Consider the following question. X is supposed to represent the sale price of a hotel room, while Y represents the cost price. Therefore the profit is given by X-Y. I am going to use the term expected average as opposed to the more standard expected value or expectation. 

There is a problem with the interpretation and I wouldn’t treat this particular exercise with much importance.

Suppose that X and Y are independent random variables with distributions

 

Find the expected average of the profit on a single room. Find the expected average of the profit on 1,000 rooms. Find the probability that the profit on 1,000 rooms is less than 20,000.

Solution : The expected average of a variable is given by:

\mathbb{E}[X]=\sum_ix_i\mathbb{P}[X=x_i]=\sum_ix_ip_i.

Now expected average is linear:

\mathbb{E}[X+\lambda Y]=\sum_i(x_i+\lambda y_i)p_i=\sum_{i}x_ip_i+\lambda\sum_i y_i p_i

=\mathbb{E}[X]+\lambda\mathbb{E}[Y].

This means that to find the expected average of X-Y we can calculate the expected values of X and Y separately — and then use linearity to write \mathbb{E}[X-Y]=\mathbb{E}[X]-\mathbb{E}[Y].

\mathbb{E}[X]=90(0.35)+100(0.5)+110(0.15)=98,

\mathbb{E}[Y]=70(0.45)+80(0.4)+90(0.15)=77.

So the expected average profit on a single room is 21.

Now this is where things get a bit more complicated. There are two possible readings of the question (to be able to answer the question at all implies that the first interpretation is correct — even if it seems a bit unrealistic):

  1. Once a price has been agreed for the cost and sale price for the very first room, the prices will be the same for the next 999 rooms.
  2. The cost and sale price of every room is random and follows the above distribution.

We will now answer the second two questions separately in case 1 and case 2.

Assuming 1.

The random variable we are interested in is 1,000(X-Y). Using linearity

\mathbb{E}[1,000(X-Y)]=1,000\mathbb{E}[X-Y]=21,000.

Now the probability that the expected average profit on 1,000 rooms is less than 20,000 is the same as the probability that the profit on the first room is less than or equal to 20 (again we have another problem with interpretation. I would read less than 20,000 as including 20,000. This is what I have followed but make sure to make a note of the assumptions you are using). The following outcomes imply this

(X,Y)=(90,70)\text{ or }(90,80)\text{ or }

(90,90)\text{ or }(100,80)\text{ or }(100,90)\text{ or }(110,90)).

We know want the probability that (X,Y) is one of these. Now the probability that X=x and Y=y is given by \mathbb{P}[X=x]\times\mathbb{P}[Y=y]:

(0.35)(0.45)+(0.35)(0.4)+(0.35)(0.15)+

(0.5)(0.4)+(0.5)(0.15)+(0.15)(0.15)=0.6475.

Assuming 2.

Now things are very different. Let P_i:=X_i-Y_i be the profit on the i-th room. The expected average profit is the sum of all these profits:

\mathbb{E}[P_1+P_2+P_3+\cdots+P_{1,000}].

Luckily we can see that the linearity of expectation simplifies things greatly:

\mathbb{E}[P_1+P_2+\cdots+P_{1,000}]=\mathbb{E}[P_1]+\cdots+\mathbb{E}[P_{1,000}].

Luckily all these expected average profits are the same and we get the same answer as above — 1,000 times the expected average profit on a single room:

\mathbb{E}[\text{Profit}]=1,000\mathbb{E}[Y-X]=21,000.

We don’t really have the tools in ST1021 to answer the question. The law of large numbers suggests that with 1,000 rooms, the expected average profit is with a high probability very close to 21,000. This means that \mathbb{P}[\text{Profit }\leq20,000]\approx0.

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