Continuous Random Variable — Custom pdf: 04-BS-2, May 2013, Q4

Question 4: Continuous Random Variable — Custom pdf (25 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. $f(y)=Ky(100-y^2)$ for $0<y<10$, zero otherwise.

Find. (a) K; (b) E(Y); (c) Var(Y); (d) F(y).

Approach. Fix K from the normalization condition $\int_0^{10}f(y)\,dy=1$, then use $E(Y)=\int yf(y)\,dy$, $E(Y^2)=\int y^2f(y)\,dy$, $\mathrm{Var}(Y)=E(Y^2)-[E(Y)]^2$, and $F(y)=\int_0^y f(t)\,dt$.

(a) Normalize to find K. $$\int_0^{10}Ky(100-y^2)\,dy=K\left[50y^2-\dfrac{y^4}{4}\right]_0^{10}=K(5000-2500)=2500K=1\ \Rightarrow\ \boxed{K=\dfrac{1}{2500}=0.0004}.$$
Fig. 4(a) — f(y) = 0.0004·y(100−y²) on 0<y<10, peaking near y ≈ 5.8.
(b) Mean. $$E(Y)=\int_0^{10}yf(y)\,dy=K\int_0^{10}y^2(100-y^2)\,dy=K\left[\dfrac{100y^3}{3}-\dfrac{y^5}{5}\right]_0^{10}=K(33{,}333.33-20{,}000)=\boxed{5.333}.$$
(c) Variance. First compute the second moment, $$E(Y^2)=K\int_0^{10}y^3(100-y^2)\,dy=K\left[25y^4-\dfrac{y^6}{6}\right]_0^{10}=K(250{,}000-166{,}666.67)=33.333.$$ Then $$\mathrm{Var}(Y)=E(Y^2)-[E(Y)]^2=33.333-(5.333)^2=\boxed{4.889}\ \ (\sigma_Y=2.211).$$
(d) Cumulative distribution function. $$F(y)=\int_0^{y}Kt(100-t^2)\,dt=K\left(50y^2-\dfrac{y^4}{4}\right),\qquad 0<y<10,$$ with $F(y)=0$ for $y\le 0$ and $F(y)=1$ for $y\ge 10$. So $$\boxed{F(y)=0.0004\left(50y^2-\dfrac{y^4}{4}\right)},\quad 0<y<10.$$
Fig. 4(d) — F(y), a monotone S–shaped curve rising from 0 at y=0 to 1 at y=10.

Question 4 — final results
Part	Quantity	Value
(a)	K	0.0004 (= 1/2500)
(b)	E(Y)	5.333
(c)	Var(Y)	4.889 (σ=2.211)
(d)	F(y), 0<y<10	0.0004(50y²−y⁴/4)

Topic: Moments and cumulative distribution function of a custom continuous random variable.

Key relations: normalization $\int f=1$; $E(Y)=\int yf\,dy$; $\mathrm{Var}(Y)=E(Y^2)-\mu^2$; $F(y)=\int_{-\infty}^{y}f(t)\,dt$, with $F'(y)=f(y)$ wherever f is continuous.

Why this works: any nonnegative function on a bounded support can be turned into a valid pdf by choosing the single normalizing constant K that makes the total area 1; every other functional (mean, variance, CDF) then follows mechanically by integration — no distributional assumption beyond the given shape is needed.

Common pitfall: computing Var(Y) as $E[(Y-\mu)^2]$ by direct (harder) integration instead of the algebraically-equivalent, easier $E(Y^2)-\mu^2$ shortcut; or forgetting the "otherwise 0" branches when stating F(y).

Source: Walpole, Myers, Myers & Ye, Ch. 3 (Continuous Random Variables, Mathematical Expectation).

Question 4 of 8: Continuous Random Variable — Custom pdf

Question 4: Continuous Random Variable — Custom pdf (25 marks)