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
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 |
Show Hint
- $\frac{193}{27}$
- $\frac{25}{27}$
- $\frac{23}{27}$
- $\frac{83}{27}$
The Correct Option is C
Solution and Explanation
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).
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