Create a distribution of random numbers that would result in average demand per period for a Monte Carlo simulation that is equivalent to the expected demand per period using the data given by the chart below.
$$\begin{array}{cccc}\text{ Demand }& \text{ Probability }& \begin{array}{c}\text{ Cumulative }\\ \text{ Probability }\end{array}& \begin{array}{c}\text{ Interval of Random }\\ \text{ Numbers }\end{array}\\ 0& .& & \\ 1& .15& & \\ 2& .4& & \\ 3& .15& & \\ 4& .2& & \end{array}$$

$$\begin{array}{cccc}\text{ Demand }& \text{ Probability }& \begin{array}{c}\text{ Cumulative }\\ \text{ Probability }\end{array}& \begin{array}{c}\text{ Interval of Random }\\ \text{ Numbers }\end{array}\\ 0& .1& .1& 01-10\\ 1& .15& .25& 11-25\\ 2& .4& .65& 26-65\\ 3& .15& .8& 66-80\\ 4& .2& 1& 81-00\end{array}$$ This problem will most likely confuse many students, however its aim is to test their true understanding and ability to work both ways. The smallest common denominator for the probabilities is .05 so 20 separate random numbers must be generated for the simple solution where each possible demand has probability/.05 random numbers representing it within the set. For example, one set would be
(0, 0, 11, 11, 11, 26, 26, 26, 26, 26, 26, 26, 26, 66, 66, 66, 81, 81, 81, 81)
Students may also create sets that draw unevenly from various demand rows, such as drawing only from 0 and 4. The expected demand is 2.4, so students would solve a relation of the form X ∗ 0 + Y ∗ 4/(X + Y) = 2.4 thus X = 2Y/3 so when Y = 3 then X = 2. Therefore another possible set of random numbers would be (00, 00, 00, 01, 01)