1. **Problem Statement:** Given the probability density function (pdf) $f(x) = 3x^2$ for $0 \leq x \leq 1$, find values of $a$ and $b$ such that: i) $P(X \leq a) = P(X > b) = 0.05$ ii) $P(X > b) = 0.05$ 2. **Formula and Rules:** The cumulative distribution function (CDF) is given by: $$F(x) = P(X \leq x) = \int_0^x 3t^2 dt = [t^3]_0^x = x^3$$ Since $f(x)$ is a valid pdf, $F(1) = 1$. 3. **Step i) Find $a$ and $b$ such that $P(X \leq a) = 0.05$ and $P(X > b) = 0.05$:** - From $P(X \leq a) = 0.05$, we have: $$F(a) = a^3 = 0.05$$ $$a = \sqrt[3]{0.05}$$ - From $P(X > b) = 0.05$, we have: $$P(X > b) = 1 - F(b) = 0.05 \implies F(b) = 0.95$$ $$b^3 = 0.95 \implies b = \sqrt[3]{0.95}$$ 4. **Step ii) Find $b$ such that $P(X > b) = 0.05$:** This is the same as above: $$b = \sqrt[3]{0.95}$$ 5. **Calculations:** $$a = 0.05^{1/3} \approx 0.368$$ $$b = 0.95^{1/3} \approx 0.983$$ 6. **Final answers:** i) $a \approx 0.368$, $b \approx 0.983$ ii) $b \approx 0.983$