Qrb 501 Assignment

Qrb 501 Assignment Words: 1751

Individual Assignment QRB/501 University of Phoenix – MBA Activity 18:1: Question 1. Toss a coin 10 times and record the proportion of heads so far. Answer. Table for Head Proportion Toss #12345678910 H or T? HTTHHHTTHT Proportion1/11/21/32/43/53/63/73/84/95/9 Question 2. Plot the proportion of heads so far, for each toss from the previous table. What does the graph show? Answer. Shown below is the graph plotted in Excel worksheet showing the first toss of H and the proportion of toss 1 at 1. 0 and the subsequent one as it increases getting to 0. 5 Question 3 Answer.

Therefore INT(2 * 0. 13061) ? INT(0. 26122) ? 0 Also INT(2 * 0. 78934) ? INT(1. 57868) ? 1 Answer to Question 4. Graph from Excel From the graph – it can be seen that as the tosses number increase, so also the head proportion moving toward 0. 5 In our simulation, the first toss has an H, which we can see because the proportion starts at 1. 0. Following the first simulated toss, the proportion changes with each new toss depending on whether we got an H or a T. Answer 5 Each time that a new set of random numbers is generated, the graph changes accordingly.

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However, in all of these graphs we can still see that the proportion of heads so far approaches the 0. 5 value as the number of tosses increases. The following is one of the graphs obtained this way: Answer 6: The following is a graph obtained with the simulation. As expected, the graph shows that the proportion of hits approaches 0. 5 as the number of free throws increases. In the simulation, the first throw was a hit, which gave us a first proportion of 1. 0. However, after that, we had a streak of misses that lower the proportion all the way to 0. 40. By the time we got to throw 100, our proportion (the hits mean) was 0. 7. Our longest streak of hits happened from throws #7 to #15, inclusive, for a streak length of 9. Our longest streak of misses happened from throws #72 to #77, inclusive, for a streak length of 6. Answer 7: The graph obtained from the simulation appears next. As the number of throws increases, the proportion of hits so far approaches 0. 75, as expected. The overall proportion of hits by throw 100 was 0. 73. The longest streak of hits was from throws #87 to #98, inclusive, for a length of 12 consecutive hits. The longest streak of misses was from throws #55 to #60, inclusive, for a length of 6 consecutive misses.

When compared against the previous simulation, the longest streak of hits was substantially longer; in the previous simulation we had Shaquille with a streak of 9 hits, while here we have a streak of 12 hits. For misses, in both cases the longest streak was of 6 misses. We can conclude that we’ll see longer streaks of hits as the probability of hits increases. Chapter 2 The following frequency distribution reports the number of frequent flier miles, reported in thousands, for employees of Brumley Statistical Consulting, Inc. , during the most recent quarter. Frequent Flier Miles (000)Number of Employees up to 35 3 up to 612 6 up to 923 9 up to 128 12 up to 152 Total50 a. How many employees were studied? Answer. A total of 50 employees were studied. c. Construct a histogram. Answer. The following graph is the histogram for the data given. Chapter 3 Exercise 85 Answer A. The life expectancy data is shown next, sorted in ascending order: CountryLife Expectancy South Africa48. 09 Nigeria51. 07 India62. 68 Brazil63. 24 Iraq66. 95 Russia67. 34 Saudi Arabia68. 09 Indonesia68. 27 Algeria69. 95 Iran69. 95 Turkey71. 24 China71. 62 Hungary71. 63 Mexico71. 76 Qatar72. 62 Venezuela73. 31

Poland73. 42 United Arab Emirates74. 29 South Korea74. 65 Czech Republic74. 73 Argentina75. 26 Libya75. 65 Portugal75. 94? Kuwait76. 27 ? Denmark76. 72 Ireland76. 99 United States77. 26 Luxembourg77. 3 Finland77. 58 Germany77. 61 United Kingdom77. 82 Austria77. 84 Belgium77. 96 New Zealand77. 99 Netherlands78. 43 Greece78. 59 Norway78. 79 France78. 9 Spain78. 93 Italy79. 14 Iceland79. 52 Canada79. 56 Sweden79. 71 Switzerland79. 73 Australia79. 87 Japan80. 8 I used the descriptive statistics function on the data and obtained the following output: Statistical Outcomes Table Mean73. 806

