Question

Under average driving conditions, the life lengths of automobile tires of a certain brand are found to follow an exponential distribution, with a mean of 30,000 miles. Find the probability that one of these tires, bought today, will last the following number of miles:a.Over 30,000 milesb.Over 30,000 miles, given that it already has gone 15,000 miles.

Answer 1

a)
`P(X>30000)=1-( 1- e^{-(30000)/(30000)})=e^(-1)=0.368`

b)
`P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-(15000)/(30000)})=e^(-0.5)=0.607`

Explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

`P(X=x)=\lambda e^(-\lambda x), x>0`

And 0 for other case. Let X the random variable that represent "life lengths of automobile tires of a certain brand" and we know that the distribution is given by:

`X \sim Exp(\lambda=(1)/(30000))`

The cumulative distribution function is given by:

`F(X) = 1- e^{-(x)/(\mu)}`

Part a

We want to find this probability:

`P(X>30000)` and for this case we can use the cumulative distribution function to find it like this:

`P(X>30000)=1-( 1- e^{-(30000)/(30000)})=e^(-1)=0.368`

Part b

For this case w want to find this probability

`P(X>30000|X>15000)`

We have an important property on the exponential distribution called "Memoryless" property and says this:

`P(X>a+t| X>t)=P(X>a)`

On this case if we use this property we have this:
`P(X>30000|X>15000)=P(X>15000+15000|X>15000)=P(X>15000)`

We can use the definition of the density function and find this probability:

`P(X>15000)=1-( 1- e^{-(15000)/(30000)})=e^(-0.5)=0.607`