Let$X~Exp(0.1).$

a. decay rate$=\_\_\_\_\_\_\_\_$

b.$\mu =\_\_\_\_\_\_\_\_$

c. Graph the probability distribution function.

d. On the graph, shade the area corresponding to $P(x<6)$and find the probability.

e. Sketch a new graph, shade the area corresponding to $P(3<x<6)$and find the probability.

f. Sketch a new graph, shade the area corresponding to $P(x<7)$and find the probability.

g. Sketch a new graph, shade the area corresponding to the $40th$percentile and find the value.

h. Find the average value of $x$.

a probability distribution is a numerical capacity that gives the probabilities of events of various potential results for an experiment. It is a numerical portrayal of an irregular peculiarity as far as its example space and the probabilities of occasions.

Given,

$\mathrm{X}~\mathrm{Exp}\left(0.1\right)$where m =$0.1$

decay rate $\lambda =0.1$

The mean of the given problem is given as:

$\mu =\frac{1}{\lambda }=\frac{1}{0.1}=10$

The probability distribution function is given by:

$f\left(x\right)=\lambda e^{-\lambda x}$

On substituting the value of decay in the above equation, we get

$f\left(x\right)=0.1e^{-(0.1)x}$

The graph is plotted for the above probability density function as:

The probability $P(x<6)$is given by,

$P(x<6)={\int }_0^6 \lambda e^{-\lambda x}dx=\lambda {\left|\frac{e^{-\lambda x}}{-\lambda }\right|}_0^6=-e^{-0.1\times 6}+1=0.4512$

The area of probability in the graph is given by,

The probability $P(3<x<6)$is given by

$\begin{array}{r}P(3<x<6)=F\left(6\right)-F\left(3\right)\\ =\left(1-e^{-(0.1)(6}\right)-\left(1-e^{-(0.1)\left(3\right)}\right)\\ =e^{-(0.3)}-e^{-(0.6)}=0.74080-0.5488\\ P(3<x<6)=0.192\end{array}$

The shaded area corresponding to $P(3<x<6)$is given by,

The probability $P(x<7)$is given by,

$P(x<7)={\int }_0^7 \lambda e^{-\lambda x}dx=\lambda {\left|\frac{e^{-\lambda x}}{-\lambda }\right|}_0^7=-e^{-0.1\times 7}+1=0.5034$

The shaded area corresponding to $P(x<7)$is given by,

we know that, $0.4=e^{-0.1x}$

Now taking $\ln$on both sides, we get

localid="1651519319294" $$\begin{gathered} -(0.1)x=\ln (0.4) \\ x=\frac{0.916}{0.1} \\ x=9.16 \end{gathered}$$

The probability of localid="1651518188604" $40th$percentile means, localid="1651519109843" $P(x<9.16)$is given by,

localid="1651519117914" $P(x<9.16)={\int }_0^{9.16} \lambda e^{-\lambda x}dx=\lambda {\left|\frac{e^{-\lambda x}}{-\lambda }\right|}_0^{9.16}=-e^{-0.1\times 9.16}+1=0.6$

The shaded area corresponding to localid="1651519126351" $P(x<9.16)$is given by,

The average value of $x$is given by,

$$\begin{gathered} x=\frac{0+1+2+3+4+5+6+7+8+9+10}{11} \\ x=\frac{55}{11}=5 \end{gathered}$$