# Homework Help: Estimate how many books profit over 200k$? 1. Nov 18, 2011 ### Mixer 1. The problem statement, all variables and given/known data In the year 2010 a profit of book in x dollars follows the distribution which density function is f(x) = cx-3. There were 96 books in that year which profit exceeded 105$. Estimate how many books profit over 2*105 ?

2. Relevant equations

For cumulative distribution function F(x) = $\int f(x)$

3. 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-

2. Nov 18, 2011

### Harrisonized

∫ 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: Nov 18, 2011
3. Nov 18, 2011

### Ray Vickson

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. Nov 18, 2011

### Mixer

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. Nov 18, 2011

### SteamKing

Staff Emeritus
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. Nov 19, 2011 ### Mixer Hi, I got c in terms of x: $\int cf(x) dx$ from 1 to x = 105 c/2 - c/x2 = 105 c = (2x2105)/(x2 - 2) Now what? 7. Nov 19, 2011 ### Ray Vickson 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. Nov 19, 2011 ### Mixer 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: $\int$ 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. Nov 19, 2011 ### Ray Vickson 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. Nov 20, 2011 ### Mixer Ok thanks :) So my first attempt was correct. If I now calculate the probability to get 105$ (or more) profit for one book:

$\int$ 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: $\int$ 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: Nov 20, 2011
11. Nov 20, 2011

### Ray Vickson

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: Nov 20, 2011
12. Nov 20, 2011

### Mixer

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. Nov 21, 2011

### Mixer

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

$\int$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

$\int$(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. Nov 21, 2011

### Ray Vickson

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