Question

If the probability distribution of a random variable X is as follows, then the mean of X is

X = xi	-1	0	1	2
P(X = xi)	k3	2k3 + k	4k - 10k2	4k - 1

• A. $\frac{193}{27}$
• B. $\frac{25}{27}$
• C. $\frac{23}{27}$
• D. $\frac{83}{27}$

Hint:

Ensure the sum of probabilities equals 1 to find $k$. Compute the mean as $E(X) = \sum x_i P(x_i)$, and verify probabilities are consistent.

Answer 1

The Correct Option is C

To solve for the mean of the random variable \( X \), we need to use the formula for the expected value of a discrete random variable:

1. Understanding the Concepts:

- The mean (or expected value) of a random variable \( X \) is given by:

\[ \mu = E(X) = \sum_{i=1}^{n} x_i \cdot P(X = x_i) \] where \( x_i \) represents the possible values of \( X \) and \( P(X = x_i) \) represents the corresponding probabilities.

2. Given Values:

The given probability distribution is:

X = \( x_i \)	-1	0	1	2
P(X = \( x_i \))	\( k^3 \)	\( 2k^3 + k \)	\( 4k - 10k^2 \)	\( 4k - 1 \)

3. Normalization Condition:

The sum of the probabilities must equal 1, as it represents a complete probability distribution. Therefore, we have:

\[ k^3 + (2k^3 + k) + (4k - 10k^2) + (4k - 1) = 1 \] Simplifying this equation: \[ k^3 + 2k^3 + k + 4k - 10k^2 + 4k - 1 = 1 \] \[ 3k^3 - 10k^2 + 9k - 1 = 1 \] \[ 3k^3 - 10k^2 + 9k - 2 = 0 \] This is a cubic equation in terms of \( k \). Solving this equation, we find \( k = \frac{1}{3} \).

4. Calculating the Mean:

Now that we have \( k = \frac{1}{3} \), we can calculate the mean \( E(X) \). Using the formula for expected value: \[ \mu = E(X) = (-1) \cdot P(X = -1) + 0 \cdot P(X = 0) + 1 \cdot P(X = 1) + 2 \cdot P(X = 2) \] Substitute the probabilities: \[ \mu = (-1) \cdot \left(\frac{1}{3}\right)^3 + 0 \cdot \left(2 \left(\frac{1}{3}\right)^3 + \frac{1}{3}\right) + 1 \cdot \left(4 \cdot \frac{1}{3} - 10 \left(\frac{1}{3}\right)^2\right) + 2 \cdot \left(4 \cdot \frac{1}{3} - 1\right) \] Simplifying each term, we calculate: \[ \mu = \frac{23}{27} \]

Final Answer:

The mean of \( X \) is \( \frac{23}{27} \) (Option 3).