Question 4 of 8: Continuous Random Variable — Custom pdf

Question 4: Continuous Random Variable — Custom pdf (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.

Note: the source prints the last two sub-parts re-using the labels (a)(b); they are treated here as (c) Var(X) and (d) the CDF, matching the marking scheme's four-part 5+5+5+5 split.

Given. $f(x)=K(-x^3+3x^2+4x)$ on $0\le x\le4$, zero elsewhere.

Find. (a) K; (b) E(X); (c) Var(X); (d) F(x) and its sketch.

Approach. Use $\int f(x)\,dx=1$ to fix K, then compute the moments $E(X)=\int xf(x)\,dx$, $E(X^2)=\int x^2f(x)\,dx$, $\mathrm{Var}(X)=E(X^2)-E(X)^2$, and $F(x)=\int_0^x f(t)\,dt$ by direct polynomial integration.

1. (a) Normalizing constant. $$\int_0^4(-x^3+3x^2+4x)\,dx=\left[-\dfrac{x^4}{4}+x^3+2x^2\right]_0^4=-64+64+32=32.$$ Setting $32K=1$ gives $K=\boxed{\tfrac{1}{32}}=0.03125$.
2. (b) Mean. $$E(X)=\dfrac{1}{32}\int_0^4(-x^4+3x^3+4x^2)\,dx=\dfrac{1}{32}\left[-\dfrac{x^5}{5}+\dfrac{3x^4}{4}+\dfrac{4x^3}{3}\right]_0^4=\dfrac{1}{32}(72.5333)=\boxed{2.2667}.$$
3. (c) Variance. $$E(X^2)=\dfrac{1}{32}\int_0^4(-x^5+3x^4+4x^3)\,dx=\dfrac{1}{32}\left[-\dfrac{x^6}{6}+\dfrac{3x^5}{5}+x^4\right]_0^4=\dfrac{1}{32}(187.7333)=5.8667.$$ $$\mathrm{Var}(X)=E(X^2)-[E(X)]^2=5.8667-(2.2667)^2=\boxed{0.7289}.$$
4. (d) Cumulative distribution function. For $0\le x\le4$, $$F(x)=\dfrac{1}{32}\int_0^x(-t^3+3t^2+4t)\,dt=\dfrac{1}{32}\left[-\dfrac{x^4}{4}+x^3+2x^2\right]=\boxed{\dfrac{x^2(-x^2+4x+8)}{128}},$$ with $F(x)=0$ for $x<0$ and $F(x)=1$ for $x>4$ (check: $F(4)=\frac{16(-16+16+8)}{128}=\frac{128}{128}=1$).
   F(x) rises monotonically from 0 at x=0 to 1 at x=4, with an inflection reflecting the underlying pdf's single interior mode.