Final answer:
The simulated demand value using of the second random number r2=0.19 is '1', matching option B of the given choices.
Step-by-step explanation:
The student is working with a discrete probability distribution and a simulation technique to determine demand values based on a series of random numbers. When using a sequence of random numbers to simulate demand values from a given probability distribution, we assign ranges of random numbers to each demand value in accordance with their frequencies. We create these ranges by cumulatively adding the frequencies.
The cumulative distribution for the given frequencies will be: 0 (0-0.15), 1 (0.15-0.45), 2 (0.45-0.70), 3 (0.70-0.85), 4 (0.85-1.00).
Since we should start with the second random number, r2=0.19, we look for the range that this value falls into. The value of 0.19 falls within the range for demand value 1 (0.15-0.45), so the simulated demand would be 1, which corresponds to option B.