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Estimate how many books profit over 200k$?

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  • #1
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Homework Statement



In the year 2010 a profit of book in x dollars follows the distribution which density function is f(x) = cx-3. There were 96 books in that year which profit exceeded 105 $. Estimate how many books profit over 2*105 ?



Homework Equations



For cumulative distribution function F(x) = [itex]\int f(x)[/itex]


The Attempt at a Solution



I don't what to do. I tried to define c but I failed. I don't want direct answer but just some guidelines which help me to get started.


-Mixer-
 

Answers and Replies

  • #2
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∫ f(x) dx

If there is a profit distribution, then integrating over some segment can give you the total profit in that segment. Since this is cumulative, you want to start the integration at -∞ and end at x, the profit you want. That's how you can calibrate c.

How is this precalc.
 
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  • #3
Ray Vickson
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Homework Statement



In the year 2010 a profit of book in x dollars follows the distribution which density function is f(x) = cx-3. There were 96 books in that year which profit exceeded 105 $. Estimate how many books profit over 2*105 ?



Homework Equations



For cumulative distribution function F(x) = [itex]\int f(x)[/itex]


The Attempt at a Solution



I don't what to do. I tried to define c but I failed. I don't want direct answer but just some guidelines which help me to get started.


-Mixer-
Is there a *positive* lower bound on x? (That is, is there a number a > 0 for which we have x >= a for all allowed profit values?) There had better be such an 'a', because otherwise your cumulative distribution function does not exist: you get a divergent integral.

RGV
 
  • #4
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Hi there!

I forgot to mention that x ≥ 1 . But if I can get c how does it help me with the original question?
 
  • #5
SteamKing
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You use the density function to determine how many books made more than $200K in profit.

Hint: the number will be less than 96.
 
  • #6
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Hi,

I got c in terms of x:

[itex]\int cf(x) dx[/itex] from 1 to x = 105

c/2 - c/x2 = 105

c = (2x2105)/(x2 - 2)

Now what?
 
  • #7
Ray Vickson
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Hi,

I got c in terms of x:

[itex]\int cf(x) dx[/itex] from 1 to x = 105

c/2 - c/x2 = 105

c = (2x2105)/(x2 - 2)

Now what?
What on Earth are you doing? The value of c is determined independently of any values like 10^5 or whatever. Go back and look in your textbook or course notes, to see what you need to do. Remember: the total probability (over all possible outcomes) must be 1.

RGV
 
  • #8
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What on Earth are you doing? The value of c is determined independently of any values like 10^5 or whatever. Go back and look in your textbook or course notes, to see what you need to do. Remember: the total probability (over all possible outcomes) must be 1.

RGV
Hi there!

I'm pretty sure that the density function given is not a probability function or is it?

My first attempt to find c was like this:

[itex]\int[/itex] cf(x) from 1 to ∞ = 1

-c/2 ( 0 - 1/1) = 1

so c = 2

Is that more correct? But total profit cannot be just 1?
 
  • #9
Ray Vickson
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c=2 is correct; that makes the total *probability* equal to 1. Of course, profit x can be any number >= 1; the PDF f(x) =2/x^3 just tells you about the probability of getting profit x.

RGV
 
  • #10
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c=2 is correct; that makes the total *probability* equal to 1. Of course, profit x can be any number >= 1; the PDF f(x) =2/x^3 just tells you about the probability of getting profit x.

RGV
Ok thanks :) So my first attempt was correct. If I now calculate the probability to get 105 $ (or more) profit for one book:


[itex]\int[/itex] 2x-3 from 105 to ∞

-1(1/∞ - (10-10) = 10-10

So that's the probability for one book. There are 96 books which exceeded that amount of dollars. So the total probability for 96 books to profit over 105 $ is 96 times the probability of one book?

P (96 books profit over 105 $) = 96*10-10

So now I calculate the probability to get 2*105 $ (or more) profit for one book:

[itex]\int [/itex] 2x-3 from 2*105 to ∞

-1(1/∞ - (2*10-10) = 2*10-10

So the probability for x books to profit over 2*10-10 is x times the probability of one book:

P (x books profit over 2*105 $) = x*2*10-10

Is my calculations correct and if they are, what should I do now?
 
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  • #11
Ray Vickson
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Ok thanks :) So my first attempt was correct. If I now calculate the probability to get 105 $ (or more) profit for one book:


[itex]\int[/itex] 2x-3 from 105 to ∞

-1(1/∞ - (10-10) = 10-10

So that's the probability for one book. There are 96 books which exceeded that amount of dollars. So the total probability for 96 books to profit over 105 $ is 96 times the probability of one book?

P (96 books profit over 105 $) = 96*10-10

So now I calculate the probability to get 2*105 $ (or more) profit for one book:

[itex]\int [/itex] 2x-3 from 2*105 to ∞

-1(1/∞ - (2*10-10) = 2*10-10

So the probability for x books to profit over 2*10-10 is x times the probability of one book:

P (x books profit over 2*105 $) = x*2*10-10

Is my calculations correct and if they are, what should I do now?
You have things exactly backwards. You start with some unknown number of books, say N. Some of these books have profit < 10^5, others have profit >= 10^5, etc. All you know is that among these N books, 96 of them have profit >= 10^5. You don't start with 96 books!

RGV
 
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  • #12
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Hi,

I'm little confused what to do now. I know that 96 books profit over 10^5 but now I need to know how many books of these 96 profit over 2*10^5. How do I do that?
 
  • #13
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Finally I got an answer! I worked the problem with my friend and he told me think the density function as a "bookdensity" function. So his solution was like this:

First we try to find c

[itex]\int[/itex]cx-3 from 10^5 to ∞ = 96

so -c/2 (1/∞ - 1/10^10 ) = 96

so c = 96*2*10^10

Now let's find the amount of books that profit over 2*10^5

[itex]\int[/itex](96*2*10^10)x-3 from 2*10^5 to ∞

-96*10^10 ( 1/∞ - 1/4*10^10) = 96/4 = 24

So 24 books profit over 200 k$

That seems to be correct. How is your opinion?
 
  • #14
Ray Vickson
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Finally I got an answer! I worked the problem with my friend and he told me think the density function as a "bookdensity" function. So his solution was like this:

First we try to find c

[itex]\int[/itex]cx-3 from 10^5 to ∞ = 96

so -c/2 (1/∞ - 1/10^10 ) = 96

so c = 96*2*10^10

Now let's find the amount of books that profit over 2*10^5

[itex]\int[/itex](96*2*10^10)x-3 from 2*10^5 to ∞

-96*10^10 ( 1/∞ - 1/4*10^10) = 96/4 = 24

So 24 books profit over 200 k$

That seems to be correct. How is your opinion?
I think you should tell your friend to go back to the drawing board. The constant c is not 96*2*10^10; we already established before that c = 2, remember? I have also told you more-or-less what you need to do, but you have not been listening. I give up.

RGV
 

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