Question

If the distribution function of $X$is given by

$F\left(b\right)=\left\{\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b<0\\ \frac{1}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le b<1\\ \frac{3}{5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\le b<2\\ \frac{4}{5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le b<3\\ \frac{9}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\le b<3.5\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\ge 3.5\end{array}\right.$

calculate the probability mass function of $X$.

Short answer

$\mathrm{p}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{1}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=0\\ \frac{1}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=1\\ \frac{1}{5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=2\\ \frac{1}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=3\\ \frac{1}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=3.5\end{array}\right.$

Step-by-step solution

Given information

$F\left(b\right)=\left\{\begin{array}{l}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b<0\\ \frac{1}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le b<1\\ \frac{3}{5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\le b<2\\ \frac{4}{5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\le b<3\\ \frac{9}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\le b<3.5\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\ge 3.5\end{array}\right.$

Explanation

The probability mass procedure is equivalent to zero for every particular where the distribution function is locally equivalent to some consistent. Therefore, we only have to think of facts in which function $F$has a brake.

Here,

$\mathrm{P}\left(0\right)=\mathrm{F}\left(0\right)-\mathrm{P}(\mathrm{b}<0)$

$=\frac{1}{2}$

$P\left(1\right)=\frac{3}{5}-\frac{1}{2}$

$=\frac{1}{10}$

$P\left(2\right)=\frac{4}{5}-\frac{3}{5}$

$=\frac{1}{5}$

$P\left(3\right)=\frac{9}{10}-\frac{4}{5}$

$=\frac{1}{10}$

$P(3.5)=1-\frac{9}{10}$

$=\frac{1}{10}$

Final answer

Probability mass function is

$\mathrm{p}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{1}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=0\\ \frac{1}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=1\\ \frac{1}{5}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=2\\ \frac{1}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=3\\ \frac{1}{10}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=3.5\end{array}\right.$