My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
March 6th, 2014, 12:23 AM   #11
Senior Member
 
Joined: Jan 2014

Posts: 196
Thanks: 3

Re: Help understanding/putting together this type of problem

I see this just plugging in values..... I see the middle does seem to be the lowest value.

Just not understanding why, or the general approach.

Do I set this = 0 an solve for c?

Also something was mentioned about finding the min between an interval. how would I find the function for this?
WWRtelescoping is offline  
 
March 6th, 2014, 02:31 AM   #12
Senior Member
 
Joined: Jun 2013
From: London, England

Posts: 1,316
Thanks: 116

Re: Help understanding/putting together this type of problem

Quote:
Originally Posted by WWRtelescoping
I see this just plugging in values..... I see the middle does seem to be the lowest value.

Just not understanding why, or the general approach.
Why not try it out with sets of numbers. For example:

2, 6, 21

1, 5, 10, 12, 13

0, 8, 9, 20, 25, 30, 33

Try the middle and see what happens when you move either side of it.
Pero is offline  
March 6th, 2014, 06:19 AM   #13
Senior Member
 
Joined: Jan 2014

Posts: 196
Thanks: 3

Re: Help understanding/putting together this type of problem

I did something like this.

e.g.







And with other random numbers, have seen the same thing, that when is the middle term, w/ an odd number of terms, it minimizes more than the others.

Though I have seen that this works w/ some examples, is there a way to show this works for all cases?

How do I know there isn't another ; that minimizes this further?

Thanks for all the help with this!
WWRtelescoping is offline  
March 6th, 2014, 06:30 AM   #14
Senior Member
 
Joined: Jun 2013
From: London, England

Posts: 1,316
Thanks: 116

Re: Help understanding/putting together this type of problem

Quote:
Originally Posted by WWRtelescoping

Though I have seen that this works w/ some examples, is there a way to show this works for all cases?

How do I know there isn't another ; that minimizes this further?
It's called logic!

Sometimes you prove things with equations and formulas; sometimes, you can prove them with a logical argument, as I did a few posts above. In a nutshell:

As you move away from the middle term (in either direction) you are moving away from more terms than you are moving towards. You should try to see this with the examples you've done.

E.g. if you move from 6 to 5.9, you move 0.1 away from 6 and 21 and 0.1 towards 2. So, the sum of differences increases by 0.1. And, it must increase by any movement away from the middle term.
Pero is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
problem, type, understanding or putting



Search tags for this page
let x be a random variable with support
,

let x be a random variable with support {1,2,3,5,15,25,50} , each point of which has the same probability 1/7 argue that c=5 is the value that minimizes E(|x-C|)

,

argue that c=5 is the value that minimizes

,

let x be the random variable with support {1,2,3,5,15,25,50}, each point which has the same probablilty 1/7.

,

value of 5/15/25

,

let x be a random support c=5 minimizes compare

,

let x be a random variable with support {1,2,3,5,15,25,50} , each point of which has the same probability 1/7 argue that c=5 is the value that minimizes E(|x-C|) answers

,

let x be a random variable with support(1,2,3,5,15,25,50,)each point of which has the same probability 1/7 argue that C =5 is the value that minimizes E(/x c/) compare this to the value of b that minimizes E /(x b)^2/

,

let X be a random variable with support (1,2,3,5,15,25,50) each point of which has the same probability 1/7.Argue that c=5 is the value that minimizes E(|x-c|). compare this to a value b that minimizes E(|x-b|^2)

,

Let X be a random variable with support {1,2,3,5,15,25,50}, each point which has the same probability 1/7 Argue that C=5 is the value that minimizes E(|x-C|). Compare this to the value of b that minimizes E[(x-b)^2].-solution

,

let X be a random variable with support (1,2,3,5,15,25,50) each point of which has the same probability 1/7 argue that c=5 is the value that minimizes E(|x-c|). Compare this to the value of b that minimizes E[(x-b)^2]

,

Let X be a random variable with support {1, 2, 3, 5, 15, 25, 50}, each point of which has the same probability 1/7. Argue that c = 5 is the value that mini- mizes h(c) = E( |X − c| ). Compare c with the value of b that minimizes g(b) = E[(X − b)2]

,

let x be a random variable with support {1,2,3,5,15,25,50} each of which has the same probability 1/7. argue that c=5 is the value that minimizes

,

Let x be a random variable with support (1,2,3,5,15,25,50),each point of which has the same probability 1 /7 Argue that c =5 is the value that minimizes E(|x-c |) compare this to the value of b that minimizes E[(x-b)^2]

,

1,2,3,5,15,25,50

Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Help applying mean and variance to this type of problem WWRtelescoping Algebra 1 March 16th, 2014 02:58 PM
what type of problem is this? xamdarb Calculus 1 March 10th, 2014 05:36 PM
Ratio type problem lfd2142 Elementary Math 2 December 30th, 2013 02:13 PM
Algebra Type Problem coolhandluke Applied Math 4 September 12th, 2010 02:36 PM
LCD type problem Bryce Algebra 6 June 17th, 2009 02:14 PM





Copyright © 2019 My Math Forum. All rights reserved.