Continuous random variable (supply and demand)

jimbobian
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Homework Statement



In the winter, the monthly demand in tonnes, for solid fuel from a coal merchant may be modeled by the continuous random variable X with probability density function given by:

f(x)=\frac{x}{30} 0≤x<6

f(x)=\frac{(12-x)^{2}}{180} 6≤x≤12

f(x)=0 otherwise

(a) Sketch y=f(x) - Done and correct
(b) State, giving a reason, whether the median demand is less than 6 tonnes - Done and correct
(c) Calculate the mean monthly demand - Done and correct
(d) Show that P(X≥8)=16/135 - Done (and being a "show that", correct)

The coal merchant has sufficient storage for 8 tonnes of solid fuel and this is replenished each month. Find the expected amount of solid fuel sold each month. - This is the question that I need help with


Homework Equations



Not sure there are any relative equations, I know how to find probabilities by integration and also how to get between PDFs and CDFs.

The Attempt at a Solution



My first thought was that it would be the same as in (c) ie. 5.4 tonnes, but then I realized that this can't be the case because if monthly demand goes above 8 tonnes, the merchant doesn't have any more coal to sell.

Second thought was that I could just chop the definition of f(x) and do:
\int^{8}_{0}xf(x)
To find the expectation of sales, but then the original definition of f(x) would not have an area of 1 so I think this would make it invalid.

Thirdly I thought I really was stuck and even looking at the answer in the book I couldn't fathom how they got to it (5.28 tonnes). Interestingly this is simply the answer to (c) [5.4 tonnes] minus the answer to (d), but this surely doesn't make sense? Imagine the merchant had sufficient storage for zero tonnes, P(X≥0)=1 but 5.4-1≠0 and so this clearly doesn't work.


Obviously I'm looking for some help with the problem and I know I have to demonstrate that I've given it a proper go myself before anyone can help me. I hope you can understand that I have given this a good deal of thought, but as I can't get past the thinking stage all I am able to write down is what I have argued with myself!

Cheers.
 
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If S is the amount sold in a month, the S = X if X ≤ 8 and 8 if X > 8. Try using the distribution of S.
 
jimbobian said:

Homework Statement



In the winter, the monthly demand in tonnes, for solid fuel from a coal merchant may be modeled by the continuous random variable X with probability density function given by:

f(x)=\frac{x}{30} 0≤x<6

f(x)=\frac{(12-x)^{2}}{180} 6≤x≤12

f(x)=0 otherwise

(a) Sketch y=f(x) - Done and correct
(b) State, giving a reason, whether the median demand is less than 6 tonnes - Done and correct
(c) Calculate the mean monthly demand - Done and correct
(d) Show that P(X≥8)=16/135 - Done (and being a "show that", correct)

The coal merchant has sufficient storage for 8 tonnes of solid fuel and this is replenished each month. Find the expected amount of solid fuel sold each month. - This is the question that I need help with


Homework Equations



Not sure there are any relative equations, I know how to find probabilities by integration and also how to get between PDFs and CDFs.

The Attempt at a Solution



My first thought was that it would be the same as in (c) ie. 5.4 tonnes, but then I realized that this can't be the case because if monthly demand goes above 8 tonnes, the merchant doesn't have any more coal to sell.

Second thought was that I could just chop the definition of f(x) and do:
\int^{8}_{0}xf(x)
To find the expectation of sales, but then the original definition of f(x) would not have an area of 1 so I think this would make it invalid.

Thirdly I thought I really was stuck and even looking at the answer in the book I couldn't fathom how they got to it (5.28 tonnes). Interestingly this is simply the answer to (c) [5.4 tonnes] minus the answer to (d), but this surely doesn't make sense? Imagine the merchant had sufficient storage for zero tonnes, P(X≥0)=1 but 5.4-1≠0 and so this clearly doesn't work.


Obviously I'm looking for some help with the problem and I know I have to demonstrate that I've given it a proper go myself before anyone can help me. I hope you can understand that I have given this a good deal of thought, but as I can't get past the thinking stage all I am able to write down is what I have argued with myself!

Cheers.

Sales S = min(C,X), where C = capacity (= starting inventory) and X = demand. That is: if X >= C you just sell all C units; if X < C you just sell X units.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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