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

Answer:

a) $P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368$

b) $P(X>30000|X>15000)=P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607$

Step-by-step 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:

$X \sim Exp(\lambda=\frac{1}{30000})$

The cumulative distribution function is given by:

$F(X) = 1- e^{-\frac{x}{\mu}}$

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:

$P(X>30000)=1-( 1- e^{-\frac{30000}{30000}})=e^{-1}=0.368$

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:

$P(X>15000)=1-( 1- e^{-\frac{15000}{30000}})=e^{-0.5}=0.607$