Mean‚ Mode and Median Ungrouped and Grouped Data Ungrouped Data refers to raw data that has been ‘processed’; so as to determine frequencies. The data‚ along with the frequencies‚ are presented individually. Grouped Data refers to values that have been analysed and arranged into groups called ‘class’. The classes are based on intervals – the range of values – being used. It is from these classes‚ are upper and lower class boundaries found. Mean Mean The ‘Mean’ is the total of all the values
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Disadvantages: Outliers can change the mean a lot... making it much lower/higher than it should be. Affected by extreme values (outliers) Median: Advantages: Finds the middle number of a set of data‚ so outliers have little or no effect. Disadvantages: If the gap between some numbers is large‚ while it is small between other numbers in the data‚ this can cause the median to be a very inaccurate way to find the middle of a set of values. Mode: Advantages: Allows you to see what value happened the
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Project in Math Statistical Data Concerning the Height of BSDAS Academy Department Students S.Y. 2013-2014 Made by: Kent Irvin Pastrana Miguel Pump Adawan Christian Yango Presented to Wayne B. Valera on the day of March 24‚ 2014 I. Abstract Many times‚ people would ask what another person’s height is‚ and joke or say wow depending on the answer. Height is important in our lives; it allows us to reach higher (yes‚ literally)‚ see farther (yes‚ literally again) and
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86% percentage. 7. Could a median be determined for the education data? If so‚ what would the median be for education for the experimental and the control groups? Provide a rationale for your answer. Answer: Yes a median can be determined for the educational data. The median for the experimental group would be some college‚ and the median for the control group would be college graduate or higher. The experimental group median value was 11 while the control group median value was 13. 8. Can the findings
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range = 20 mode = 73 variance = 324 median = 74 The coefficient of variation equals a. 0.30% b. 30% c. 5.4% d. 54% Answer: b 6. The variance of a sample of 169 observations equals 576. The standard deviation of the sample equals a. 13 b. 24 c. 576 d. 28‚461 Answer: b 7. The median of a sample will always equal the a. mode b. mean c. 50th percentile d. all of the above answers are correct Answer: c 8. The median is a measure of a. relative dispersion
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“2-7 Algebra Lab Percentiles” 1. Find the median‚ lower quartile‚ and upper quartile of the scores. Median: 16.5 Lower quartile: 10.5 Upper quartile: 23.5 2. Which performer was at the 50th percentile? the 25th percentile? the 75th percentile? 50th percentile: Arnold 25th percentile: Ingrid 75th percentile: Brooke 3. Compare and contrast the values for the median‚ lower quartile‚ upper quartile‚ and the scores for the 25th‚ 50th‚ and 75th percentiles. If you estimate the percentiles‚ then the
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e) Determine a 90% confidence interval for the true mean of the Business Account holders and interpret the result in the context of the situation. Descriptive Statistics: Quality Variable Use N N* Mean SE Mean StDev Minimum Q1 Median Q3 Quality 1 45 0 5.971 0.199 1.336 3.700 4.800 6.000 7.050 2 30 0 8.323 0.172 0.941 6.200 7.675 8.400 9.025 Variable Use Maximum Quality 1 8.400 2 10.000
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AJ DAVIS AJ DAVIS MATH 533 Project Consumer Tel – 123-456-7891 July 21st 2013 MATH 533 Project Consumer Tel – 123-456-7891 July 21st 2013 Lakshan Nanayakkara AJ DAVIS is a department store chain‚ which has many credit customers. A sample of 50 credit customers is selected with data collected on location‚ income‚ credit balance‚ number of people and years lived in the house Lakshan Nanayakkara AJ DAVIS is a department store chain‚ which has many credit customers. A sample of 50 credit
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summary indicates this. If not‚ determine if the data is skewed left or skewed right‚ and explain how the summary indicates this. Ans: The histogram would look skewed to the right. The median is much closer to the 1st quartile than it is to the 3rd quartile‚ and the maximum value is much farther away from the median than the minimum value. (b) Would the boxplot for the data indicate any outliers? Explain why or why not. Ans: IQR=9100-8850=250. 1.5 × IQR = 375. The inner fences will be 8850-375=8475
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NORTHCENTRAL UNIVERSITY ASSIGNMENT COVER SHEET Student: Marjoree Corless BTM8104-8 Hamid Jafari‚ PhD Statistics 1 Week 2 Assignment: Data Visually and through Numbers <Add student comments here> Faculty Use Only <Faculty comments here> <Faculty Name> <Grade Earned> <Date Graded> Chapter 3 and 4 Data file 2/ Data Visually and through Numbers Marjoree Corless Northcentral University Week 2 Assignment: Apply Graphs and Descriptive Stati
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