Respuesta :
Answer:
[tex] X \sim Exp (\lambda =1.5)[/tex]
For this case we want to find this probability:
[tex] P(X <1)[/tex]
And we can use the cumulative distribution given by:
[tex] F(x) = 1-e^{-\lambda x}[/tex]
And replacing we got:
[tex] P(X<1) = 1- e^{-1.5 *1}= 1-0.2231=0.7769[/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= "how long it took customers to receive their order at the drive-thru"
And the distribution for X is given by:
[tex] X \sim Exp (\lambda =1.5)[/tex]
For this case we want to find this probability:
[tex] P(X <1)[/tex]
And we can use the cumulative distribution given by:
[tex] F(x) = 1-e^{-\lambda x}[/tex]
And replacing we got:
[tex] P(X<1) = 1- e^{-1.5 *1}= 1-0.2231=0.7769[/tex]
