Finding Formulas for c, μx, and Var(X) in a Center of Mass Word Problem

Click For Summary

Homework Help Overview

The discussion revolves around a probability function f(x) = cx^p(1-x) for a random variable X defined over the interval [0, 1]. Participants are tasked with finding formulas for the constant c, the expected value μx, and the variance Var(X) in terms of the parameter p. The context is situated within a chapter on center of mass, which raises questions about the relevance of the title to the problem at hand.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the properties of probability functions, including the requirement that the integral of f(x) over its domain equals 1. There is also exploration of the relationship between μx and the expected value E(X), with some participants questioning how to derive these values from the given function.

Discussion Status

Some participants have provided hints and guidance on how to approach finding the constant c and calculating μx and Var(X). There is an ongoing exploration of the integration process and the implications of treating c as a constant. Multiple interpretations of the problem's context are being discussed, particularly regarding its classification as a center of mass problem.

Contextual Notes

Participants express confusion about the placement of this problem within the center of mass chapter, indicating a potential disconnect between the problem's content and its categorization. There is also mention of the need for further clarification on the definitions and relationships between the variables involved.

PCSL
Messages
146
Reaction score
0
Let p be a positive constant. Suppose that a random variable X has probability function f(x)=cxp(1-x) for 0≤x≤1. Find formulas for c, μx, and Var(X) in terms of p.

I know Var(X)=∫(x-μx)2f(x)dx

I know that I did not show any work so please just give me a hint (I'm not asking you to solve it for me). Thanks and I would have provided work if I even had a guess on where to start.

P.S. I'm kind of confused why this is at the end of the center of mass chapter since I totally understood everything up to this problem.
 
Physics news on Phys.org
PCSL said:
Let p be a positive constant. Suppose that a random variable X has probability function f(x)=cxp(1-x) for 0≤x≤1. Find formulas for c, μx, and Var(X) in terms of p.

I know Var(X)=∫(x-μx)2f(x)dx
Since f(x) is a probability function, isn't its integral over the interval equal to 1?
Also, isn't μX the same as E(X)? And isn't this an integral as well?
PCSL said:
I know that I did not show any work so please just give me a hint (I'm not asking you to solve it for me). Thanks and I would have provided work if I even had a guess on where to start.

P.S. I'm kind of confused why this is at the end of the center of mass chapter since I totally understood everything up to this problem.
 
PCSL said:
Let p be a positive constant. Suppose that a random variable X has probability function f(x)=cxp(1-x) for 0≤x≤1. Find formulas for c, μx, and Var(X) in terms of p.

I know Var(X)=∫(x-μx)2f(x)dx

I know that I did not show any work so please just give me a hint (I'm not asking you to solve it for me). Thanks and I would have provided work if I even had a guess on where to start.

P.S. I'm kind of confused why this is at the end of the center of mass chapter since I totally understood everything up to this problem.

Why have you titled this as a "Center of Mass Word Problem"?
 
LCKurtz said:
Why have you titled this as a "Center of Mass Word Problem"?

Because it is a word problem at the end of the section titled center of mass. The solution involves finding the second moment also, I believe.

@Mark I did not know that Mu sub x is equivalent to E(X). I'll go on that assumption and see if I get the right answer.
 
LCKurtz said:
Why have you titled this as a "Center of Mass Word Problem"?

Oh, finding the expectation value of f(x) could be considered as a 'center of mass' problem. The formula looks similar. It's a first moment. But the title is not that important. The first thing PCSL should do is find c. Then the rest should be easy.
 
Alright, so as Mark said

\int_{0}^{1} cx^p(1-x)dx=1
because it is a PDF

When solving this for c do I assume that c is a constant and pull it out of the integral? How do I solve this for μx when it isn't even in the formula? Thanks, and sorry I'm not providing more work but I'm pretty lost...
 
PCSL said:
Alright, so as Mark said

\int_{0}^{1} cx^p(1-x)dx=1
because it is a PDF

When solving this for c do I assume that c is a constant and pull it out of the integral? How do I solve this for μx when it isn't even in the formula? Thanks, and sorry I'm not providing more work but I'm pretty lost...

Sure, c is a constant. Pull it out and solve for it in term of p. Then use that value of c to find mu_x. Etc.
 
Last edited:
I got [\frac{cx^{p+1}}{p+1}-\frac{cx^{p+2}}{p+2}]_{0}^{1}
which I simplified to
(p+1)(p+2)=c

What do you mean I use this value of c to find mu sub x. Actually, it says above mu_x is equal to x so is that relevant...?
 
PCSL said:
I got (p+1)(p+2)=c

What do you mean I use this value of c to find mu sub x. Actually, it says above mu_x is equal to x so is that relevant...?

Ok, so f(x)=(p+1)(p+2)x^p(1-x). Now you just integrate f(x)*x to find E(x). Then use that find Var(f(x)). It should be routine from here on.
 
  • #10
Dick said:
Ok, so f(x)=(p+1)(p+2)x^p(1-x). Now you just integrate f(x)*x to find E(x). Then use that find Var(f(x)). It should be routine from here on.

Thank you so much! I assume I continue to integrate from 0 to 1 and I can pull (p+1)(p+2) out of the integral? Thanks again.
 
  • #11
Yes, the limits of integration for all integrals are 0 and 1. And for the integration, p is a constant.
 

Similar threads

Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 19 ·
Replies
19
Views
7K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K