Answer:
Step-by-step explanation:
a) We know that the cummulative distribution function of a continous random variable X is defined as:
[tex]F(x)=\int\limits^x_{-\infty} {f(t)} \, dt[/tex]
given: [tex]f(x)=-\frac{2}{25} x + \frac{2}{5}[/tex]
hence its cdf will be
[tex]F(x)=\int\limits^x_0 (-\frac{2}{25}t + \frac{2}{5} )\, dt\\\\=\frac{2}{5}[-\frac{t^2}{10} + t]\limits^x_0 \\\\F(x)=-\frac{x^2}{25} +\frac{2}{5}x[/tex]
b) Mean of X is given as:
[tex]E(x)=\int\limits^{\infty}_{-\infty} {f(x)} \, dx\\\\=\int\limits^{5}_{0} (-\frac{2}{25}x+\frac{2}{5})x \, dx\\\\\frac{2}{5}\int\limits^{5}_{0} (-\frac{1}{5}x^2+x) \, dx\\\\=\frac{2}{5}(-\frac{x^3}{15}+\frac{x^2}{2})\limits^5_0\\\\E(x)=\frac{5}{3}[/tex]
c) For second moment
[tex]E(x^2)=?\\\\E(x^2)=int\limits^{\infty}_{-\infty} {x^2f(x)} \, dx\\\\=\int\limits^{5}_{0} {(-\frac{2}{25}x + \frac{2}{5})} \, dx\\\\=\frac{25}{6}[/tex]
