91Ó°ÊÓ

a. As per the information that has been given in the problem, $X$can be explained as;

$X=$The useful life of the car battery in months

b. As per information given, the random variable $X$follows an Exponential Distribution. Thus it can be commented that $X$will be Continuous.

c. According to the information that has been shared in the question, the random variable $X$is distributed in an Exponential manner and the decay parameter of it is 0.025. Thus, distribution of $X$can be written as;

$X~Exp(0.025)$

where,

$m=0.025$

d. The average battery life of the car battery can be calculated as explained below;

$\begin{array}{rcl}\mathrm{μ}& =& \frac{1}{m}\\ & =& \frac{1}{0.025}\\ & =& 40\end{array}$

Thus, a car battery will last for around 40 months.

e. As per information received from Step 4, one battery lasts for 40 months.

Thus, if 9 batteries are used one after another then the average life of the car battery would be;

$$\begin{gathered} =9×40 \\ =360months \end{gathered}$$

f. It can be commented that cumulative distribution function of exponential distribution can be explained as

$P(X<x)=1-e^{-mx}$

As per the above equation, the probability can be calculated as;

$\begin{array}{rcl}P(x& >& 36)=1-(1-e^{-0.025×36})\\ & =& e^{-0.025×36}\\ & =& 0.4066\end{array}$

Thus, the value of $P(x>36)$is $0.4066$

g. Calculation of the 70^th Percentile can be represented as;

$$\begin{gathered} P(x>k)=1-(1-e^{-mk}) \\ 0.70=e^{-0.025×k} \end{gathered}$$

Applying logarithms on both sides of the equation we get,

$$\begin{gathered} \ln \left(e^{-0.025×k}\right)=\ln (0.70) \\ 0.025×k=0.3567 \\ k=\frac{0.3567}{0.025} \\ =14.27 \end{gathered}$$

Thus, it can be commented that 70^th batteries will last for$14.27$Months