EBK PRACTICAL MANAGEMENT SCIENCE
5th Edition
ISBN: 9780100655065
Author: ALBRIGHT
Publisher: YUZU
expand_more
expand_more
format_list_bulleted
Concept explainers
Textbook Question
Chapter 10.2, Problem 2P
Use Excel’s functions (not @RISK) to generate 1000 random numbers from a
- a. Calculate the mean and standard deviation of these random numbers. Are they approximately what you would expect?
- b. What fraction of these random numbers are within k standard deviations of the mean? Answer for k = 1; for k = 2; for k = 3. Are the answers close to what they should be (about 68% for k = 1, about 95% for k = 2, and over 99% for k = 3)?
- c. Create a histogram of the random numbers using about 10 bins of your choice. Does this histogram have approximately the shape you would expect?
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
4.1 TYPES OF RANDOM VARIABLES. Which of the following describe continuous random variables? Which describe discrete random variables?
The number of newspapers sold by the New York Times each month
The amount of ink used in printing a Sunday edition of the New York Times
The actual number of ounces in a 1-gallon bottle of laundry detergent
The number of defective parts in a shipment of nuts and bolts
The number of people collecting unemployment insurance each month
The mean annual earning for U.S. workers with advanced degrees is $80,977 with a standard deviation of $100,895; the distribution is highly right-skewed. Use this information to answer questions 7 to 10 below. A random sample of n = 86 workers is drawn, and their annual earnings recorded.
What is the expected value of the sample mean of their annual earnings?
What is the standard deviation of the sample mean of their annual earnings?
what is the probability that the sample mean is at least $100,000?
The actual sample mean earnings for the random sample of 86 workers with advanced degrees was $82,105. What is the sampling error?
Given a normal distribution with μ= 105 and σ = 20, and given you select a sample of n = 16, complete parts (a)
through (d).
a. What is the probability that X is less than 92?
P(X<92)
0.0047
(Type an integer or decimal rounded to four decimal places as needed.)
b. What is the probability that X is between 92 and 93.5?
P(92 X 93.5)= 0.0061
(Type an integer or decimal rounded to four decimal places as needed.)
c. What is the probability that X is above 105.2?
P(X 105.2)= 0.4840
(Type an integer or decimal rounded to four decimal places as needed.)
d. There is a 67% chance that X is above what value?
X =
(Type an integer or decimal rounded to two decimal places as needed.)
Chapter 10 Solutions
EBK 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 - Prob. 34PCh. 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...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, operations-management and related others by exploring similar questions and additional content below.Similar questions
- 4. ABC Dog Food Company located in Ottawa sells large bags of dog food to warehouse clubs. ABC uses an automatic filling process to fill the bags. Weights of the filled bags are approximately normally distributed with a mean of 50 kilograms and a standard deviation of 1.25 kilograms. (a) What is the probability that a filled bag will weigh less than 49.5 kilograms? (b) What is the probability that a randomly sampled filled bag will weigh between 48.5 and 51 kilograms? (c) What is the minimum weight a bag of dog food could be and remain in the top 15% of all bags filled? (d) ABC is unable to adjust the mean of the filling process. However, it is able to adjust the standard deviation of the filling process. What would the standard deviation need to be so that 2% of all filled bags weigh more than 52 kilograms?arrow_forwardSuppose you sample one value from a uniform distribution with a = 0 and b=20. a. What is the probability that the value will be between 12 and 172 b. What is the probability that the value will be between 4 and 72 c. What is the mean? d. What is the standard deviation? a. The probability that the value will be between 12 and 17 is (Type an integer or a decimal)arrow_forwardChoose the correct statement with regard to the given distributions: a. Poisson: If mean is u, standard deviation will beu b. Exponential: If mean is u, standard deviation will be c. Standard Normal distribution: If mean is 0, standard deviation will be All of the above Answer A B C Darrow_forward
- SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. What is the probability that the sample mean will be less than 1230? 0.999 0.222 0.899 1 Answer Sample: n= Mean= SD= Probability[P(X<)]= z= Probability[P(Z<)]/Normal Distribution= Grapharrow_forwardPlease do not give solution in image format thankuarrow_forwardA bank has kept records of the checking balances of its customers and determined that the average daily balance of its customers is $300 with a standard deviation of $48. A random sample of 144 checking accounts is selected. What is the probability that the sample mean will be more than $306.60? 0.049 1 0.55 0.88 Answer Sample: n= Mean= SD= Probability[P(X>)]= z= Probability[P(Z)]= Grapharrow_forward
- Nancy believes that the average running time of movies is equal to 180 minutes. A sample of 9 movies was taken, and the following running times were obtained. Assume the distribution of the population is normally distributed with the mean of 175 and standard deviation of 10 minutes. Using a p-value, test the hypothesis at the 10% level of significance. p value=0.13 Therefore, DO Not reject H0. There is sufficient evidence at α = .01 to conclude that average running time of movies is 180 minutes. p value=0.17 Therefore, DO Not reject H0. There is NO sufficient evidence at α = .05 to conclude that average running time of movies is 180 minutes. p value=0.13 Therefore, reject H0. There is sufficient evidence at α = .01 to conclude that average running time of movies is 180 minutes. p value=0.17 Therefore, reject H0. There is No sufficient evidence at α = .05 to conclude that average running time of movies is 180 minutes.arrow_forwardA researcher is studying whether being a smoker causes people to get colds more often. Classify each method as a survey, an experiment, or an observational study. Then explain which method would be the most reliable. Method A: Choose people who already smoke and who do not and monitor how many colds they get in a period of years. Method B: Choose 100 people at random. Have 50 people smoke regularly for six months and 50 people not smoke for six months. Then monitor how many of the 100 people get colds within the next six months. Method C: Ask people who currently have a cold whether or not they smoke. Tell whether each method of gathering data is a survey, an observational study, or an experiment. Which option would give the most reliable results?arrow_forwardA normal population has σ = 4. A random samplehas: n = 50 and X-bar = 22. Find the 90% PI for asingle new value from the population.arrow_forward
- 6arrow_forwardModels that have no random input are called iconic models. Select one: True Falsearrow_forwardDiscuss in 300 words the different types of probability sampling designs and briefly explain how these differ from non-probability sampling approaches. Provide at least two relevant references.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Practical Management ScienceOperations ManagementISBN:9781337406659Author:WINSTON, Wayne L.Publisher:Cengage,
Practical Management Science
Operations Management
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:Cengage,
Single Exponential Smoothing & Weighted Moving Average Time Series Forecasting; Author: Matt Macarty;https://www.youtube.com/watch?v=IjETktmL4Kg;License: Standard YouTube License, CC-BY
Introduction to Forecasting - with Examples; Author: Dr. Bharatendra Rai;https://www.youtube.com/watch?v=98K7AG32qv8;License: Standard Youtube License