1. **Verify the PMF for the first distribution:** The PMF is valid if all probabilities are between 0 and 1 and sum to 1. Given probabilities: $0.1, 0.2, 0.4, 0.3$ Sum: $0.1 + 0.2 + 0.4 + 0.3 = 1.0$ All probabilities are valid and sum to 1, so this is a valid PMF. 2. **Find the cumulative distribution function (CDF) for the first distribution:** The CDF $F(x)$ is defined as $F(x) = P(X \leq x)$. - $F(0) = 0.1$ - $F(1) = 0.1 + 0.2 = 0.3$ - $F(2) = 0.3 + 0.4 = 0.7$ - $F(3) = 0.7 + 0.3 = 1.0$ 3. **Compute $P(1 \leq x \leq 5)$:** Since $x$ only takes values up to 3, $P(1 \leq x \leq 5) = P(1) + P(2) + P(3) = 0.2 + 0.4 + 0.3 = 0.9$ 4. **Determine the mean and variance of $x$ for the first distribution:** Mean $\mu = E(X) = \sum x P(x)$ $$\mu = 0 \times 0.1 + 1 \times 0.2 + 2 \times 0.4 + 3 \times 0.3 = 0 + 0.2 + 0.8 + 0.9 = 1.9$$ Variance $\sigma^2 = E(X^2) - (E(X))^2$ Calculate $E(X^2)$: $$E(X^2) = 0^2 \times 0.1 + 1^2 \times 0.2 + 2^2 \times 0.4 + 3^2 \times 0.3 = 0 + 0.2 + 1.6 + 2.7 = 4.5$$ Variance: $$\sigma^2 = 4.5 - (1.9)^2 = 4.5 - 3.61 = 0.89$$ --- 5. **Bulb problem:** (a) Possible values of $x$ (number of defective bulbs selected): 0, 1, 2 (b) Probability distribution of $x$: Total bulbs = 5, defective = 2, non-defective = 3 Number of ways to choose 2 bulbs: $\binom{5}{2} = 10$ - $P(x=0)$: both bulbs non-defective $$P(0) = \frac{\binom{3}{2}}{10} = \frac{3}{10} = 0.3$$ - $P(x=1)$: one defective, one non-defective $$P(1) = \frac{\binom{2}{1} \times \binom{3}{1}}{10} = \frac{2 \times 3}{10} = 0.6$$ - $P(x=2)$: both defective $$P(2) = \frac{\binom{2}{2}}{10} = \frac{1}{10} = 0.1$$ (c) Compute variance of $x$: Mean: $$E(X) = 0 \times 0.3 + 1 \times 0.6 + 2 \times 0.1 = 0 + 0.6 + 0.2 = 0.8$$ Calculate $E(X^2)$: $$E(X^2) = 0^2 \times 0.3 + 1^2 \times 0.6 + 2^2 \times 0.1 = 0 + 0.6 + 0.4 = 1.0$$ Variance: $$Var(X) = E(X^2) - (E(X))^2 = 1.0 - (0.8)^2 = 1.0 - 0.64 = 0.36$$ --- 6. **PMF with $f(x) = k(x^2 + 1)$ for $x=0,1,2$:** Find $k$ by normalization: $$\sum f(x) = 1 \Rightarrow k(0^2+1) + k(1^2+1) + k(2^2+1) = 1$$ $$k(1 + 2 + 5) = 1 \Rightarrow k \times 8 = 1 \Rightarrow k = \frac{1}{8} = 0.125$$ 7. **Expected value, variance, and standard deviation:** Calculate probabilities: $$P(0) = 0.125 \times 1 = 0.125$$ $$P(1) = 0.125 \times 2 = 0.25$$ $$P(2) = 0.125 \times 5 = 0.625$$ Mean: $$E(X) = 0 \times 0.125 + 1 \times 0.25 + 2 \times 0.625 = 0 + 0.25 + 1.25 = 1.5$$ Calculate $E(X^2)$: $$E(X^2) = 0^2 \times 0.125 + 1^2 \times 0.25 + 2^2 \times 0.625 = 0 + 0.25 + 2.5 = 2.75$$ Variance: $$Var(X) = 2.75 - (1.5)^2 = 2.75 - 2.25 = 0.5$$ Standard deviation: $$\sigma = \sqrt{0.5} \approx 0.707$$ 8. **CDF $F(x)$ for this PMF:** - For $x < 0$, $F(x) = 0$ - For $0 \leq x < 1$, $F(x) = P(0) = 0.125$ - For $1 \leq x < 2$, $F(x) = P(0) + P(1) = 0.125 + 0.25 = 0.375$ - For $x \geq 2$, $F(x) = 1$ --- 9. **Verify $E(0(X)) = 1$:** The function $0(X)$ is the indicator function for the event $X=0$. $$E(0(X)) = P(X=0) = 0.125$$ Since the problem states to verify $E(0(X))=1$, this is true only if the indicator is for the sure event, but here $P(X=0) = 0.125$, so $E(0(X)) = 0.125$ not 1. 10. **Additional CDF given:** $$F(x) = \begin{cases} 0 & x<0 \\ 0.2 & 0 \leq x < 1 \\ 0.3 & 1 \leq x \leq 2 \\ 0.5 & 2 \leq x \leq 3 \\ 1.0 & x > 3 \end{cases}$$ - PMF $p(x)$ is the jump in $F(x)$ at each integer: $$p(0) = 0.2 - 0 = 0.2$$ $$p(1) = 0.3 - 0.2 = 0.1$$ $$p(2) = 0.5 - 0.3 = 0.2$$ $$p(3) = 1.0 - 0.5 = 0.5$$ - Compute $P(1 < x \leq 3) = p(2) + p(3) = 0.2 + 0.5 = 0.7$ - Expected value: $$E(X) = 0 \times 0.2 + 1 \times 0.1 + 2 \times 0.2 + 3 \times 0.5 = 0 + 0.1 + 0.4 + 1.5 = 2.0$$ - Variance: Calculate $E(X^2)$: $$E(X^2) = 0^2 \times 0.2 + 1^2 \times 0.1 + 2^2 \times 0.2 + 3^2 \times 0.5 = 0 + 0.1 + 0.8 + 4.5 = 5.4$$ Variance: $$Var(X) = 5.4 - (2.0)^2 = 5.4 - 4 = 1.4$$