Question

The time between breakdowns of an alarm system is exponentially distributed with mean 10 days. What is the probability that there are no breakdowns on a given day?

Answer 1

Answer:

$P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048$

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:

$P(X=x)=\lambda e^{-\lambda x}$

Solution to the problem

For this case the time between breakdowns representing our random variable T is exponentially distirbuted $T \sim Exp (\mu = 10)$

So on this case we can find the value of $\lambda$ like this:

$\lambda = \frac{1}{\mu} = \frac{1}{10}$

So then our density function would be given by:

$P(T)=\lambda e^{-\frac{t}{10}}$

The exponential distribution is useful when we want to describe the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time between two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:

$P(T>t)= e^{-\lambda t}$

And on this case we are looking for this probability:

$P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048$