Math Problem Statement
A discrete random variable X has the distribution function: P(X = x) = k(1 - x)^2, where x = -1, 0, 1, 2. Find k, E(X), E(X^2), and Var(X).
Solution
Let's walk through the problem step-by-step.
The given probability mass function is:
\begin{cases} k(1 - x)^2 & \text{for } x = -1, 0, 1, 2, \\ 0 & \text{otherwise}. \end{cases}$$ --- ### (i) Show that $$k = \frac{1}{6}$$. Since this is a probability mass function, the total sum of all probabilities must be 1: $$\sum_{x = -1}^{2} P(X = x) = 1.$$ We substitute the function $$P(X = x) = k(1 - x)^2$$ for the valid values of $$x$$: $$k(1 - (-1))^2 + k(1 - 0)^2 + k(1 - 1)^2 + k(1 - 2)^2 = 1.$$ Evaluating each term: - When $$x = -1$$, $$(1 - (-1))^2 = (2)^2 = 4$$. - When $$x = 0$$, $$(1 - 0)^2 = 1^2 = 1$$. - When $$x = 1$$, $$(1 - 1)^2 = 0^2 = 0$$. - When $$x = 2$$, $$(1 - 2)^2 = (-1)^2 = 1$$. Thus, the equation becomes: $$k(4 + 1 + 0 + 1) = 1.$$ $$k(6) = 1 \quad \Rightarrow \quad k = \frac{1}{6}.$$ So, $$k = \frac{1}{6}$$ is correct. --- ### (ii) Find $$E(X)$$. The expected value $$E(X)$$ is defined as: $$E(X) = \sum_{x = -1}^{2} x \cdot P(X = x).$$ Using $$P(X = x) = \frac{1}{6}(1 - x)^2$$, we compute each term: - For $$x = -1$$: $$-1 \cdot \frac{1}{6}(4) = -\frac{4}{6} = -\frac{2}{3}.$$ - For $$x = 0$$: $$0 \cdot \frac{1}{6}(1) = 0.$$ - For $$x = 1$$: $$1 \cdot \frac{1}{6}(0) = 0.$$ - For $$x = 2$$: $$2 \cdot \frac{1}{6}(1) = \frac{2}{6} = \frac{1}{3}.$$ Thus: $$E(X) = -\frac{2}{3} + 0 + 0 + \frac{1}{3} = -\frac{1}{3}.$$ --- ### (iii) Find $$E(X^2)$$. The expected value $$E(X^2)$$ is defined as: $$E(X^2) = \sum_{x = -1}^{2} x^2 \cdot P(X = x).$$ We compute each term: - For $$x = -1$$: $$(-1)^2 \cdot \frac{1}{6}(4) = 1 \cdot \frac{4}{6} = \frac{2}{3}.$$ - For $$x = 0$$: $$(0)^2 \cdot \frac{1}{6}(1) = 0.$$ - For $$x = 1$$: $$(1)^2 \cdot \frac{1}{6}(0) = 0.$$ - For $$x = 2$$: $$(2)^2 \cdot \frac{1}{6}(1) = 4 \cdot \frac{1}{6} = \frac{2}{3}.$$ Thus: $$E(X^2) = \frac{2}{3} + 0 + 0 + \frac{2}{3} = \frac{4}{3}.$$ --- ### (iv) Find the Variance $$\text{Var}(X)$$. The variance $$\text{Var}(X)$$ is given by: $$\text{Var}(X) = E(X^2) - \left( E(X) \right)^2.$$ From the previous parts, we know: - $$E(X^2) = \frac{4}{3}$$, - $$E(X) = -\frac{1}{3}$$. Thus: $$\text{Var}(X) = \frac{4}{3} - \left( -\frac{1}{3} \right)^2 = \frac{4}{3} - \frac{1}{9}.$$ Simplifying: $$\frac{4}{3} = \frac{12}{9}, \quad \frac{12}{9} - \frac{1}{9} = \frac{11}{9}.$$ So: $$\text{Var}(X) = \frac{11}{9}.$$ --- ### Final Answers: 1. $$k = \frac{1}{6}$$. 2. $$E(X) = -\frac{1}{3}$$. 3. $$E(X^2) = \frac{4}{3}$$. 4. $$\text{Var}(X) = \frac{11}{9}$$. --- Would you like further details or have any questions? Here are some related questions for further exploration: 1. How does the variance measure the spread of a probability distribution? 2. What is the significance of finding $$E(X)$$ in a distribution? 3. How would the distribution change if we modified the function to $$k(1 - x)^3$$? 4. Can the value of $$E(X)$$ be used to predict outcomes? In what situations? 5. How is the variance affected if the random variable is shifted or scaled? **Tip:** When solving probability problems, always ensure the probabilities sum to 1 to confirm you’ve used the correct constant $$k$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Discrete Random Variables
Expected Value
Variance
Formulas
P(X = x) = k(1 - x)^2 for x = -1, 0, 1, 2
E(X) = Σ x * P(X = x)
E(X^2) = Σ x^2 * P(X = x)
Var(X) = E(X^2) - (E(X))^2
Theorems
Basic Properties of Probability Distributions
Suitable Grade Level
University level (Undergraduate Probability/Statistics)
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