiply each by the corresponding probability:

$\text{Variance} = (4)(0.3) + (1)(0.1) + (0)(0.2) + (1)(0.1) + (4)(0.3)$ $\text{Variance} = 1.2 + 0.1 + 0 + 0.1 + 1.2 = 2.6$

Final Answer:

The variance of this distribution is 2.6 (rounded to one decimal place).

---

Would you like me to explain any step further or help with any other questions?

Here are some related questions:

1. How do you calculate the standard deviation from the variance?
2. How can the expected value (mean) change if the probabilities are adjusted?
3. What is the significance of variance in probability distributions?
4. How would the variance change if $x = 6$ had a higher probability?
5. How do you interpret a variance value in the context of real-life data?

Tip: Variance gives you an idea of how spread out the data is, but for a more intuitive measure, the standard deviation is often used, as it’s in the same units as the data.