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{\bf Question}

Suppose that the pdf of a r.v. $X$ is given by

$$f(x)=\left\{ \begin{array} {ll} c(9-x^2), & {\rm for}\ -3\leq x
\leq 3 ,\\ 0, & {\rm otherwise} \end{array} \right.$$

Determine the value of the constant $c$.  Find the cdf of $X$, and
sketch the pdf and cdf of $X$.  Find the values of the following
probabilities:

$$P\{X<0\},\ P\{-1\leq X \leq 1\},\ P\{X>2\}.$$


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{\bf Answer}

For $f(x)$ to be a pdf, it is necessary that

$$\int_{-\infty}^{\infty} f(u) \,du=\int_{-3}^{3} c(9-u^2) \,du =
1$$

and so $\displaystyle c=\frac{1}{36}$.  Using the relationship
between cdf and pdf

\begin{eqnarray*} & & F(x) = \displaystyle \int_{-\infty}^{x}
f(u) \,du\\ & = & \displaystyle \int_{-3}^x \displaystyle
\frac{9-u^2}{36} \,du \ \ x \in (-3,3)\\& = & \displaystyle
\frac{(18+9x-\frac{x^3}{3})}{36} \ \ x \in (-3,3)
\end{eqnarray*}

Consequently,

\hspace{1in}$\displaystyle P\{X<0\} = F(0) = \frac{1}{2},$

\hspace{1in}$\displaystyle P\{-1\le X\le
1\}=F(1)-F(-1)=\frac{13}{27}$

\hspace{1in}$\displaystyle P\{X>2\}=1-F(2)=\frac{2}{27}$


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