Density function

[barabbas]

Hi, would be thankful if someone could help me out here

Density function for random variable X i give by

f(x) = cx^2, -1<x<1

a) Find c.
b) Find Pr(X>1/2).
c) Find E(X).

a) Here I set the density function to equal 1 and got that c= 3/2. I'm not sure if it is correct since the boundaries on the density function is -1<x<1 and not -1≤x≤1, anyone can explain what the difference is?

b) Here the integrals boundaries are set from 1>1/2 where f(x) = (3/2)x^2 and I get 7/18. Any thoughts?

c) Integrated with c=3/2 and multiplied the function with x and got 3/8-3/8=0. Is that correct or what am I missing?

 

[pka]

Hi, would be thankful if someone could help me out here

Density function for random variable X i give by

f(x) = cx^2, -1<x<1

a) Find c.
b) Find Pr(X>1/2).
c) Find E(X).
a) Here I set the density function to equal 1 and got that c= 3/2. I'm not sure if it is correct since the boundaries on the density function is -1<x<1 and not -1≤x≤1, anyone can explain what the difference is?
b) Here the integrals boundaries are set from 1>1/2 where f(x) = (3/2)x^2 and I get 7/18. Any thoughts?
c) Integrated with c=3/2 and multiplied the function with x and got 3/8-3/8=0. Is that correct or what am I missing?

Please show us how you did part b).

 

[Dr.Peterson]

Density function for random variable X is given by
f(x) = cx^2, -1<x<1

a) Find c.
b) Find Pr(X>1/2).
c) Find E(X).

a) Here I set the density function to equal 1 and got that c= 3/2. I'm not sure if it is correct since the boundaries on the density function is -1<x<1 and not -1≤x≤1, anyone can explain what the difference is?

b) Here the integrals boundaries are set from 1>1/2 where f(x) = (3/2)x^2 and I get 7/18. Any thoughts?

c) Integrated with c=3/2 and multiplied the function with x and got 3/8-3/8=0. Is that correct or what am I missing?

a) The value of a PDF at an individual point has no effect. Don't worry about it.
b) You probably either made a tiny error or typed one digit wrong. Show your work or correct your answer.
c) Graph the function and think about whether 0 makes sense as an answer. (Think symmetry.)

 

[barabbas]

a) The value of a PDF at an individual point has no effect. Don't worry about it.
b) You probably either made a tiny error or typed one digit wrong. Show your work or correct your answer.
c) Graph the function and think about whether 0 makes sense as an answer. (Think symmetry.)

b) Integral from 1(b) to 1/2(a),where 1 is the upper boundary. Setting the function to (3/2)x^2 dx => take out constant = (3/2)[(x^3)/3] = F(b)-F(a) = (3/2)[(1/3)-(1/24)] = [(1/2)-(1/16)] = 7/16.

Yes I missed something in my first calculation. Is this one correct?

c) Yeah 0 seems to be the mean when i'm graphing it.

Thank you for replying.

 

[Dr.Peterson]

Now it's correct.