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$\begingroup$

To part b:

Someone please help me write a proper piece wise density function:

Given the PDF: $\Big( \frac{c(1-x^2 -1 \lt x \lt 1}{0, otherwise}\Big)$

a) Find the value of c:

$\int_{-1}^1 c(1-x^2)dx=1$

$\int_{-1}^{1} cx-\frac{cx^3}{3}dx=1$

$2c-\frac{2c}{3}=1$

$-\frac{1}{3}x^2+c-\frac{1}{2}=0$

Using the quadratic formula we obtain:

$x=\frac{3\sqrt{3}+3}{3}=4.732050808$

b) Find the Cumulative Distribution Function:

$f(x)= c-cx^2 $

By definition: $F(x)= \int_{-\infty}^x c-ct^2 dt$

$\int_{-1}^x \frac{3t}{4} - \frac{3t^3}{12} \vert_{t=-1}^{t=x}=$

$=\frac{3x}{4}-\frac{3x^3}{12} - \frac{3}{4} + \frac{3}{12}$

$= \frac{3}{4}x-\frac{3}{12}x^3+\frac{1}{2}$

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    **Tip:** Rather than distributing the constant into the polynomial then integrating, it would instead be best to move to constant safely out of the way.$$c\cdot \int^1_{−1} (1−x^2 )\;\mathrm dx=1$$ That way you will not get your $c$ and $x$ muddled. – 2017-02-22

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The integral $\int_{-1}^{1}c(1-x^2)dx=\frac{4c}{3}$ so that $c=\frac{3}{4}$.

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    Thank you I had made an arithmetic mistake when integrating – 2017-02-22