Rubric for Online Discussion and discussion 2 answers

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Purdue University *

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57000

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Feb 20, 2024

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Uploaded by AdmiralWolverine1375

Question 1 (10 points)  No initial posting exists to evaluate. 0 Points  Your final answer is incorrect but the work shows one or more steps that are correct. 1-6 Points  The final answer is incorrect but most of the work is correct. 7 Points  The final answer is incorrect with only a minor calculation error, or the final answer is correct but the work needs more detail. 8 Points  The final answer is correct and the work is complete. 10 points Question 2 (10 points)  No initial posting exists to evaluate. 0 Points  Your final answer is incorrect but the work shows one or more steps that are correct. 1-6 Points  The final answer is incorrect but most of the work is correct. 7 Points  The final answer is incorrect with only a minor calculation error, or the final answer is correct but the work needs more detail. 8 Points  The final answer is correct and the work is complete. 10 points Participation in Discussion (10 points)  No responses to other classmates were posted in this discussion forum. 0 Points  May include one or more of the following: 1-5 Points  Comments to only one other student's post.  Comments are not substantive, such as just one line or saying, “Good job” or “I agree.  Comments are off topic.  Comments to two or more classmates’ initial posts but only on one day of the week. Comments are substantive, meaning they reflect and expand on what the other student wrote. 6-7 Points  Comments to two or more classmates’ initial posts on more than one day. Comments are substantive, meaning they reflect and expand on what the other student wrote. 8-9 Points.  Comments to two or more classmates’ initial posts and to the instructor's comment (if applicable) on two or more days. Responses demonstrate an analysis of peers’ comments, building on previous posts. Comments extend and deepen meaningful conversation and may include a follow-up question. 10 Points

Q4 pg 735 Use @Risk’s define distribution tool to draw a normal distribution with mean 500 and standard deviation 100. Then answer the following questions: I will be honest I struggled with using @Risk for the below questions. I am more familiar with using excel to find the answers to the below questions. To create the above table I selected @Risk, clicked on distribution than define, selected Normal for my distribution, entered 500 into the mean/µ box, entered 100 into the std. dev/σ box, and then hit okay. What is the probability that a random number from this distribution is less than 450? I used =NORM.DIST(x,mean,std dev,TRUE) = (450,500,100,TRUE) = 0.3085 or 30.85% of random numbers are less than 450. What is the probability that a random number from this distribution is greater than 650? I used =1-NORM.DIST(x,C4,C5,TRUE) = 1-(650,500,100,TRUE) = 0.0668 or 6.68% of random numbers are greater than 650. What is the probability that a random number from this distribution is between 500 and 700?

I used =RiskOutput()+NORM.DIST(x,mean,std.dev,TRUE)-NORM.DIST(x,mean,std.dev,TRUE) = RiskOutput()+NORM.DIST(700,500,100,TRUE)-NORM.DIST(500,500,100,TRUE) = 0.4772 or 47.72% of random numbers are between 500 to 700. Prob9 pg 735 A company is about to develop and then market a new product. It wants to build a simulation modelfor the entire process, and one key uncertain input is the development cost. For each of the following scenarios, choose an appropriate distribution together with its parameters, justify your choice in words, and use @Risk’s define distribution tools to show your chosen distribution. a.Company experts have no idea what the distribution of the development cost is. All they can state is “we are 95% sure it will be at least $450,000, and we are 95% sure it will be no more than $650,000”. To create the below table I selected @Risk, clicked on distribution than define, selected Uniform for my distribution, after completing the below math, entered 339.89 into the min box, entered 761.11 into the max box, and then hit okay. We have to complete a little math for this question. The interval length is 650-450 = 200. The probability of the random number is 90%. So the total region has an interval length of (200*100/90) = 222.22 The to compute the overall interval length first take 222.22/2 = 111.11 Lower limit 450-111.11 = 338.89 Upper limit 650+111.11 = 761.11 Cost is a continuous variable, meaning between the two numbers there can be infinite numbers, making the distribution uniform.

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b.Company experts can still make the same statement as in part a, but now they can also state: “We believe the distribution is symmetric, reasonably bell-shaped, and its most likely value is about $550,000”. To create the below table I selected @Risk, clicked on distribution than define, selected normal for my distribution, after completing the below math, entered 550,000 into the mean/µ box, entered 60,800 into the std. dev/σ box, and then hit okay. By stating symmetric distribution and bell shaped, with a most likely value of 550,000, suggests a normal distribution should be used. In a normal distribution, 90% of the area under the curve lies within 1.645 standard deviations of the mean. Given that this range is 100,000, we can calculate the standard deviation as 100,000 / 1.645, which is approximately 60,800. The mean in this situation being 550,000.

c.Company experts can still make the same statement as in part a, but now they can also state: “We believe the distribution is skewed to the right, and its most likely value is about $500,000”. To create the below table I selected @Risk, clicked on distribution than define, selected triangular for my distribution, entered 500,000 into the mid. likely box, entered 450,000 into the minimum box, entered 650,000 in to the maximum box, I then decreased 450,000 and increased 650,000 in increments of ten, I got to 420,000 and 700,000 and could see that I was close based on my std. dev, so I continued to decrease and increase in increments of 1,000 until I got to a minimum of 415,000 and maximum of 705,000 and std. dev of 60,800, and then hit okay. The statement about a distribution that is skewed to the right with a most likely value of 500,000 suggests a triangular distribution. The minimum and maximum values of the distribution can be determined through trial and error to satisfy the conditions about 450,000 and 650,000. Using @RISK, the approximate minimum and maximum values are 415,000 and 705,000. By adjusting the min and max it brings the standard deviation to approximately 60,800.

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