Question

The time (in minute) for a lab assistant to prepare the equipment for a certain experiment is a random variable $X$ taking values between $25$ and $35$ minutes with p.d.f.

$f\left(x\right)=\{\begin{array}{c}\frac{1}{10},25\le x\le 35\\ \\ 0,\text{otherwise}\end{array}$.
What is the probability that preparation time exeeds $33$ minutes? Also find the c.d.f. of $X$.

Answer 1

$f\left(x\right)=\frac{1}{10}$

Required probability $=P(X>33)$

$={\int }_{33}^{\infty }f\left(x\right)dx$

$={\int }_{33}^{35}f\left(x\right)dx+{\int }_{35}^{\infty }f\left(x\right)dx$

${\int }_{33}^{35}f\left(x\right)dx+0[\because f(x)=0$ when $x>35]$

$={\int }_{33}^{35}\frac{1}{10}dx$.

$=\frac{1}{10}{\int }_{33}^{35}1dx$.

$=\frac{1}{10}[x{]}_{33}^{35}$

$=\frac{1}{10}[35-33]=\frac{1}{5}$

Let $F\left(x\right)$ be the c.d.f. of $X$.

$\therefore F\left(x\right)=P[X\le x]$

$={\int }_{-\infty }^{\infty }f\left(x\right)dx$

$={\int }_{-\infty }^{25}f\left(x\right)dx+{\int }_{25}^{x}f\left(x\right)dx$

$=0+{\int }_{25}^{x}f\left(x\right)dx[\because f(x)=0$, when $f\left(x\right)<25]$

$={\int }_{25}^x\frac{1}{10}dx$

$=\frac{1}{10}[x{]}_{25}^x$

$=\frac{1}{10}[x-25]$

$\therefore F\left(x\right)=\frac{x-25}{10}$.