Respuesta :
Answer:
a) [tex] C=e^1=e[/tex]
b) [tex]P(X<2)=\int_{1}^2 \frac{1}{x} dx = ln(x) \Big|_1^2 =ln(2)-ln(1)=ln(2)=0.693[/tex]
c) [tex]E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1[/tex]
[tex]E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)[/tex]
[tex]Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242[/tex]
Step-by-step explanation:
a) what must the value of C be so that f(x) is a probability density function?
In order to be a probability function we need this condition:
[tex]\int_{1}^C \frac{1}{x} dx =1[/tex]
And solving the left part of the integral we have:
[tex]ln(x) \Big|_1^C \ =1[/tex]
[tex]ln(C)-ln(1)=1[/tex], so then [tex] C=e^1=e[/tex]
b) find P(X<2)
We can find this probability on this way using the density function:
[tex]P(X<2)=\int_{1}^2 \frac{1}{x} dx = ln(x) \Big|_1^2 =ln(2)-ln(1)=ln(2)=0.693[/tex]
c) find E(X) and Var(X)
We can find the expected value on this way:
[tex]E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1[/tex]
In order to find the Var(X) we need to find the second moment given by:
[tex]E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)[/tex]
And now we can use the following definition:
[tex]Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242[/tex]
