Estimate how many books profit over 200k$?

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Homework Help Overview

The discussion revolves around estimating the number of books that made a profit exceeding $200,000 based on a given probability density function for profits in 2010. The density function is defined as f(x) = cx-3, with a known number of books (96) that exceeded a profit of $105.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss how to determine the constant c in the density function and its implications for calculating probabilities. There are attempts to integrate the density function over specified ranges to find probabilities associated with different profit thresholds. Some participants express confusion about the relationship between the number of books and the calculated probabilities.

Discussion Status

Several participants have provided insights into the integration process and the importance of the constant c, with some reaching different conclusions about its value. There is ongoing exploration of how to apply the density function to estimate the number of books exceeding the $200,000 profit threshold, but no consensus has been reached on the correct approach or calculations.

Contextual Notes

Participants are working under the assumption that the density function must integrate to 1 over its range, and there are discussions about the implications of the lower bound for profit values. The original poster has indicated a desire for guidance rather than direct answers, which shapes the nature of the responses.

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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-
 
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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.
 
Last edited:
Mixer said:

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
 
Hi there!

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

Hint: the number will be less than 96.
 
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?
 
Mixer said:
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
 
Ray Vickson said:
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?
 
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
Ray Vickson said:
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?
 
Last edited:
  • #11
Mixer said:
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
 
Last edited:
  • #12
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
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
Mixer said:
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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