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Tasya [4]
3 years ago
7

The time between breakdowns of an alarm system is exponentially distributed with mean 10 days. What is the probability that ther

e are no breakdowns on a given day?
Mathematics
1 answer:
Nimfa-mama [501]3 years ago
5 0

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

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