# Central limit theorem

1. Dec 13, 2007

### Gott_ist_tot

1. The problem statement, all variables and given/known data
A newsagent finds that the probability of selling 50n copies of a certain magazine is 2^{-n-1} for n = 1,2,3,... What is the largest sensible number of copies of the magazine that they should stock.

2. Relevant equations

3. The attempt at a solution

I really don't know how to approach this one. I hate not to provide anything in the attempt, but I am really lost on this one. Any suggestions are appreciated

2. Dec 13, 2007

### Dick

I suspect they want you to compute the expectation value of the number of copies you can sell. Beyond that, what's the definition of "sensible"? Stocking "infinity" copies is always safe against demand, but you haven't stated any cost for stockpiling them.

3. Dec 13, 2007

### Gott_ist_tot

Oh, yes. He forgot to write the cost to sell them and it was amended. It is $1 for him to buy it and$1.50 is what he sells it for.

The "sensible" through me off also. Hopefully the price will help. I will start looking at the expectation.

4. Dec 14, 2007

### Gott_ist_tot

I must be missing/not understanding something here.

E[X] = 0.5(50n*2^(-n-1)) - 1.0(50n*2^(-n-1))
= -25n * 2^(-n-1)

I tried finding a maximum but I had no luck. The 50n is the number of magazines. 2^(-n-1) is the probability. Then the 0.5 is how much he would make from a sell and the -1 is how much he would lose from overstock.

5. Dec 14, 2007

### Dick

If he buys 50*N magazines the profit is 150*(50*s(n))-100*50N, where s(n) is the number he actually sells. s(n)=n if n<=N, s(n)=N for n>N. That's the expectation value you want to maximize as a function of N. Also 2^(-n-1) doesn't sum to unit probability for n=1,2,3... I think you want 2^(-n).