Standard Error1. 018 Median76. 105 Mode69. 950 Standard Deviation6. 905 Sample Variance47. 675 Kurtosis5. 202 Skewness-2. 100 Range32. 710 Minimum48. 090 Maximum80. 800 Sum3395. 060 Count46. 000 Largest(1)80. 800 Smallest(1)48. 090 Confidence Level (95. 0%)2. 050 From here, we have that: Mean: 73. 81 Median: = 76. 10 Standard Deviation (seen the 46 countries as a sample): 6. 90 Distribution Summary: The median is higher than the mean, so we can conclude that this distribution will be negatively skewed. This is also indicated by the skewness of -2. 1. Also the range is 80. 8 – 48. 09 ? 32. 1 years, almost 67% of the lowest life expectancy; Therefore the range is then very large. However, in reality most countries have a life expectancy that is at least 62 years. We can consider countries like South Africa (48 years) and Nigeria (51 years) to be outliers. If those two outliers were eliminated from the data, the mean would increase to 74. 90 and, most notably, the standard deviation would decrease to 4. 56. We could speculate that these outliers could be the result of unusual health or economic conditions in these countries, such as the effect of AIDS or internal strife. . Select the variable GDP/cap. Find the mean, median, and standard deviation. b. Write a brief summary of the distribution of GDP/cap. Answer. The per-capita GDP data is shown next, sorted in ascending order: CountryGDP/cap Nigeria0. 95 India2. 2 Iraq2. 5 Indonesia2. 9 China3. 6 Algeria5. 5 Venezuela6. 2 Iran6. 3 Brazil6. 5 Turkey6. 8 Russia7. 7 Poland8. 5 South Africa8. 5 Libya8. 9 Mexico9. 1 Saudi Arabia10. 5 Hungary11. 2 Argentina12. 9 Czech Republic12. 9 Kuwait15 Portugal15. 8 South Korea16. 1 Greece17. 2c New Zealand17. 7c Spain18 Qatar20. 3 Ireland21. 6 Italy22. 1 Sweden22. 2

United Arab Emirates22. 8 United Kingdom22. 8 Finland22. 9 Australia23. 2 Germany23. 4 France24. 4 Netherlands24. 4 Canada24. 8 Iceland24. 8 Japan24. 9 Austria25 Belgium25. 3 Denmark25. 5 Norway27. 7 Switzerland28. 6 United States36. 2 Luxembourg36. 4 I used the descriptive statistics function on the data and obtained the following output: Statistical Outcomes Table Mean16. 5815 Standard Error1. 3674 Median17. 4500 Mode12. 9000 Standard Deviation9. 2743 Sample Variance86. 0126 Kurtosis-0. 8939 Skewness0. 05676 Range35. 4500 Minimum0. 9500 Maximum36. 400 Sum762. 7500 Count46 Largest(1)36. 4000

Smallest(1)0. 9500 Confidence Level (95. 0%)2. 7541 From here, we have that: Mean ? 16. 58 Median ? 17. 45 Standard Deviation ? 9. 17 Distribution Description. This distribution is very symmetric; notice that the skewness is very small, at 0. 056. Also, the standard deviation is fairly large with respect to the mean and the range; hence, the data in this distribution is widely spread. It’s interesting to see that we have a very large range of 35. 45, which is more than 90% the highest value. The highest per-capita GDP corresponds to Luxemburg, at 36. 4, followed by the United States with 36. . The next per-capita GDP, Switzerland, is at 28. 6. So, we might be inclined to consider the values for Luxemburg and the United States as outliers, since there is a substantial gap between them and the countries with the next lower per-capita GDPs. The lowest per-capita GDP is for Nigeria, with only 0. 95. However, we don’t really have outliers to the lower end of per-capita GDP. The per-capita GDP becomes progressively smaller and even Nigeria, with the lowest per-capita GDP, can’t be consider an outlier. Activity 17. 1 Answer a and b TossesNumber of HeadsTossesNumber of Heads 0136 15143 27153 33168 45179 56182 62191 76204 86214 97228 108233 112243 123255 Answer c Number of HeadsFrequency 00 11 21 34 46 510 66 75 82 90 100 Total35 Combined 60 data points table: Number of HeadsFrequency 01 11 24 310 48 513 610 77 85 91 100 Total60 Answer D: This is the scatterplot: The data appears to be centered around 5 heads; the peak frequency happens at 5 heads. The curve extends or “spreads” from 0 to 9. We could describe the shape as somewhat similar to a bell-shape. Answer F. This is the bar-graph: This is the bar-graph with a bell-curve approximated to the data.

