Question

The probability distribution of a random variable $X$ is as follows.

Find $K$, the mean, and variance of $X.

A $[\frac{1}{8}, \frac{22}{27}]$ B $[\frac{1}{8}, \frac{23}{27}]$ C $[\frac{1}{8}, \frac{24}{27}]$ D $[\frac{1}{8}, \frac{3}{4}]$

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Answer

To solve the given problem, we need to determine the value of $K$, the mean, and the variance of the random variable $X$.

Given Probability Distribution

$$ \begin{array}{c|cccc} X = x & 0 & 1 & 2 & 3 \ \hline P(X = x) & K & 3K & 3K & K \ \end{array} $$

Step 1: Finding $K$

The sum of all probabilities must equal 1: $$ K + 3K + 3K + K = 1 $$ Simplify and solve for $K$: $$ 8K = 1 \ K = \frac{1}{8} $$

Step 2: Mean (Expected Value) of $X$

The mean, $E(X)$, is calculated as: $$ E(X) = \sum{X \cdot P(X)} $$ Using the given probabilities: $$ E(X) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8} $$ Calculate the terms: $$ E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} \ E(X) = \frac{12}{8} \ E(X) = \frac{3}{2} $$

Step 3: Variance of $X$

Variance, $Var(X)$, is calculated using: $$ Var(X) = E(X^2) - [E(X)]^2 $$ First, find $E(X^2)$: $$ E(X^2) = \sum{X^2 \cdot P(X)} $$ Using the given probabilities: $$ E(X^2) = 0^2 \cdot \frac{1}{8} + 1^2 \cdot \frac{3}{8} + 2^2 \cdot \frac{3}{8} + 3^2 \cdot \frac{1}{8} $$ Calculate the terms: $$ E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} \ E(X^2) = \frac{24}{8} \ E(X^2) = 3 $$

Now calculate the variance: $$ Var(X) = E(X^2) - [E(X)]^2 \ Var(X) = 3 - \left(\frac{3}{2}\right)^2 \ Var(X) = 3 - \frac{9}{4} \ Var(X) = \frac{12}{4} - \frac{9}{4} \ Var(X) = \frac{3}{4} $$

Summary

  • Value of $K$: $$ K = \frac{1}{8} $$
  • Mean: $$ E(X) = \frac{3}{2} $$
  • Variance: $$ Var(X) = \frac{3}{4} $$

Final Answer

Therefore, the correct answer is: D (\left[\frac{1}{8}, \frac{3}{4}\right])


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