..................................................................................... 2 3.3 Measure of Variation................................................................................2 3.0 MEAN................................................................................................................. 3 3.1 Arithmetic Mean...................................................................................... 3 3.2 Arithmetic Mean of grouped data............................................................ 4 3.3 Properties of the mean............................................................................. 5 3.4 Geometric Mean....................................................................................... 6 3.5 Harmonic Mean........................................................................................6 4.0 MODE.........................................................................................................…… 7 4.1 Locating Repeated values........................................................................ 8 4.2 Estimating the Mode by Interpolation..................................................... 8 4.3 Locating the Mode by Graphical method............................................... 9 4.4 Estimating the Mode from the Mean and Median................................... 9 5.0 STANDARD DEVIATION….......................................................................... 10 5.1 Merits of Standard...
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...Mean, median, and mode are differing values that furnish information regarding a set of observations. The mean is used when one desires to determine the average value for data ranked in intervals. The median is used to learn the middle of graded information, and the mode is used to summarize non-numeric data. The mean is equal to the amount of all the data in a set divided by the number of values in that set. It is typically used with continuous figures. The result will probably not be one of the values in the data set, but is a representation of all those values. In other words, if I want to find the mean salary at a particular company, I would add together all the salaries and divide by the total number of salaries added: $50,000 + $56,000 + $54,500 = $53,500. The problem with mean figures is they are easily slanted by one figure that stands far above or below the others. In the previous example, if I have three annual salaries of $50,000, $56,000, and $54,500, and then the company president’s salary of $260,000, I will derive an average salary of $105,125. This mean is double the actual salaries of the lower paid workers. In this case it would be more appropriate to find the median salary. To find the median salary in the previous example, we arrange the data according to value: $50,000, $54,500, $56,000, and $260,000 and find the middle which would be $55,250. If I wanted to know the breakdown of salaries in the company, I would use mode. Using this method, I could compile...
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...and desires can make communication difficult. Use the Thomas-Kilmann questionnaire to learn what others are doing in those situations and learn to understand your own behavior during tense moments. You can master these challenges with knowledge and practice. The Five Conflict-Handling Modes The Thomas-Kilmann Conflict Mode Instrument (TKI) assesses an individual’s behavior in conflict situations—that is, situations in which the concerns of two people appear to be incompatible. In conflict situations, we can describe a person’s behavior along two basic dimensions*: (1) assertiveness, the extent to which the individual attempts to satisfy his or her own concerns, and (2) cooperativeness, the extent to which the individual attempts to satisfy the other person’s concerns. These two dimensions of behavior can be used to define five methods of dealing with conflict. These five conflict-handling modes are shown below: C O M P E T I N G Competing is assertive and uncooperative, a power-oriented mode. When competing, an individual pursues his or her own concerns at the other person’s expense, using whatever power seems appropriate to win his or her position. Competing might mean standing up for your rights, defending a position you believe is correct, or simply trying to win. C O L L A B O R A T I N G Collaborating is both assertive and cooperative. When collaborating, an individual attempts to work with the other person to find a solution that fully satisfies...
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...Running Head: A VARK Analysis A VARK Analysis of One Student’s Learning Style LorieAnn T Dailey Grand Canyon University: NRS-429v April 22, 2012 A VARK Analysis of One Student’s Learning Style This paper will discuss Fleming and Bonwell’s VARK analysis of learning styles. (2002). It will also discuss the results from this student’s use of the VARK instrument and will compare those results to this student’s own observations about her preferred learning styles or methods. Next, this paper will discuss possible changes this student could make in her learning methods that might tend to make her a more successful student in view of her results on VARK questionnaire. Finally, this paper will briefly address the analysis of this student’s learning style in the specific context of e-learning or distance learning. VARK, an acronym for Visual, Aural, Read/write, and Kinesthetic, is a system developed in its current form by Neil Fleming which uses a sixteen-question instrument to evaluate a part of student’s learning styles. (Fleming & Bonwell, 2002, FAQs (Frequently Asked Questions)). While VARK is commonly referred to as an analysis of learning styles, that is technically inaccurate. While “learning style”, as the term is generally understood, refers to several different factors such as working with others vs. working alone, physical conditions of the learning environment, and even biorhythms, VARK analyzes only one aspect of learning styles: the way a student receives...
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...1, 1, 5, 0, 2, 0, 3, 1, 1, 0, 4, 0, 0, 1, 2, 1, 1, 2, 0, 1, 0, 3, 0, 0, 1, 5, 3, 0, 0, 1, 0 In an order 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 6 Mean of data is 63/50=1.26. Median of data is 1+1/2=1. Mode of data is 0. For the normal distribution of data, there must be * mean = median = mode * symmetry about the center * 50% value less than mean and 50% value greater than mean. The given distribution not even verifying one of the criteria for the normal distribution. Therefore sample does not represent a normal distribution. 41, 52, 46, 42, 46, 36, 46, 61, 58, 44, 49, 48, 48, 52, 50, 45, 68, 45, 48, 47, 49, 57, 44, 48, 49, 45, 47, 48, 43, 45, 45, 56, 48, 54, 51, 47, 42, 53, 48, 41 In an order 36, 41, 41, 42, 42, 43, 44, 44, 45, 45, 45, 45, 45, 46, 46, 46, 47, 47, 47, 48, 48, 48, 48, 48, 48, 48, 49, 49, 49, 50, 51, 52, 52, 53, 54, 56, 57, 58, 61, 68 Mean of data is 1932/40=48.3. Median of data is 48. Mode of data is 48. For the normal distribution of data, there must be * mean = median = mode * symmetry about the center * 50% value less than mean and 50% value greater than mean. The given distribution not even verifying one of the criteria for the normal distribution. This given data is very close, but…… The sample does not represent a normal...
