91Ó°ÊÓ

To find the value of c, we need to use the property that the integral of a pdf over its entire range is equal to 1, i.e. $$ \int_{-\infty}^{\infty} f(x) dx = 1 $$ Substitute the given pdf and solve for \(c\): $$ \int_{-1}^{1} c(1-x^{2}) dx = 1 $$ $$ c \int_{-1}^{1} (1-x^{2}) dx = 1 $$ $$ c \left[ x - \frac{x^3}{3}\right]_{-1}^{1} = 1 $$ $$ c \left[(1-\frac{1}{3}) - (-1-\frac{-1}{3}) \right] = 1 $$ $$ c (2\cdot \frac{2}{3}) = 1 $$ Then, we have: $$ c = \frac{3}{4} $$ So the pdf of \(X\) is: $$ f(x)=\left\\{\begin{array}{ll} \frac{3}{4}(1-x^{2}), & -1<x<1 \\ 0, & \text { otherwise } \end{array}\right. $$

The cumulative distribution function (CDF) of a random variable is defined as the integral of its pdf from the lower bound of the domain to a given value x. So, for \(X\), the CDF, denoted by \(F(x)\), is: $$ F(x) = \int_{-\infty}^{x} f(t) dt $$ For \(x<-1\), the integral is 0, because the pdf is 0 for those values. For \(-1<x<1\), we have: $$ F(x) = \int_{-1}^{x} \frac{3}{4}(1-t^2) dt $$ Integrating, we get: $$ F(x) = \left[\frac{3}{4}(t-\frac{t^3}{3})\right]_{-1}^{x} $$ $$ F(x) = \frac{3}{4}\left( (x - \frac{x^3}{3}) - (-1+\frac{1}{3}) \right) $$ For \(x\ge1\), the integral is 1, because we have covered the whole range where the pdf is non-zero. So, the final CDF is: $$ F(x)=\left\\{\begin{array}{ll} 0, & x \le -1 \\ \\ \frac{3}{4}\left( (x - \frac{x^3}{3}) - (-1+\frac{1}{3}) \right), & -1<x<1 \\ \\ 1, & x \ge 1 \end{array}\right. $$ So the cumulative distribution function of \(X\) is given by the above piecewise function.