anonymous
  • anonymous
A small market orders copies of a certain magazine for its magazine rack each week. Let X = demand for the magazine, with the following pmf. x 1 2 3 4 5 6 p(x) 1/14 1/14 4/14 3/14 3/14 2/14 Suppose the store owner actually pays $2.00 for each copy of the magazine and the price to customers is $4.00. If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand X, and then compute the expected revenue.] Please explain what steps you took and why. I really want to learn the material but I'm having trouble interpreting what information the question is giving me, what it means, and how to set it up. Any help would be greatly appreciated!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
suppose he buys \(n\) magazines, costing him \(2n\) dollars, and the quantity demanded (i.e. the # of magazines people are willing to buy), is denoted \(X\). if \(n\le X\), then he can can sell all \(n\) of the magazines he bought, so the revenue is \(4n\), giving a profit/net-revenue of \(4n-2n=2n\). on the other hand, if \(n> X\), i.e. he bought more than were demanded, then he only sells \(X\) many, so the revenue is \(4X\) and the profit or net-revenue is \(4X-2n\). so we have two cases: $$X\ge n\implies \text{the net revenue is }2n\\X< n\implies\text{the net revenue is }4X-2n$$ so if he buys \(n=3\), then we see that: $$\begin{array}{c|c} X & \text{Net Revenue }(\$)\\ \hline 1 & -2\\ 2 & 2\\ 3 & 6\\ 4 & 6\\ 5 & 6\\ 6 & 6 \end{array}$$ whereas if he buys \(n=4\) we see $$\begin{array}{c|c} X & \text{Net Revenue }(\$)\\ \hline 1 & -4\\ 2 & 0\\ 3 & 4\\ 4 & 8\\ 5 & 8\\ 6 & 8 \end{array}$$
anonymous
  • anonymous
now, compute the expected net revenue for \(n=3\) and \(n=4\) using the given probabilities for \(X=1,2,3,\dots,6\): $$\begin{array}{c||cc}X&1&2&3&4&5&6\\\hline p &\frac1{14}&\frac1{14}&\frac4{14}&\frac3{14}&\frac3{14}&\frac2{14}\end{array}$$ so the expected net revenue for \(n=3\) is given by $$\frac1{14}\cdot(-2)+\frac1{14}\cdot2+\frac4{14}\cdot6+\frac3{14}\cdot6+\frac3{14}\cdot6+\frac2{14}\cdot6=\frac{36}7\approx5.143$$whereas for \(n=4\) we get $$\frac1{14}\cdot(-4)+\frac1{14}\cdot0+\frac4{14}\cdot4+\frac3{14}\cdot8+\frac3{14}\cdot8+\frac2{14}\cdot8=\frac{38}7\approx5.43$$ so on average he can expect more profit from buying \(n=4\) magazines rather than only \(3\)
anonymous
  • anonymous
Ahh, thank you!! That makes sense

Looking for something else?

Not the answer you are looking for? Search for more explanations.