Practical Management Science
6th Edition
ISBN: 9781337406659
Author: WINSTON, Wayne L.
Publisher: Cengage,
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Textbook Question
Chapter 10.2, Problem 6P
Use @RISK to draw a binomial distribution that results from 50 trials with probability of success 0.3 on each trial, and use it to answer the following questions.
- a. What are the mean and standard deviation of this distribution?
- b. You have to be more careful in interpreting @RISK probabilities with a discrete distribution such as this binomial. For example, if you move the left slider to 11, you find a probability of 0.139 to the left of it. But is this the probability of “less than 11” or “less than or equal to 11”? One way to check is to use Excel’s BINOM.DIST function. Use this function to interpret the 0.139 value from @RISK.
- c. Using part b to guide you, use @RISK to find the probability that a random number from this distribution will be greater than 17. Check your answer by using the BINOM.DIST function appropriately in Excel.
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Chapter 10 Solutions
Practical Management Science
Ch. 10.2 - Use the RAND function and the Copy command to...Ch. 10.2 - Use Excels functions (not @RISK) to generate 1000...Ch. 10.2 - Use @RISK to draw a uniform distribution from 400...Ch. 10.2 - Use @RISK to draw a normal distribution with mean...Ch. 10.2 - Use @RISK to draw a triangular distribution with...Ch. 10.2 - Use @RISK to draw a binomial distribution that...Ch. 10.2 - Use @RISK to draw a triangular distribution with...Ch. 10.2 - We all hate to keep track of small change. By...Ch. 10.4 - Prob. 11PCh. 10.4 - In August of the current year, a car dealer is...
Ch. 10.4 - Prob. 13PCh. 10.4 - Prob. 14PCh. 10.4 - Prob. 15PCh. 10.5 - If you add several normally distributed random...Ch. 10.5 - In Problem 11 from the previous section, we stated...Ch. 10.5 - Continuing the previous problem, assume, as in...Ch. 10.5 - In Problem 12 of the previous section, suppose...Ch. 10.5 - Use @RISK to analyze the sweatshirt situation in...Ch. 10.5 - Although the normal distribution is a reasonable...Ch. 10.6 - When you use @RISKs correlation feature to...Ch. 10.6 - Prob. 24PCh. 10.6 - Prob. 25PCh. 10.6 - Prob. 28PCh. 10 - Six months before its annual convention, the...Ch. 10 - Prob. 30PCh. 10 - A new edition of a very popular textbook will be...Ch. 10 - Prob. 32PCh. 10 - W. L. Brown, a direct marketer of womens clothing,...Ch. 10 - Assume that all of a companys job applicants must...Ch. 10 - Lemingtons is trying to determine how many Jean...Ch. 10 - Dilberts Department Store is trying to determine...Ch. 10 - It is surprising (but true) that if 23 people are...Ch. 10 - Prob. 40PCh. 10 - At the beginning of each week, a machine is in one...Ch. 10 - Simulation can be used to illustrate a number of...Ch. 10 - Prob. 43PCh. 10 - Prob. 46PCh. 10 - If you want to replicate the results of a...Ch. 10 - Suppose you simulate a gambling situation where...Ch. 10 - Prob. 49PCh. 10 - Big Hit Video must determine how many copies of a...Ch. 10 - Prob. 51PCh. 10 - Prob. 52PCh. 10 - Why is the RISKCORRMAT function necessary? How...Ch. 10 - Consider the claim that normally distributed...Ch. 10 - Prob. 55PCh. 10 - When you use a RISKSIMTABLE function for a...Ch. 10 - Consider a situation where there is a cost that is...
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- If you add several normally distributed random numbers, the result is normally distributed, where the mean of the sum is the sum of the individual means, and the variance of the sum is the sum of the individual variances. (Remember that variance is the square of standard deviation.) This is a difficult result to prove mathematically, but it is easy to demonstrate with simulation. To do so, run a simulation where you add three normally distributed random numbers, each with mean 100 and standard deviation 10. Your single output variable should be the sum of these three numbers. Verify with @RISK that the distribution of this output is approximately normal with mean 300 and variance 300 (hence, standard deviation 300=17.32).arrow_forwardContinuing the previous problem, assume, as in Problem 11, that the damage amount is normally distributed with mean 3000 and standard deviation 750. Run @RISK with 5000 iterations to simulate the amount you pay for damage. Compare your results with those in the previous problem. Does it appear to matter whether you assume a triangular distribution or a normal distribution for damage amounts? Why isnt this a totally fair comparison? (Hint: Use @RISKs Define Distributions tool to find the standard deviation for the triangular distribution.)arrow_forwardIn Example 11.2, the gamma distribution was used to model the skewness to the right of the lifetime distribution. Experiment to see whether the triangular distribution could have been used instead. Let its minimum value be 0, and choose its most likely and maximum values so that this triangular distribution has approximately the same mean and standard deviation as the gamma distribution in the example. (Use @RISKs Define Distributions window and trial and error to do this.) Then run the simulation and comment on similarities or differences between your outputs and the outputs in the example.arrow_forward
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