Question

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

a. Define the random variable.$X=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

b.$X~=\_\_\_\_\_\_\_\_$

c.$\mu =\_\_\_\_\_\_\_\_$

d.$\sigma =\_\_\_\_\_\_\_\_$

e. Draw a graph of the probability distribution. Label the axes.

f. Find the probability that a car required over $\backslash )300$for maintenance during its first year.

Short answer

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

b. $X~Exp\left(\frac{1}{150}\right)$

c. $\mu =150$

d. $\sigma =150$

e. the graph is drawn

f.$0.1340$

Step-by-step solution

introduction

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

Explanation (part a)

Defining the random variable as

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

Explanation (part b)

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

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

Explanation (part c)

The mean can be represented as below:

$\mu =150$

Explanation (part d)

The standard deviation is represented as follows:

$\sigma =\mu =150$

Explanation (part e)

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:

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

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

$x$	$f\left(x\right)$
-4	0.006882
-3	0.006836
-2	0.00679
-1	0.006745
0	0.0067
1	0.006655
2	0.006611
3	0.006567
4	0.006523

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

Explanation (part f)

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" $P={\int }_{150}^{\mathrm{\infty }} \lambda e^{-\lambda x}dx=\lambda {\left|\frac{e^{-\lambda x}}{-\lambda }\right|}_{150}^{\mathrm{\infty }}=e^{-0.0067\times 150}-0=0.1340$