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

Here x is a random variable with parameters$胃=2$

The pdf of x is given by,

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

Here,$胃=2$

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

The mean of the random variable is given by,

$$\begin{gathered} 蟽=\sqrt{{胃}^2} \\ =2 \end{gathered}$$

The standard deviation is given by,

$$\begin{gathered} 蟽=\sqrt{{胃}^2} \\ =\sqrt{2^2} \\ =2 \end{gathered}$$

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

a.

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

is,

$$\begin{gathered} P\left(x>渭+3蟽\right)=P\left(x>2+3脳2\right) \\ =P\left(x>8\right) \\ =1-P\left(x鈮�8\right) \\ =1-F\left(8\right) \\ =1-\left[1-exp\left(-0.5脳8\right)\right] \\ =exp\left(-4\right) \\ =0.0183 \end{gathered}$$

Thus, the probability is 0.0183.

b.

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

is,

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

Thus, the probability is 0.

c.

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

is,

$$\begin{gathered} P\left(渭-0.5蟽<x<渭+0.5蟽\right)=P\left(2-0.5脳2<x<2+0.5脳2\right) \\ =P\left(1<x<3\right) \\ =P\left(x<3\right)-P\left(x鈮�\right) \\ =F\left(3-\right)-F\left(1\right) \\ =\left[1-exp\left(-0.5脳3\right)\right]-\left[1-exp\left(-0.5脳1\right)\right] \\ =exp\left(-0.5\right)-exp\left(-1.5\right) \\ =0.606531-0.22313 \\ =0.3834 \end{gathered}$$

Thus, the required probability is 0.3834.