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If the probability density function of a continuous random variable is $f(x)=\frac{x^{3}}{3}$ if $-1 < x < 2$
$=0$, otherwise,
then the cumulative distribution function of $X$ is
Options:
$=0$, otherwise,
then the cumulative distribution function of $X$ is
Solution:
1693 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{12}\left[x^{4}-1\right]$
c.d.f. of $x$ is given by
$\begin{aligned}
\mathrm{f}(\mathrm{x}) &=\int_{-1}^{\mathrm{x}} \mathrm{f}(\mathrm{y}) \mathrm{dy} \quad=\int_{-1}^{\mathrm{x}} \frac{\mathrm{y}^{3}}{3} \mathrm{dy} \\
&=\left[\frac{\mathrm{y}^{4}}{12}\right]_{-1}^{\mathrm{x}}=\frac{1}{12}\left(\mathrm{x}^{4}-1\right)
\end{aligned}$
$\begin{aligned}
\mathrm{f}(\mathrm{x}) &=\int_{-1}^{\mathrm{x}} \mathrm{f}(\mathrm{y}) \mathrm{dy} \quad=\int_{-1}^{\mathrm{x}} \frac{\mathrm{y}^{3}}{3} \mathrm{dy} \\
&=\left[\frac{\mathrm{y}^{4}}{12}\right]_{-1}^{\mathrm{x}}=\frac{1}{12}\left(\mathrm{x}^{4}-1\right)
\end{aligned}$
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