Question

Let the probability distribution of a random variable X be given by the following table: {cccccc} X & -1 & 0 & 1 & 2 & 3
$p(X)$ & a & 2a & 3a & 4a & 5a
Then the expectation of X is

• A. $\frac{1}{5}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $\frac{4}{15}$
• E. $\frac{5}{3}$

Hint:

Shortcut Tip: You don't have to evaluate $a$ in the very first step. It's often mathematically cleaner to find the expectation purely in terms of $a$ (like getting $25a$), and only substitute the fraction $a = 1/15$ at the very end to avoid messy intermediate fractions.

Answer 1

Answer: (E)

Concept:
For any valid probability distribution, the sum of all individual probabilities must equal exactly 1 ($\sum P(X) = 1$). The expectation (or expected value) of a discrete random variable $X$ is defined as the sum of the products of each possible outcome and its corresponding probability: $E(X) = \sum [x \cdot P(x)]$.

Step 1: Set the sum of all probabilities to 1.
Sum the values in the $P(X)$ row of the table: $$a + 2a + 3a + 4a + 5a = 1$$

Step 2: Solve for the constant a.
Combine the like terms on the left side: $$15a = 1$$ $$a = \frac{1}{15}$$

Step 3: Set up the expected value formula.
Multiply each $X$ value by its corresponding $P(X)$ probability: $$E(X) = (-1)(a) + (0)(2a) + (1)(3a) + (2)(4a) + (3)(5a)$$

Step 4: Simplify the expectation expression in terms of a.
Perform the individual multiplications and sum them up: $$E(X) = -a + 0 + 3a + 8a + 15a$$ $$E(X) = 25a$$

Step 5: Substitute the value of a to find the final numerical answer.
Plug in $a = \frac{1}{15}$ from Step 2 into the expectation expression: $$E(X) = 25 \left( \frac{1}{15} \right)$$ $$E(X) = \frac{25}{15}$$ Divide the top and bottom by 5 to simplify the fraction: $$E(X) = \frac{5}{3}$$ Hence the correct answer is (E) $\frac{5{3}$}.