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...a normal distribution as it is depicted graphically? A. Square B. Bell C. Triangle D. Star E. Hat For questions 2 through 4, consider the following array of numbers: 5 6 7 7 7 8 8 9 9 9 10 15 19 20 21 2. In the array provided, what is the mode? A. 7 B. 9 C. 10 D. 15 E. Both A and B 3. In the array provided, what is the median? A. 7 B. 9 C. 10 D. 15 E. Both A and B 4. In the array provided, what is the mean? A. 7 B. 9 C. 10 D. 15 E. Both A and B 5. The difference between the smallest and the largest values in a distribution is the _____. A. mean B. median C. mode D. range E. deviation 6. Which of the following is a bar chart arranged in increasing order by size? A. Control chart B. Simple bar chart C. Pareto diagram D. Histogram 7. Which of the following hypotheses is a null hypothesis? A. There is no difference in the monthly grocery bills of families with one child and families with two children B. Grocery bills vary according to the number of meals eaten outside the home C. Families with two children have significantly higher grocery bills than families with just one child D. There is a relationship between grocery bills and the number of people in a household E. The mean age in a household is predictive of the amount of money sent on food each month 8. _____ occurs when the sample results lead to the rejection of a null hypothesis that is in fact true. A. Type I error B. Two-tailed...
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...of Submission: 27th February (10 marks will be deducted if anyone cross this deadline) 1. The following data give the weekly amounts spent on groceries for a sample of households. 271 |363 |159 |76 |227 |337 |295 |319 |250 | |279 |205 |279 |266 |199 |177 |162 |232 |303 | |192 |181 |321 |309 |246 |278 |50 |41 |335 | |116 |100 |151 |240 |474 |297 |170 |188 |320 | |429 |294 |570 |342 |279 |235 |434 |123 |325 | | a) How many class do you Suggest? b) What class interval do you suggest? c) Organize data in to a frequency distribution. d) Calculate relative frequency, cumulative frequency e) Draw frequency histogram. f) Draw frequency polygon. g) Calculate mean, median and mode for random data h) Calculate mean and Median for grouped data according to the frequency distribution of question no (c) 2. A student is taking two courses, history and math. The probability that the student will pass the history course is 0.60 and the probability of passing math course is 0.70. The probability of passing both is 0.50. What is the probability of passing at least one? 3. The Law firm of Hagel and Hagel is located in downtown Cincinnati. There are ten partners in the firm; seven live in Ohio and three in northern Kentucky. Ms. Wendy Hagel, the managing partner; wants to appoint a committee of three partners to look into moving the firm to northern Kentucky. If the committee is selected at random from the ten partners...
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...ANALYSIS OF A LESSON WITH AN INTERN PERSPECTIVE Episode: On Tuesday, When I enter the 7B class in a Public School with the teacher, the students welcomed me with a smile face and they raced with each other to share their seats with me. I also smiled them and went back to seat near a girl sitting alone. The lesson was about central tendency and spread measurement units. First of all, the teacher asked the class some questions about central tendency measurement such as what is the meaning of mean, mode and median. She tried to catch up the students who were not interested in the lesson, have misconceptions and seem not to understand basic part of the topic. She controlled them by using question answer method. She wrote a question from the student work book about bar graph interpretation. After drawing the graph, she asked them how many people who gain the least salary there are by reading the graph. Then, she asked them to calculate mean of the total salaries which number of people given. One or two students listed each salary as only one person gains it. However, on the graph there were two people who gains 500 and five people who gains 1500, for example. The teacher could not aware of the students’ faults. She said “yes right”. After a while, she asked the class there was anybody who was aware of the fault and asked them to explain why they had fault. But nobody was aware of this fact. One student said that “May be our prospective teacher can say our fault”. Thankfully...
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...on a sample of 50 credit customers with AJ DAVIS is selected with data collected base on five variables as following: • Location (Rural, Urban, Suburban) • Income (in $1,000's) • Size (Household Size, meaning number of people living in the household) • Years (the number of years that the customer has lived in the current location) • Credit Balance (the customers current credit card balance on the store's credit card, in $). To have more understand of what the data truly mean, three of individual variable and three of pairing variable are analyzed by using numerical techniques of summarization as and graphical such as stem-leaf diagram, histogram, boxplot, and bar chart on this report. Since a bar graph is useful for comparing facts and help us to see relationships quickly, the graphs for all three of the pairing variable are applied to bar chart graph. 2.0 Individual Variable: Income Descriptive Statistics: INCOME($1000) Variable Mean StDev Variance Minimum Q1 Median Q3 Maximum INCOME($1000) 43.48 14.55 211.72 21.00 30.00 42.00 55.00 67.00...
