Question 4 of 8

Question 4 (20 marks: (a) 5, (b) 5, (c) 5, (d) 5)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f(y)=K(y^3-9y^2+18y)$ on $0\le y\le 3$, zero elsewhere.

Find. (a) K and the graph of f(y). (b) E(Y). (c) Var(Y). (d) F(y) and its graph.

Approach. Use $\int f(y)\,dy=1$ to solve for K, then the standard moment integrals $E(Y)=\int y f(y)\,dy$, $E(Y^2)=\int y^2 f(y)\,dy$, and $F(y)=\int_0^y f(u)\,du$.

1. (a) Solve for K. $$\int_0^3 (y^3-9y^2+18y)\,dy = \left[\frac{y^4}{4}-3y^3+9y^2\right]_0^3 = \frac{81}{4}-81+81 = \frac{81}{4}.$$ Setting $K\cdot\frac{81}{4}=1$ gives $$K = \frac{4}{81} = \boxed{0.04938}.$$ The graph (below) shows a single hump rising from $f(0)=0$ to a peak near $y\approx 1$, then falling back to $f(3)=0$.
2. (b) Find E(Y). $$E(Y) = K\int_0^3 y(y^3-9y^2+18y)\,dy = K\int_0^3 (y^4-9y^3+18y^2)\,dy = K\left[\frac{y^5}{5}-\frac{9y^4}{4}+6y^3\right]_0^3 = K(28.35) = \boxed{1.40}.$$
3. (c) Find Var(Y). First $E(Y^2)=K\int_0^3(y^5-9y^4+18y^3)\,dy = K\left[\frac{y^6}{6}-\frac{9y^5}{5}+\frac{18y^4}{4}\right]_0^3 = K(48.6)=2.40$. Then $$\text{Var}(Y) = E(Y^2)-[E(Y)]^2 = 2.40-(1.40)^2 = \boxed{0.44}.$$
4. (d) Find F(y). Integrating $f$ from 0 to $y$ (for $0\le y\le 3$): $$F(y) = K\left[\frac{y^4}{4}-3y^3+9y^2\right] = \frac{4}{81}\left(\frac{y^4}{4}-3y^3+9y^2\right), \qquad F(y)=0\ (y<0),\ F(y)=1\ (y>3).$$ Check: $F(0)=0$, $F(3)=\boxed{1.000}$ — the graph rises monotonically from 0 to 1 with the steepest slope where $f(y)$ peaks (near $y=1$).