Question

4.135 Suppose xhas an exponential distribution with $\mathit{\theta }\mathbf{=}\mathbf{1}$. Find

the following probabilities:

$$\begin{gathered} \mathit{a}\mathbf{.}\mathbf{}\mathit{P}\mathbf{(}\mathit{x}\mathbf{>}\mathbf{1}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{b}\mathbf{.}\mathit{P}\mathbf{(}\mathit{x}\mathbf{\le }\mathbf{3}\mathbf{)} \\ \mathit{c}\mathbf{}\mathit{P}\mathbf{(}\mathit{x}\mathbf{>}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{d}\mathbf{.}\mathbf{}\mathit{P}\mathbf{(}\mathit{x}\mathbf{\le }\mathbf{5}\mathbf{)} \end{gathered}$$

Short answer

1. .The probability is 0.3679
2. The probability is 0.9502
3. The probability is 0.2231
4. The probability is 0.9933

Step-by-step solution

Given Information

The random variable x has an exponential distribution with$\theta =1$

The probability density function (PDF) of x

Here x is a random variable with parameters$\theta =1$

The pdf of x is given by,

$f(x,\theta )=\frac{1}{\theta }exp\left(-\frac{x}{\theta }\right),x>0$

Here, $\theta =1$

$f\left(x\right)=exp(-x);x>0$

Finding cdf of x

$$\begin{gathered} F\left(x\right)=P(X\le x) \\ ={\int }_0^{x}f\left(t\right)dt \\ ={\int }_0^{x}exp(-t)dt \\ ={\left[\frac{exp(-t)}{-1}\right]}_0^x \\ =-exp(-x)+1 \\ =1-exp(-x) \\ F\left(x\right)=1-exp(-x) \end{gathered}$$

Finding the probability when P(x > 1)

$$\begin{gathered} a. \\ P(x>1)=1-P(x\le 1) \\ =1-F\left(1\right) \\ =1-\left[1-exp(-1)\right] \\ =exp(-1) \\ =0.37879 \\ =0.3679 \end{gathered}$$

Thus, the required probability is 0.3679.

Finding the probability when P(x≤3)

$$\begin{gathered} b. \\ P(x\le 3)=F\left(3\right) \\ =1-exp(-3) \\ =1-0.049787 \\ =0.950213 \\ =0.9502 \end{gathered}$$

Thus, the required probability is 0.9502.

Finding the probability when P(x>1.5)

$$\begin{gathered} c. \\ P(x>1.5)=1-P(x\le 1.5) \\ =1-F(1.5) \\ =1-\left[1-\exp (-1.5)\right] \\ =\exp (-1.5) \\ =0.22313 \\ =0.2231 \end{gathered}$$

The required probability is 0.2231.

Finding the probability when P(x≤5)

$$\begin{gathered} d. \\ P(x\le 5)=F\left(5\right) \\ =1-exp(-5) \\ =1-0.006738 \\ =0.993262 \\ =0.9933 \end{gathered}$$

The required probability is0.9933.