4.138 The random variable xcan be adequately approximated by an exponential probability distribution with$\mathit{\theta }\mathbf{=}\mathbf{2}$ . Find the probability that xassumes a value

a. More than 3 standard deviations from$\mathit{\mu }$

b. Less than 2 standard deviations from$\mathit{\mu }$

c. Within half a standard deviation of$\mathit{\mu }$

The random variable x has an exponential distribution with$\theta =2$

Here x is a random variable with parameters$\theta =2$

The pdf of x is given by,

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

Here,$\theta =2$

$f\left(x\right)=\frac{1}{2}exp\left(-\frac{x}{2}\right);x>0$

The mean of the random variable is given by,

$$\begin{gathered} \sigma =\sqrt{{\theta }^2} \\ =2 \end{gathered}$$

The standard deviation is given by,

$$\begin{gathered} \sigma =\sqrt{{\theta }^2} \\ =\sqrt{2^2} \\ =2 \end{gathered}$$

$$\begin{gathered} F\left(x\right)=P\left(X\le x\right) \\ ={\int }_0^{x}f\left(t\right)dt \\ ={\int }_0^x\frac{1}{2}exp\left(-\frac{t}{2}\right)dt \\ =\frac{1}{2}{\left[\frac{exp\left(-{\displaystyle \frac{t}{2}}\right)}{-{\displaystyle \frac{1}{2}}}\right]}_0^x \\ =-exp\left(-\frac{x}{2}\right)+1 \\ =1-exp\left(-\frac{x}{2}\right) \\ =1-exp\left(-0.5x\right) \\ F\left(x\right)=1-exp\left(-0.5x\right) \end{gathered}$$

a.

The probability of x assumes a value more than 3 standard deviations from its mean

is,

$$\begin{gathered} P\left(x>\mu +3\sigma \right)=P\left(x>2+3\times 2\right) \\ =P\left(x>8\right) \\ =1-P\left(x\le 8\right) \\ =1-F\left(8\right) \\ =1-\left[1-exp\left(-0.5\times 8\right)\right] \\ =exp\left(-4\right) \\ =0.0183 \end{gathered}$$

Thus, the probability is 0.0183.

b.

The probability of x assumes a value less than 2 standard deviations from its mean

is,

$$\begin{gathered} P\left(x<\mu -2\sigma \right)=P\left(x<2-2\times 2\right) \\ =P\left(x<-2\right) \\ =F\left(-2\right) \\ =0\left[\sin ce,x>0\right] \end{gathered}$$

Thus, the probability is 0.

c.

The probability of x assumes value withinhalf standard deviations from its mean

is,

$$\begin{gathered} P\left(\mu -0.5\sigma <x<\mu +0.5\sigma \right)=P\left(2-0.5\times 2<x<2+0.5\times 2\right) \\ =P\left(1<x<3\right) \\ =P\left(x<3\right)-P\left(x\le \right) \\ =F\left(3-\right)-F\left(1\right) \\ =\left[1-exp\left(-0.5\times 3\right)\right]-\left[1-exp\left(-0.5\times 1\right)\right] \\ =exp\left(-0.5\right)-exp\left(-1.5\right) \\ =0.606531-0.22313 \\ =0.3834 \end{gathered}$$

Thus, the required probability is 0.3834.