1. **Problem statement:** We have a continuous random variable $X$ with probability density function (pdf): $$f(x) = \begin{cases} k(x - 3)^2 & 2 \leq x \leq 6 \\ 0 & \text{otherwise} \end{cases}$$ We need to: (a) Sketch the graph of $y = f(x)$. (b) Find the mode of $X$. (c) (i) Show that $k = \frac{3}{28}$ using integration. (ii) Verify the upper quartile lies between 5.71 and 5.72. 2. **Sketching the graph:** The function $f(x) = k(x-3)^2$ is a parabola opening upwards with vertex at $x=3$. Since $k > 0$, the parabola has a minimum at $x=3$ and increases as $x$ moves away from 3. The pdf is zero outside $[2,6]$. 3. **Mode of $X$:** The mode is the value of $x$ where $f(x)$ is maximum. Since $f(x)$ is a parabola opening upwards with minimum at $x=3$, the maximum on $[2,6]$ occurs at the endpoints. Calculate $f(2) = k(2-3)^2 = k(1)^2 = k$ and $f(6) = k(6-3)^2 = k(3)^2 = 9k$. So the maximum is at $x=6$ with value $9k$. 4. **Finding $k$ by integration:** Since $f(x)$ is a pdf, total area under curve must be 1: $$\int_{2}^{6} k(x-3)^2 dx = 1$$ Calculate the integral: $$\int_{2}^{6} (x-3)^2 dx = \int_{2}^{6} (x^2 - 6x + 9) dx$$ $$= \left[ \frac{x^3}{3} - 3x^2 + 9x \right]_2^6$$ Evaluate at bounds: At $x=6$: $$\frac{6^3}{3} - 3(6^2) + 9(6) = \frac{216}{3} - 3(36) + 54 = 72 - 108 + 54 = 18$$ At $x=2$: $$\frac{2^3}{3} - 3(2^2) + 9(2) = \frac{8}{3} - 12 + 18 = \frac{8}{3} + 6 = \frac{8}{3} + \frac{18}{3} = \frac{26}{3}$$ Subtract: $$18 - \frac{26}{3} = \frac{54}{3} - \frac{26}{3} = \frac{28}{3}$$ So: $$k \times \frac{28}{3} = 1 \implies k = \frac{3}{28}$$ 5. **Verifying the upper quartile $Q_3$:** The upper quartile $Q_3$ satisfies: $$P(X \leq Q_3) = 0.75$$ Calculate: $$\int_2^{Q_3} f(x) dx = 0.75$$ Substitute $f(x)$ and $k$: $$\int_2^{Q_3} \frac{3}{28} (x-3)^2 dx = 0.75$$ Calculate the integral: $$\frac{3}{28} \int_2^{Q_3} (x-3)^2 dx = 0.75$$ From previous integral formula: $$\int_2^{Q_3} (x-3)^2 dx = \left[ \frac{(Q_3)^3}{3} - 3(Q_3)^2 + 9Q_3 \right] - \left[ \frac{8}{3} - 12 + 18 \right] = \left[ \frac{(Q_3)^3}{3} - 3(Q_3)^2 + 9Q_3 \right] - \frac{26}{3}$$ So: $$\frac{3}{28} \left( \frac{(Q_3)^3}{3} - 3(Q_3)^2 + 9Q_3 - \frac{26}{3} \right) = 0.75$$ Multiply both sides by $\frac{28}{3}$: $$\frac{(Q_3)^3}{3} - 3(Q_3)^2 + 9Q_3 - \frac{26}{3} = 0.75 \times \frac{28}{3} = 7$$ Add $\frac{26}{3}$ to both sides: $$\frac{(Q_3)^3}{3} - 3(Q_3)^2 + 9Q_3 = 7 + \frac{26}{3} = \frac{21}{3} + \frac{26}{3} = \frac{47}{3}$$ Multiply both sides by 3: $$(Q_3)^3 - 9(Q_3)^2 + 27Q_3 = 47$$ Rewrite: $$(Q_3)^3 - 9(Q_3)^2 + 27Q_3 - 47 = 0$$ Check values between 5.71 and 5.72: At $x=5.71$: $$5.71^3 - 9(5.71)^2 + 27(5.71) - 47 \approx -0.07$$ At $x=5.72$: $$5.72^3 - 9(5.72)^2 + 27(5.72) - 47 \approx 0.20$$ Since the function changes sign between 5.71 and 5.72, the root (upper quartile) lies in this interval. **Final answers:** - (a) Graph is an upward parabola segment on $[2,6]$ with minimum at $x=3$. - (b) Mode is at $x=6$. - (c)(i) $k = \frac{3}{28}$. - (c)(ii) Upper quartile lies between 5.71 and 5.72.