A process for refining sugar yields up to 1 ton of pure sugar per day, but the a

Question

A process for refining sugar yields up to 1 ton of pure sugar per day, but the actual amount produced, Y, is a random variable because of machine breakdowns and other slowdowns. Suppose that Y has density function given by f(y) = {2y 0 lessthanorequalto y lessthanorequalto 1 0 elsewhere The company is paid at the rate of $300 per ton for the refined sugar, but it also has a fixed overhead cost of $100 per day. (a) Find the mean and variance of Y. (b) Express the daily profit, P say, as a function of Y. (c) Hence, find the probability density function of P. (d) Find E(P), the average daily profit (in hundreds of dollars).

Explanation / Answer

We are given that the Y is a random variable which denotes the amount of pure sugar yielded in one day. The density function of the random variable Y is given as:

   f(y) = 2y ; 0y1

             0 elsewhere

(a)The mean of the random variable Y is given by

E(Y) =yf(y).dy=10y*2y.dy = 102y2.dy =[2/3y3]10 =2/3

The variance of the random variable Y is given by

V(Y)=E(Y2)-[E(Y)]2

Now, E(Y2)=y2f(y).dy = 10y2*2y.dy = 10 2y3.dy = [1/2y4]10 = ½

So the variance is given by V(Y)=1/2 – (2/3)2 = 1/18

(b) Now it is given that the company is paid at the rate of $300 per ton for the refined sugar and has a fixed overhead cost of $100 per day. Now, if we take P as the random variable which denotes the profit for a day then if on any given day the company produces Y tons of refined sugar then the profit for that day will be P= $300Y-$100. Since we know that 0Y1 so by putting the limit values in the above equation we get the value range of profit i.e when Y=1 P=$200 and when Y=0 P=-$100

So, we have the following to denote profit as a function of Y i.e P(y) =$300Y-$100 and -$100P(y)$300

(c)The probability density function of P is given as follows:

Working in hundreds of dollars we have P =3Y-1 Y = (P+1)/3 and we already know the range for P from (b) as worked above -1P3 (working in hundreds of dollars).

Differentiating Y = (P+1)/3 with respect to P we have d/dpY = 1/3

So, by using the definition of functional transformations in random variables we have the probability density function of P i.e f(p) = 1/3 * 2(P+1)/3 =2/9(p+1)

Thus, f(p) = 2/9(p+1) where -1P3 here P is in hundreds of dollars.

(d)The average daily profit (in hundreds of dollars) is given by

E(P) =p*f(p).dp =3-1p*2/9(p+1).dp =2/93-1(p2+p).dp = 2/9[p3/3 + p2/2]3-1 =2/9*[(9+1/3) + (4.5-1/2)] = 2/9*(28/3 + 4) =2/9 * 40/3 =80/27 = 2.963(working in hundreds of dollars)

Thus, the average daily profit is $296.3 per day.

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