Continuous random variable (supply and demand)

In summary, the conversation discusses a continuous random variable X that models the monthly demand for solid fuel from a coal merchant. The probability density function for X is given and various questions are asked, with the final one being about the expected amount of solid fuel sold each month. The solution involves finding the distribution of sales, which is the minimum between the capacity and the demand. This leads to the answer of 5.28 tonnes, which is the difference between the mean demand (5.4 tonnes) and the probability that the demand exceeds the capacity (16/135 tonnes).
  • #1
jimbobian
52
0

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:

[itex]f(x)=\frac{x}{30}[/itex] 0≤x<6

[itex]f(x)=\frac{(12-x)^{2}}{180}[/itex] 6≤x≤12

[itex]f(x)=0 [/itex] 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:
[tex][/tex][itex]\int^{8}_{0}xf(x)[/itex]
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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  • #2
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.
 
  • #3
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:

[itex]f(x)=\frac{x}{30}[/itex] 0≤x<6

[itex]f(x)=\frac{(12-x)^{2}}{180}[/itex] 6≤x≤12

[itex]f(x)=0 [/itex] 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:
[tex][/tex][itex]\int^{8}_{0}xf(x)[/itex]
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
 

1. What is a continuous random variable?

A continuous random variable is a type of random variable that can take on any value within a given range. This is in contrast to a discrete random variable, which can only take on specific, distinct values. Continuous random variables are often used to model real-world phenomena, such as measurements or quantities that can have infinite possible values.

2. How is supply and demand related to continuous random variables?

In economics, supply and demand can be represented by continuous random variables. The supply curve represents the quantity of a good or service that producers are willing to offer at different prices, while the demand curve represents the quantity that consumers are willing to buy at different prices. The intersection of these two curves determines the equilibrium price and quantity for a given market.

3. What is the difference between a random variable and a continuous random variable?

A random variable is a variable whose value is determined by chance. It can be either discrete or continuous. A continuous random variable is one that can take on an infinite number of values within a given range, while a discrete random variable can only take on a finite or countably infinite number of values.

4. How are probabilities calculated for continuous random variables?

The probability of a specific value occurring for a continuous random variable is always zero, since there are infinite possible values. Instead, probabilities are calculated for ranges of values. This is done using the probability density function (PDF) or cumulative distribution function (CDF) of the continuous random variable.

5. What are some examples of continuous random variables in everyday life?

Some examples of continuous random variables in everyday life include height, weight, time, and temperature. These variables can take on a wide range of values and are often used in statistical models to make predictions and inform decision-making.

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