Answer :
Answer:
[tex] P(X <11)[/tex]
And using the cdf we got:
[tex] P(X <11)=1-e^{-\frac{11}{256}}= 1-0.958= 0.042[/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}, x>0[/tex]
And 0 for other case. Let X the random variable that represent the random variable of interest and we know that the distribution is given by:
[tex]X \sim Exp(\lambda)[/tex]
We know the variance on this case given by :
[tex] Var(X) = \frac{1}{\lambda^2}[/tex]
So then the deviation is given by:
[tex] Sd(X) = \frac{1}{\sqrt{\lambda}}[/tex]
And if we solve for [tex]\lambda[/tex] we got:
[tex] \lambda = (\frac{1}{16})^2 = \frac{1}{256}[/tex]
The cumulative distribution function for the exponential distribution is given by:
[tex] F(x) = 1- e^{-\lambda x}[/tex]
Solution to the problem
And for this case we want to find this probability:
[tex] P(X <11)[/tex]
And using the cdf we got:
[tex] P(X <11)=1-e^{-\frac{11}{256}}= 1-0.958= 0.042[/tex]