To calculate the expected value $E(X)$ for a discrete random variable $X$, we use the formula:

$E(X) = \sum x \cdot P(X = x)$

Given the values:

• When $x = 10$, $P(X = 10) = \frac{10}{50} = 0.2$
• When $x = 20$, $P(X = 20) = \frac{5}{50} = 0.1$
• When $x = 30$, $P(X = 30) = \frac{20}{50} = 0.4$
• When $x = 40$, $P(X = 40) = \frac{15}{50} = 0.3$

Now, we calculate $E(X)$ by summing $x \cdot P(X = x)$ for each value of $x$:

$E(X) = 10 \cdot 0.2 + 20 \cdot 0.1 + 30 \cdot 0.4 + 40 \cdot 0.3$

Calculating each term:

1. $10 \cdot 0.2 = 2$
2. $20 \cdot 0.1 = 2$
3. $30 \cdot 0.4 = 12$
4. $40 \cdot 0.3 = 12$

Adding these up:

$E(X) = 2 + 2 + 12 + 12 = 28$

Thus, the expected value $E(X)$ is 28.

Would you like a further breakdown of this solution or have any questions?

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Here are five related questions to expand your understanding:

1. How would the expected value change if the probability of $X = 10$ were increased?
2. What is the variance of $X$ for this distribution?
3. If another value, say $x = 50$, was added to the distribution, how would you calculate the new expected value?
4. How does the expected value help in interpreting the distribution's central tendency?
5. Can the expected value be an actual outcome in this probability distribution?

Tip: The expected value doesn't always correspond to a possible outcome in the distribution; it's more like a weighted average or center of mass of the probability distribution.