Answer:
a) 1
b) 1
c) [tex]P(X\leq 4)=0.982[/tex]
[tex]P(X\geq 4)=0.018[/tex]
d) [tex]P(3\leq X\leq 5)=0.043[/tex]
Explanation:
An exponential distribution is a probability distribution of the time between events which occur continuously and independently at a constant average rate [tex]\lambda[/tex]. The probability function is
[tex]f(X) = \lambda e^{-\lambda X} ; x\geq 0[/tex]
a) The expected value of an exponentially distributed variable X with rate [tex]\lambda[/tex] (in this case 1) is given by
[tex]E[X]=1/\lambda[/tex]
E[X] = 1/1 = 1
b) The variance of an exponentially distributed variable X with rate [tex]\lambda[/tex] is
[tex]Var[X]=1/\lambda^2[/tex]
so the standard deviation is
[tex]\sigma=\sqrt{\frac{1}{\lambda^2}} = 1/\lambda[/tex]
[tex]\sigma= 1/1 = 1[/tex]
c) the cumulative distribution function is
[tex]F(X) = 1-e^{-\lambda X} ; x\geq 0[/tex]
with that function we can calculate the probability of time between events be lower or equal than determined number.
So [tex]P(X\leq4) = 1 - e^{-1*4}=0.982[/tex]
If you need to calcuate the probability of time between events be higher than determined number you can do the following operation
[tex]P(X \geq 4) = 1 - P(X\leq4) = 1-0.982=0.018[/tex]
d) [tex]P(3\leq X\leq 5)[/tex] we need to make a substraction
[tex]P(3\leq X\leq 5)= P(X \leq 5 ) - P(X \leq 3 ) [/tex]
[tex]P(3\leq X\leq 5)= 1-e^{-1*3} - (1-e^{-1*5}) = e^{-1*5} - e^{-1*3}=0.050-0.007= 0.043[/tex]
