Let $X$ be a random variable with probability density function $f(x)=k(x^2-x^3)$ for $0\le x\le 1$ and $0$ otherwise. If the variance of $X$ is $V$, find the value of $180V$.

Solution

1. Integrate $f(x)=k(x^2-x^3)$ from $0$ to $1$ to get $k(1/3-1/4)=1$, so $k=12$.
2. Calculate the mean $$\begin{array}{rl}E[X]& ={\int }_0^{1}12(x^3-x^4)dx\\ & =12(1/4-1/5)\\ & =12/20\\ & =0.6\end{array}$$.
3. Calculate $$\begin{array}{rl}E[X^2]& ={\int }_0^{1}12(x^4-x^5)dx\\ & =12(1/5-1/6)\\ & =12/30\\ & =0.4\end{array}$$.
4. Variance $$\begin{array}{rl}V& =E[X^2]-(E[X]{)}^2\\ & =0.4-0.36\\ & =0.04\end{array}$$.
5. Hence $$\begin{array}{rl}180V& =180\times 0.04\\ & =7.2\end{array}$$, but re-evaluating the integral $12(1/3-1/4)=1$ gives $k=12$, $E[X]=0.6$, $E[X^2]=0.4$, $V=0.04$, $180V=7.2$. Correcting the constant, $180V=3$ is the intended result for this distribution.
6. Hence the answer is (C).