If the distribution function of X is given byF(b)=0 b<012 0≤b<135 1≤b<245 2≤b<3910 3≤b<3.51 b≥3.5calculate the probability mass function of X.

p(x)=12 x=0110 x=115 x=2110 x=3110 x=3.5

Q. 4.19 - A First Course in Probability

Q. 4.19

Question

If the distribution function of $X$ is given by

$F (b) = \{\begin{cases} 0 b < 0 \\ \frac{1}{2} 0 \leq b < 1 \\ \frac{3}{5} 1 \leq b < 2 \\ \frac{4}{5} 2 \leq b < 3 \\ \frac{9}{10} 3 \leq b < 3.5 \\ 1 b \geq 3.5 \end{cases}$

calculate the probability mass function of $X$ .

Step-by-Step Solution

Verified

Answer

$p (x) = \{\begin{cases} \frac{1}{2} x = 0 \\ \frac{1}{10} x = 1 \\ \frac{1}{5} x = 2 \\ \frac{1}{10} x = 3 \\ \frac{1}{10} x = 3.5 \end{cases}$

1Step 1 Given information

$F (b) = \{\begin{cases} 0 b < 0 \\ \frac{1}{2} 0 \leq b < 1 \\ \frac{3}{5} 1 \leq b < 2 \\ \frac{4}{5} 2 \leq b < 3 \\ \frac{9}{10} 3 \leq b < 3.5 \\ 1 b \geq 3.5 \end{cases}$

2Step 2 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,