Question

4.135 Suppose xhas an exponential distribution with$\mathit{\theta }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}$ . Find

the following probabilities:

$$\begin{gathered} \mathit{a}\mathbf{.}\mathit{P}\left(x\le 4\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{b}\mathbf{.}\mathit{P}\left(x>5\right) \\ \mathit{c}\mathbf{.}\mathit{P}\left(x\le 2\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{d}\mathbf{.}\mathit{P}\left(x>3\right) \end{gathered}$$

Short answer

a. The probability is 0.7981

b. The probability is 0.1353

c. The probability is 0.5507

d. The probability is 0.3012

Step-by-step solution

Given Information

The random variable x has an exponential distribution with$\theta =2.5$

The probability density function of x

Here x is a random variable with parameters$\theta =2.5$

The pdf of x is given by,

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

Here,

$f\left(x\right)=\frac{1}{2.5}exp\left(-\frac{x}{2.5}\right);x>0$

Finding cdf of x

$$\begin{gathered} F\left(x\right)=P\left(X\le x\right) \\ ={\int }_0^{x}f\left(t\right)dt \\ ={\int }_0^x\frac{1}{2.5}exp\left(-\frac{t}{2.5}\right)dt \\ =\frac{1}{2.5}{\left[\frac{exp\left(-\frac{t}{2.5}\right)}{\frac{1}{2.5}}\right]}_0^x \\ =exp\left(-\frac{x}{2.5}\right)+1 \\ =1-exp\left(-\frac{x}{2.5}\right) \\ =1-exp\left(-0.4x\right) \\ F\left(x\right)=1-exp\left(-0.4x\right) \end{gathered}$$

Finding the probability when P(x≤4)

a.

$$\begin{gathered} P\left(x>4\right)=F\left(4\right) \\ =1-exp\left(-0.4\times 4\right) \\ =1-exp-\left(-1.6\right) \\ =1-0.201897 \\ =0.798103 \\ ~0.7981 \end{gathered}$$

Thus, the required probability is 0.7981.

Finding the probability when (x>5)

b.

$$\begin{gathered} P\left(x>5\right)=1-P\left(x\le 5\right) \\ =1-F\left(5\right) \\ =1-\left[1-exp\left(-0.4\times 5\right)\right] \\ =exp\left(-2\right) \\ =0.135335 \\ ~0.1353 \end{gathered}$$

Thus, the required probability is 0.1353.

Finding the probability when P(x≤2)

c.

$$\begin{gathered} P\left(x\le 2\right)=F\left(2\right) \\ =1-exp\left(-0.4\times 2\right) \\ =1-exp\left(-0.8\right) \\ =1-0.449329 \\ =0.550671 \\ ~0.5507 \end{gathered}$$

Thus, the required probability is 0.5507.

Finding the probability when (x>3)

d.

$$\begin{gathered} P\left(x>3\right)=1-P\left(x\le 3\right) \\ =1-F\left(3\right) \\ =1-\left[1-exp\left(-0.4\times 3\right)\right] \\ =exp\left(-1.2\right) \\ =0.301194 \\ ~0.3012 \end{gathered}$$

Thus, the required probability is 0.3012.