Respuesta :
Answer:
[tex] X \sim Exp (\lambda =\frac{1}{6.5})[/tex]
For this case we want to find this probability:
[tex] P(X>11)[/tex]
And we can use the cumulative distribution given by:
[tex] F(x) = 1-e^{-\lambda x}[/tex]
The mean is given by:
[tex]\mu = \frac{1}{\lambda}[/tex]
And then [tex] \lambda = \frac{1}{\mu}= \frac{1}{6.5}= 0.1538[/tex]
And replacing we got:
[tex] P(X>11) = 1-P(X<11) = 1- [1- e^{-0.1538 *11}]= e^{-0.1538*11}= 0.18409[/tex]
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:
[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]
Solution to the problem
For this case we can define the random variable X= "checkout times"
And the distribution for X is given by:
[tex] X \sim Exp (\lambda =\frac{1}{6.5})[/tex]
For this case we want to find this probability:
[tex] P(X>11)[/tex]
And we can use the cumulative distribution given by:
[tex] F(x) = 1-e^{-\lambda x}[/tex]
The mean is given by:
[tex]\mu = \frac{1}{\lambda}[/tex]
And then [tex] \lambda = \frac{1}{\mu}= \frac{1}{6.5}= 0.1538[/tex]
And replacing we got:
[tex] P(X>11) = 1-P(X<11) = 1- [1- e^{-0.1538 *11}]= e^{-0.1538*11}= 0.18409[/tex]
