Answer:
a. E(X) = 0.333 minutes
b. [tex]\mathbf{\sigma_x =0.333 \ minutes}[/tex]
c. 0.9986
Step-by-step explanation:
a.
To find the mean time between counts
Let X be the random variable that illustrates the time between the successive log-ons, then X is exponential with the rate [tex]\lambda = 3[/tex] log-ons per minute.
i.e.
[tex]E(X) = \dfrac{1}{\lambda}[/tex]
[tex]E(X) = \dfrac{1}{3}[/tex]
E(X) = 0.333 minutes
b. Since X is attributed to an exponential distribution, The standard deviation between counts is:
[tex]\sigma_x = \dfrac{1}{3}[/tex]
[tex]\mathbf{\sigma_x =0.333 \ minutes}[/tex]
c.
Given that;
The average number of count per minute = 3
The value of X such that P(X < x) = 0.95 can be calculated as;
[tex]\int \limits ^x_0 \lambda e^{-\lambda t}= 0.95[/tex]
[tex]3\int \limits ^x_0 \lambda e^{3 t}= 0.95[/tex]
[tex]\bigg [ -e^{3t}\bigg ]^x_0 = 0.95[/tex]
[tex]e^{3t} = 1-0.95[/tex]
-3x = ㏑ (0.05)
-3x = - 2.9957
x = -2.9957/ -3
x = 0.9986
The value of x = 0.9986
Answer:
a) 3
b)3
c) 8.9872
Step-by-step explanation:
a) According to the question, mean is 3.
b) Since it is an exponential distribution, standard deviation equals to mean. Therefore, the answer is 3.
c) P[x<X]=0.95
f(x)= λ*e^(-λ*x)
∫f(x)dx, 0 to x
mean = 1/λ =3. So, λ= 1/3.
∫(1/3)*e^(-x/3)dx= 1/3*e^(-x/3)/(-1/3)=-e^(-x/3) 0 to x
-e^(-x/3)+1 = 0.95
-e^(-x/3)=-0.05
x/3=ln(0.05)
x=ln(0.05)*3= 8.9872
