1. **State the problem:** We have a continuous random variable $X$ with probability density function (pdf) $$f(x) = k(2 - x)(2 + x), \quad 0 \leq x \leq 2.$$ We need to find: (i) The value of constant $k$. (ii) $P(X=2)$. (iii) $P(X \leq 1)$. (iv) $P(X \geq 1)$. (v) $P(0.5 \leq X \leq 1.5)$. 2. **Find $k$ using the property of pdf:** The total probability must be 1, so $$\int_0^2 f(x) \, dx = 1.$$ Substitute $f(x)$: $$\int_0^2 k(2 - x)(2 + x) \, dx = 1.$$ Expand the product: $$(2 - x)(2 + x) = 4 - x^2,$$ so $$k \int_0^2 (4 - x^2) \, dx = 1.$$ Calculate the integral: $$k \left[4x - \frac{x^3}{3}\right]_0^2 = 1.$$ Evaluate at limits: $$k \left(4 \times 2 - \frac{2^3}{3} - 0\right) = 1,$$ $$k \left(8 - \frac{8}{3}\right) = 1,$$ $$k \times \frac{24 - 8}{3} = 1,$$ $$k \times \frac{16}{3} = 1,$$ Divide both sides by $\frac{16}{3}$: $$k = \frac{1}{\frac{16}{3}} = \frac{3}{16}.$$ 3. **Calculate $P(X=2)$:** For continuous variables, $P(X = a) = 0$ for any $a$, so $$P(X=2) = 0.$$ 4. **Calculate $P(X \leq 1)$:** $$P(X \leq 1) = \int_0^1 f(x) \, dx = \int_0^1 \frac{3}{16}(4 - x^2) \, dx.$$ Calculate the integral: $$\frac{3}{16} \left[4x - \frac{x^3}{3}\right]_0^1 = \frac{3}{16} \left(4 \times 1 - \frac{1}{3}\right) = \frac{3}{16} \times \frac{12 - 1}{3} = \frac{3}{16} \times \frac{11}{3} = \frac{11}{16}.$$ 5. **Calculate $P(X \geq 1)$:** Since total probability is 1, $$P(X \geq 1) = 1 - P(X < 1) = 1 - P(X \leq 1) = 1 - \frac{11}{16} = \frac{5}{16}.$$ 6. **Calculate $P(0.5 \leq X \leq 1.5)$:** $$P(0.5 \leq X \leq 1.5) = \int_{0.5}^{1.5} \frac{3}{16}(4 - x^2) \, dx = \frac{3}{16} \left[4x - \frac{x^3}{3}\right]_{0.5}^{1.5}.$$ Evaluate at limits: At $x=1.5$: $$4 \times 1.5 - \frac{(1.5)^3}{3} = 6 - \frac{3.375}{3} = 6 - 1.125 = 4.875,$$ At $x=0.5$: $$4 \times 0.5 - \frac{(0.5)^3}{3} = 2 - \frac{0.125}{3} = 2 - 0.0416667 = 1.9583333,$$ Subtract: $$4.875 - 1.9583333 = 2.9166667,$$ Multiply by $\frac{3}{16}$: $$\frac{3}{16} \times 2.9166667 = 0.546875 = \frac{35}{64}.$$ **Final answers:** (i) $k = \frac{3}{16}$ (ii) $P(X=2) = 0$ (iii) $P(X \leq 1) = \frac{11}{16}$ (iv) $P(X \geq 1) = \frac{5}{16}$ (v) $P(0.5 \leq X \leq 1.5) = \frac{35}{64}$