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Suppose the random variable $X$ has PDF
$$f_X(x) = 2x^2~, \quad\text{when}~ 0 < x < 1~,$$ and zero otherwise.
Suppose $Z = 4X+9$, what is the CDF of $Z$? Then find the PDF of $Z$.

I began by finding $X=(Z-9)/4$, and then finding the integral of $f_X(x)$. I'm not sure of next steps.

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  • $\begingroup$ What trouble are you having with this? Please show your thoughts on , and attempts at , finding a solution. $\endgroup$ – Graham Kemp Apr 26 at 2:56
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    $\begingroup$ It looks like $f_x(x)$ should be $3x^2$, since $\int_0^1f_x(x)dx$ must $=1$. $\endgroup$ – herb steinberg Apr 26 at 3:34
  • $\begingroup$ @herbsteinberg Indeed. Good catch. $\endgroup$ – Graham Kemp Apr 26 at 3:35
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I began by finding X=(Z-9)/4, and then finding the integral of fx(x). I'm not sure of next steps.

That will give you the CDF of $Z$.

To find the pdf, differentiate with respect to $z$.$$\begin{align}{F}_Z(z)&={F}_X\left[(z-9)/4\right]\\[1ex]&=\int_{-\infty}^{(z-9)/4} f_X(x)~\mathrm d x\\[1ex]&=\mathbf 1_{0\leq (z-9)\le 4}~\int_0^{(z-9)/4} \require{cancel}\cancelto3{\color{red}2}x^2~\mathrm d x+\mathbf 1_{4\leq (z-9)}\\[1ex]&=\lower{2ex}\ddots\\[3ex] {f}_Z(x)&=\dfrac{\mathrm d~~}{\mathrm d z}F_Z(z)\\[1ex]&=\lower{2ex}\ddots\end{align}$$

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    $\begingroup$ This result did not take into account the error in the original statement. $\endgroup$ – herb steinberg Apr 26 at 3:50
  • $\begingroup$ I see the correction. $\endgroup$ – herb steinberg Apr 26 at 16:02
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$P(Z\le z)=P(4X+9\le z)=P(X\le \frac{z-9}{4})=(\frac{z-9}{4})^3$ (the CDF of Z) for $9\le z \le 13$ The CDF $=0$ for smaller $z$ and $=1$ for larger $z$.

The PDF of $z$ is $\frac{3}{4}(\frac{z-9}{4})^2$ for $9\le z \le 13$ and $=0$ otherwise.

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