Question

The time required to repair a machine is exponentially distributed with the mean 87 minutes. Find the probability that the repair time is more than 2 hours and less than 3 hours.

Answer 1

Answer:

$P(X>2) = 1-P(X\leq 2) =1- [ 1-e^{-\frac{1}{87} x}]= e^{-\frac{1}{87}*120}=0.252$  

$P(X$  

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 "The number of years a radio functions" and we know that the distribution is given by:

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

Where $\lambda = \frac{1}{87}$

Solution to the problem

The cumulative distribution function for the exponential distribution is given by:

$P(X\leq x) = 1-e^{-\frac{1}{87} x}$

We need to take in count that 2hours = 120 minutes and 3 hours = 180 minutes

And we want to find these probabilities:

$P(X>2) = 1-P(X\leq 2) =1- [ 1-e^{-\frac{1}{87} x}]= e^{-\frac{1}{87}*120}=0.252$  

$P(X$