The curve fits closely the description given in part d): it has a peak and is centered at 5, and it extends from 0 to 9, approximately. Chapter 17 – Exercise 20 Outside Diameter (millimeters) Time12345 8:0087. 187. 387. 987. 087. 0 8:3086. 988. 587. 687. 587. 4 9:0087. 588. 486. 987. 688. 2 9:3086. 088. 087. 287. 687. 1 10:0087. 187. 187. 187. 187. 1 10:3088. 086. 287. 487. 387. 8 a. Control Limits for the Means and Range. To calculate control limits ?Get the mean and Range of each time slot Time12345MeanRange 8:0087. 187. 387. 987. 087. 087. 30. 9 8:3086. 988. 587. 687. 587. 487. 61. 6 :0087. 588. 486. 987. 688. 287. 71. 5 9:3086. 088. 087. 287. 687. 187. 22. 0 10:0087. 187. 187. 187. 187. 187. 10. 0 10:3088. 086. 287. 487. 387. 887. 31. 8 Total524. 187. 8 Mean of means: ? 524. 18 / 6 ? 87. 36 Mean of ranges: ? 7. 8 / 6 ? 1. 3 Therefore the control limits for the means and the range. size n=5, from Appendix B.. 8 we have A2 = 0. 577, D3 = 0, and D4 = 2. 115. Control Limits for the means are: Upper Control Limit (UCL) = + A2 ? 87. 36 + 0. 577 * 1. 3 ? 88. 11 Lower Control Limit (LCL) = – A2 ? 87. 36 – 0. 577 * 1. 3 ? 86. 61 Cntrol limits for the range are:

Upper Control Limit (UCL) = D4 ? 2. 115 * 1. 3 ? 2. 75 Lower Control Limit (LCL) = D3 ? 0 * 1. 3 ? 0 b. Graph with control limits for the mean: Graph with control limits for the range: Topic 17 Exploration 2 Question: The following tables provide information about the top women-owned businesses in the U. S. The first table gives year 2000 revenue in millions of dollars and number of employees for Pennsylvania women-owned companies with year 2000 revenues of $70 million or higher. The second table gives the same information for Michigan women-owned companies with year 2000 revenues of $70 million or higher.

Pennsylvania Companies2000 Revenue ($ million)Number of Employees 84 Lumber20004400 Charming Shoppes160012000 Rodale5001300 Mothers Work366. 33800 Harmelin Media200105 McGettigan Partners175400 Wetherill Associates135500 Hanna Holdings73. 71284 Michigan Companies2000 Revenue ($ millions)Number of Employees Ilitch Holdings800700 Plastech Engineered Products4203500 Jerome-Duncan Ford350300 Elder Ford287. 6108 Patsy Lou Williamson Auto Gp. 221240 Mexican Industries1741463 Rush Trucking1532000 Jaguar-Saab of Troy143134 Manpower Metro Detroit118250 Continental Plastics107650

Leco82800 Rodgers Chevrolet GEO7580 Strategic Staffing Solutions75600 Two Men & a Truck Internat’l7540 Systrand Manufacturing72230 a. Use a calculator or computer to compute the mean and standard deviation of the year 2000 revenue for the Pennsylvania companies in the first table. Answer: Using Excel (calculation in the attached excel sheet), the mean is $631. 25 million, and the standard deviation is $741. 68 million. b. Use a calculator or computer to compute the mean and standard deviation of the year 2000 revenue for the Michigan companies in the second table.

Answer: Using Excel (calculation in the attached excel sheet), the mean is $210. 17 million, and the standard deviation is $195. 70 million. c. Explain what the values you calculated in parts a and b tell you about the data sets. Answer: There is higher revenues from the Pennsylvania businesses from the mean which is more than than the revenue of Michigan. d. How would the mean and standard deviations change if the largest data value in each set were removed? Answer: If we remove the largest data value in each set, we get: Pennsylvania Businesses Mean: $435. 71 million Pennsylvania Businesses Standard Deviation: $533. 8 Michigan Businesses Mean: $168. 04 Michigan Businesses Standard Deviation: $112. 11 e. Find the mean and the standard deviation of the number of employees for the Pennsylvania companies in the first table. Answer: Using Excel (calculation in the attached excel sheet), the mean is 2973. 63 employees, and the standard deviation is 3978. 87 employees. f. Find the mean and the standard deviation of the number of employees for the Michigan companies in the first table. Compare against the previous results. Answer: Using Excel (calculation in the attached excel sheet), the mean is 739. employees, and the standard deviation is 941. 0 employees. From the 2 result, it can be deducted that Pennsylvania companies have, on average, almost 4 times as many employees as Michigan companies. The mean number of employees in the Pennsylvania companies was 2974 compared to only 740 for Michigan companies. Also, there is a much higher dispersion about the mean in the number of employees for Pennsylvania companies than Michigan companies. As we can see, the standard deviation for Pennsylvania companies is almost 4000, but for Michigan companies is only 941.

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