1. **Problem statement:** Given the pmf $p(x)$ for $x=0,1,2,3$ as $p(0)=0.2$, $p(1)=0.3$, $p(2)=0.4$, $p(3)=0.1$, we need to verify it is a pmf, find the cdf values, compute the mean $\mu_X$, the expected value of $X^2$, and the variance $\sigma_X^2$. 2. **Verify pmf:** A pmf must satisfy two conditions: - $p(x) \geq 0$ for all $x$ - $\sum_x p(x) = 1$ Check sum: $$0.2 + 0.3 + 0.4 + 0.1 = 1.0$$ All probabilities are non-negative and sum to 1, so $p(x)$ is a valid pmf. 3. **Find cdf $F(x)$:** The cdf is defined as $F(x) = P(X \leq x) = \sum_{t \leq x} p(t)$. - $F(0) = p(0) = 0.2$ - $F(1) = p(0) + p(1) = 0.2 + 0.3 = 0.5$ - $F(1.5)$: Since $1.5$ is between $1$ and $2$, $F(1.5) = F(1) = 0.5$ - $F(2) = p(0) + p(1) + p(2) = 0.2 + 0.3 + 0.4 = 0.9$ - $F(3) = 1$ (sum of all probabilities) 4. **Find mean $\mu_X = E(X)$:** $$E(X) = \sum_x x p(x) = 0 \times 0.2 + 1 \times 0.3 + 2 \times 0.4 + 3 \times 0.1 = 0 + 0.3 + 0.8 + 0.3 = 1.4$$ 5. **Compute $E(X^2)$:** $$E(X^2) = \sum_x x^2 p(x) = 0^2 \times 0.2 + 1^2 \times 0.3 + 2^2 \times 0.4 + 3^2 \times 0.1 = 0 + 0.3 + 1.6 + 0.9 = 2.8$$ 6. **Find variance $\sigma_X^2$:** $$\sigma_X^2 = E(X^2) - (E(X))^2 = 2.8 - (1.4)^2 = 2.8 - 1.96 = 0.84$$ **Final answers:** - pmf is valid - $F(0)=0.2$, $F(1)=0.5$, $F(1.5)=0.5$, $F(2)=0.9$, $F(3)=1$ - $\mu_X = 1.4$ - $E(X^2) = 2.8$ - $\sigma_X^2 = 0.84$