To fill in the last line of the table, we need to find the probability $P(X)$ for each outcome. The probability is calculated by dividing the frequency of each outcome by the total number of observations.

Step 1: Calculate the Total Frequency

$\text{Total Frequency} = 26 + 22 + 27 + 25 + 12 = 112$

Step 2: Find $P(X)$ for Each Outcome

$P(X = -2) = \frac{26}{112} = \frac{13}{56} \approx 0.2321$ $P(X = -1) = \frac{22}{112} = \frac{11}{56} \approx 0.1964$ $P(X = 0) = \frac{27}{112} = \frac{27}{112} \approx 0.2411$ $P(X = 1) = \frac{25}{112} = \frac{25}{112} \approx 0.2232$ $P(X = 2) = \frac{12}{112} = \frac{3}{28} \approx 0.1071$

The completed table with probabilities is:

Outcome	-2	-1	0	1	2
Frequency	26	22	27	25	12
$P(X)$	0.2321	0.1964	0.2411	0.2232	0.1071

Step 3: Calculate the Expected Value $E(X)$

$E(X) = \sum (x \cdot P(x))$ $E(X) = (-2 \cdot 0.2321) + (-1 \cdot 0.1964) + (0 \cdot 0.2411) + (1 \cdot 0.2232) + (2 \cdot 0.1071)$ $E(X) = -0.4642 - 0.1964 + 0 + 0.2232 + 0.2142$ $E(X) = -0.2232$

Therefore, the expected value of $X$ is: $E(X) = -0.2232$

Would you like a more detailed explanation or have any questions?

Here are 5 related questions you might find interesting:

1. How do you calculate variance and standard deviation for a discrete random variable?
2. What would be the expected value if the outcomes had different weights?
3. How does the expected value change if the frequency distribution is modified?
4. Can the expected value be a non-integer even if all outcomes are integers?
5. How can we use expected value in real-life situations, such as in gambling or insurance?

Tip: When finding probabilities, always ensure the sum of $P(X)$ values equals 1, as a quick way to check your calculations.