Question 4 of 8: Custom Probability Density Function

Question 4: Custom Probability Density Function (20 marks)

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. $f(y)=Ky(9-y^2)$ on $0\le y\le3$, zero elsewhere.

Find. (a) K and a sketch of f(y). (b) E[Y]. (c) Var(Y). (d) F(y), and a sketch.

Approach. Normalize the total area to 1 to get K, then integrate $y\,f(y)$ and $y^2f(y)$ for the moments, and integrate f(y) from 0 to y for the CDF.

1. (a) Normalizing constant K. $$1=\int_0^3 Ky(9-y^2)\,dy=K\int_0^3(9y-y^3)\,dy=K\left[\dfrac{9y^2}{2}-\dfrac{y^4}{4}\right]_0^3=K(40.5-20.25)=20.25K$$ $$\boxed{K=\dfrac{1}{20.25}=\dfrac{4}{81}\approx0.04938}$$
Fig. 3 — f(y) = (4/81)·y(9−y²) on [0,3]; peak (red dot) at y=√3≈1.73.
2. (b) Mean. $$E[Y]=\int_0^3 y\cdot\dfrac{4}{81}y(9-y^2)\,dy=\dfrac{4}{81}\int_0^3(9y^2-y^4)\,dy=\dfrac{4}{81}\left[3y^3-\dfrac{y^5}{5}\right]_0^3=\dfrac{4}{81}(81-48.6)$$ $$\boxed{E[Y]=\dfrac{4}{81}(32.4)=1.6}$$
3. (c) Variance. First $E[Y^2]=\int_0^3 y^2\cdot\dfrac{4}{81}y(9-y^2)\,dy=\dfrac{4}{81}\left[\dfrac{9y^4}{4}-\dfrac{y^6}{6}\right]_0^3=\dfrac{4}{81}(182.25-121.5)=3.0$. Then $$\text{Var}(Y)=E[Y^2]-(E[Y])^2=3.0-1.6^2=3.0-2.56$$ $$\boxed{\text{Var}(Y)=0.44}$$
4. (d) Cumulative distribution function. For $0\le y\le3$: $$F(y)=\int_0^y\dfrac{4}{81}t(9-t^2)\,dt=\dfrac{4}{81}\left[\dfrac{9t^2}{2}-\dfrac{t^4}{4}\right]_0^y=\dfrac{4}{81}\left(4.5y^2-\dfrac{y^4}{4}\right)$$ $$\boxed{F(y)=\dfrac{18y^{2}-y^{4}}{81}\ (0\le y\le3);\quad F(y)=0\ (y\lt 0);\quad F(y)=1\ (y\gt 3)}$$