1. **State the problem:** We have a probability density function (pdf) $f(x)$ defined on $[0,k]$ with a triangular shape. The height at $x=0$ is $\frac{1}{2}k$ and it decreases linearly to 0 at $x=k$. We need to find: (a) the constant $k$, (b) the function $f(x)$ and the expected value $E(X)$, (c) the value $p$ such that $P(p < X < 1) = 0.25$. 2. **Find $k$ (part a):** Since $f(x)$ is a pdf, the total area under the curve must be 1. The graph is a triangle with base $k$ and height $\frac{1}{2}k$. Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times k \times \frac{1}{2}k = \frac{k^2}{4}$. Set area equal to 1: $$\frac{k^2}{4} = 1 \implies k^2 = 4 \implies k = 2$$ 3. **Find $f(x)$ and $E(X)$ (part b):** Since $k=2$, the height at $x=0$ is $\frac{1}{2} \times 2 = 1$. The function $f(x)$ decreases linearly from 1 at $x=0$ to 0 at $x=2$. Equation of the line: $$f(x) = 1 - \frac{1}{2}x$$ for $0 \leq x \leq 2$. To find $E(X)$: $$E(X) = \int_0^2 x f(x) \, dx = \int_0^2 x \left(1 - \frac{x}{2}\right) dx = \int_0^2 \left(x - \frac{x^2}{2}\right) dx$$ Calculate the integral: $$\int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = 2$$ $$\int_0^2 \frac{x^2}{2} \, dx = \frac{1}{2} \left[ \frac{x^3}{3} \right]_0^2 = \frac{1}{2} \times \frac{8}{3} = \frac{4}{3}$$ So, $$E(X) = 2 - \frac{4}{3} = \frac{6}{3} - \frac{4}{3} = \frac{2}{3}$$ 4. **Find $p$ such that $P(p < X < 1) = 0.25$ (part c):** We want: $$P(p < X < 1) = \int_p^1 f(x) \, dx = 0.25$$ Recall $f(x) = 1 - \frac{x}{2}$. Calculate the integral: $$\int_p^1 \left(1 - \frac{x}{2}\right) dx = \left[ x - \frac{x^2}{4} \right]_p^1 = \left(1 - \frac{1}{4}\right) - \left(p - \frac{p^2}{4}\right) = \frac{3}{4} - p + \frac{p^2}{4}$$ Set equal to 0.25: $$\frac{3}{4} - p + \frac{p^2}{4} = \frac{1}{4}$$ Multiply both sides by 4: $$3 - 4p + p^2 = 1$$ Rearranged: $$p^2 - 4p + 2 = 0$$ Solve quadratic: $$p = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$$ Since $p$ must be between 0 and 1, choose: $$p = 2 - \sqrt{2} \approx 0.586$$ **Final answers:** - (a) $k = 2$ - (b) $f(x) = 1 - \frac{x}{2}$ for $0 \leq x \leq 2$, and $E(X) = \frac{2}{3}$ - (c) $p = 2 - \sqrt{2}$