91Ó°ÊÓ

The cost of all maintenance for a car during its first year is approximately exponentially distributed with a mean of$\$150$.

Defining the random variable as

$X=$The cost of all maintenance of a car during its first year.

According to the provided information on the problem, the random variable $x$is exponentially distributed and can be defined as below:

localid="1651917369361" $$\begin{gathered} X~Exp\left(\frac{1}{\mathrm{μ}}\right) \\ X~Exp\left(\frac{1}{150}\right) \\ X~Exp\left(0.0067\right)where,m=0.0067 \end{gathered}$$

The mean can be represented as below:

$\mathrm{μ}=150$

The standard deviation is represented as follows:

$\mathrm{σ}=\mathrm{μ}=150$

The general form of the probability density function of the exponential distribution is given below,

$$\begin{gathered} f\left(x\right)=m{e}^{-mx} \\ f\left(x\right)=(0.0067)e^{-(0.0067)x} \end{gathered}$$

The maximum value of $f\left(x\right)$which will lie on the y-axis and at $x=0$will be:

$f\left(x\right)=(0.0067)e^{-(0.0067×0)}=0.0067$

The value of $f\left(x\right)$for different values of $x$, we get

From the above table, the graph of the probability distribution is given as:

It is provided that a car required over $300$dollars for maintenance during its first year.

Then the car required for maintenance is, $300-150=150$dollars.

The probability of $X>150$is as follows:

localid="1651917491624" $P={∫}_{150}^{\mathrm{∞}} \mathrm{λ}{e}^{−\mathrm{λ}x}dx=\mathrm{λ}{\left|\frac{e^{−\mathrm{λ}x}}{−\mathrm{λ}}\right|}_{150}^{\mathrm{∞}}=e^{−0.0067×150}−0=0.1340$