Please help with density function question solving for sigma(x)?

Question

Homwork says:
find E(X), Var(X), sigma(X) of a random variable X
let x be given by its density function F(x) such that

f(x) = 0, if x <= 1
f(x) = 1/4, if 1 < x <= 5
f(x) = 0, if x > 5

I think I solved E(X) and Var(x) correctly. I found that E(x)= 5/2 and Var(x)=11/4. Am I correct? I need to solve for sigma(x) too. Please help.

Answer 1

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I'm assuming x is continuous.

E[x] = ∫xf(x)dx
E[x²] = ∫x²f(x)dx

Your density is just the constant 1/4 from 1 to 5. That makes this the uniform distribution from 1 to 5.

First verify that this is a pdf - the integral of f(x) from 1 to 5 must be 1

5 ........... 5
∫f(x)dx = ∫dx/4 = 1/4(5 - 1) = 1
1 .......... 1

So this is a pdf.

......... 5
E[x] = ∫xf(x)dx
......... 1

.......... 5
E[x] = ∫x/4dx
......... 1

E[x] = (1/4){(1/2)[5² - 1²]} = 24/8 = 3

For a uniform distribution from a to b the expected value is (a+b)/2 and in this case (1+5)/2=3

Var[x] = E[x²] - (E[x])²

.......... 5
E[x²] = ∫x²f(x)dx
.......... 1

.......... 5
E[x²] = ∫x²/4dx
.......... 1

E[x²] = (1/4){(1/3)[5³ - 1³]} = 124/12 = 31/3

Var[x] = E[x²] - (E[x])² = 31/3 - 3² = 4/3

The variance of the uniform distribution from a to b is just
(b-a)²/12 and in this case (5-1)²/12 = 16/12 = 4/3.

How did you get your answers? You might want to check where you made a mistake or if you think your answer is right - point out mine. :)

For sigma(x) I assume you mean the standard deviation which is just the square root of the variance.

Answer 2

E(X)=int[from 1 to5] x*f(x) dx=
=(1/4)*(x^2)/2 [from 1 to 5]=
=3

Var(X)=int[from 1 to 5](x-E(x))^2*f(x) dx=
=(1/4)int[from 1 to 5] (x-3)^2 dx=
=(1/4)(1/3)(x-3)^3 [from 1 to 5]=
=(1/12)(2^3-(-2)^3)=16/12=4/3

sigma(x)=sqrt(Var(X))=
=2/sqrt3