Continuous Random variable Problem

[Disar]

Hi,I'm new to the site.

I come with a question I was hoping soemone out there can help me set up.

"The manager of a bakery knows that the number of chocalate cakes he can sell on any given day is a random variable with the probability mass function:

p(x)=1/6 For x=0,1,2,3,4,5

Also, there is a profit of 1$ on each cake sold and a loss of .40$ on each cake he doesn't sell (on a given day). Assuming each cake can only be sold on the day it was made, Find the baker's profit for a day when he bakes 5 chocalate cakes"

The hint is: Let the random vairable Y be the actual daily profit or loss. Using the probability distribution for X (number of customers asking for cakes) find the probability distribution for Y. The probability distribution for Y will depend on the number of cakes you decide to bake that day. You can answer the question using the random variable Y. Be careful to consider the possibility that the demand for your cakes, X, may exceed your supply.

With that said, my problem is this. I don't quite understand the problem and can't set it up. I have tried using the binomial distribtution. But don't understand how to actually get the profit.

Any help with this would be greatly appreciated.

Thanks,
Disar

 

[zwtipp05]

Do you know how to find expected value? If you find the expected value, you can find the profit. There appears to be no need to use the binomial distribution, in fact, you cannot use it here since you do not have bernoulli trials.

 

[HallsofIvy]

Unfortunately, there appear to be a number of things you are not sure of- starting with the title of this post! This is not a "continuous" probability distribution, it is discrete. Also, as zwtippo05 said, there is no reason to use the binomial distribution- there is nothing "binomial" about it.

Also, did the problem really say "Find the baker's profit"?? You can't really do that since this is a probability problem. All you can do is find the expected profit.

Suppose the baker makes 5 cakes.

case 1) He sells 0 cakes. What is his "profit" for that day? (it will, of course, be negative.) What is the probability that will happen?

case 2) He sells 1 cake. What is his profit for that day? What is the probability that will happen?

cases 3), 4), 5), 6) same questions for 2, 3, 4, 5 cakes sold.

"Expected value of profit" is (I am sure you have the definition in your text) the sum of the profit in each of those cases time the probability that case happens.


Strictly speaking, that is not using the "hint" (which I think is a bit confused anyway). Using the hint, you would argue, for example, that if the baker sells no cakes, he loses 5(0.40)= $2.00 ("profit" is -$2.00) and the probability of that is 1/6. If he sells 1 cake he gains $1 but loses 4(0.40)= $1.60 so he loses $0.60. His profit is -$0.60 and the probability of that is 1/6. Continue for 2, 3, 4, 5 cakes sold.
The "probability distribution" for his profit is:
-$2.00 1/6
-$0.60 1/6
.
.
.
$5.00 1/6

 

[Disar]

Thank you both for your replies. I am really struglling to keep my head above water in this class. As far as the topic heading I was just using the heading of the lectures notes given for the section we are currently covering. And yes the way I wrote the question is the way it was written in the problem my professor gave to me.

Thank you again