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...Stress James Farrar From the moment of birth to the end of life one way or another we all have to deal with stress. Each of us cope with stress in our own way because there are some many different types of stress from self- imposed to environmental, there seems to be no end to it. What determines the extant of the damage done by the effects of whatever stress that is imposed is the ability to cope with, process and or deal with whichever type of stress one has to handle. It is my belief that most stress is self- imposed by that I mean a lot of the problems we face every day are brought on by poor decisions which can lead to undue amounts of stress. Through long periods of isolation and dehumanization I have personally had to deal with some stresses that would have rendered the average person delirious. It is also my belief as previously stated that each and every one of us copes with stress in our own way, what may drive one person crazy may not bother another. There are some people I know that crumble when the slightest thing goes wrong and then there are others who step up when put under pressure. The above really only covers the obvious, there are instances such as some children born with debilitating diseases or say someone who was born blind, they would still have to deal with the disability but learning to cope with it or maybe just absence of stimuli makes it easier to handle, it may also...
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...Question 1 a) EMPLOYMENT CATEGORY Frequency Percent Valid Percent Cumulative Percent Valid CLERICAL 227 47.9 47.9 47.9 OFFICE TRAINEE 136 28.7 28.7 76.6 SECURITY OFFICER 27 5.7 5.7 82.3 COLLEGE TRAINEE 41 8.6 8.6 90.9 EXEMPT EMPLOYEE 32 6.8 6.8 97.7 MBA TRAINEE 5 1.1 1.1 98.7 TECHNICAL 6 1.3 1.3 100.0 Total 474 100.0 100.0 In the left most column, all job category are listed. The Frequency column records the number of observations that fall within a particular job category. As I have sorted our using SPSS, Clerical, Office Trainee, Security Officer, College Trainee, Exempt Employee, MBA Trainee, and Technical job category represent 227, 136, 27, 41, 32, 5, and 6 respectively out of the total employees of 474. The percent column shows how much each job category possesses of total employees. As you can see, as Clerical, Office Trainee, Security Officer, College Trainee, Exempt Employee, MBA Trainee, and Technical employment category represent 47.9%, 28.7%, 5.7%, 8.6%, 6.8%, 1.1% and 1.1% respectively. The Cumulative frequency column lists the total of each frequency added to its predecessor, such as, Clerical job category itself represents 47.9% of all employees and then Office Trainee job category represents 28.7%, but both Clerical and Office Trainee represents 76.6% of the total population. If we consider, Clerical, Office Trainee, and Security Officer; all of them represent 82.3% of the total employees. This is what cumulative percent...
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...previously undertook the test can be plotted in a graph that will resemble a “normal” curve. (DISCUSS PECENTILES AND STANDARDS SCORES ETC). The various forms of averages and how they are calculated Mean The mean is calculated by adding up all the numbers in a dataset and then dividing that by the number of values in that dataset. An example is given below of a dataset of six students and their test scores out of 20: S1 S2 S3 S4 S5 S6 Total Divided by: Mean 18 16 17 13 14 19 97 6 16.17 In order to find the mean test score, the total amount scored (97) is divided by the number of students (6) which equals 16.17. The formula for calculation the mean is as follows: _ X is the symbol for the mean and is referred to as bar X (ex) Σ is the Greek symbol sigma and simply means sum or add up X refers to each of the individual values that make up the dataset n is the number of values that make up the dataset The use of mean is effective when the values of the dataset are evenly spread with no extreme high or low values. If the dataset does contains one or two extremely high or low values the result will be adversely affected by this. An example of this is given below, where S1 and S4 are much lower than the other variables. S1 S2 S3 S4 S5 S6 Total Divided by: Mean 2 16 17 3 14 19 71 6 11.83 If a dataset does containing extremely low or extremely high values, then the median is a better way of measuring average value. Median This refers to the middle...
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...| Table [ 1 ]. Ages of 10 randomly selected students in a Judo school in Brisbane. (a) Mean (x) : 19.5 Standard Deviation (S) : 7.24568 (b) Median: 16+182=17 When comparing the mean and the median it tells the distribution is skewed to the right, which is positively skewed. The skewness of the dataset is positive due to bigger mean value than the median, as mean pulls the dataset to the right. (c) Coefficient of Variation: CV= Sx= 7.2519.5=0.37157 (d) Using Excel: * First Quartile (Q1) = 13.5 [=QUARTILE.EXC(array,quart)] * Third Quartile (Q3) = 25.75 [=QUARTILE.EXC(array,quart)] * 80th Percentile = 27.4 [=PERCENTILE.EXC(array,k)] Working out by common sense: * First Quartile (Q1) = 12+142=13 * Third Quartile (Q3) = 25+282=26.5 * 80th Percentile = Lp=n+1×P100=10+1×80100=8.8 ≈9th value * Therefore, the approximate of 80th percentile is 28 (9th value in data set). (e) Range: highest value – lowest value = 32 – 11 = 21 IQR: Q3-Q1=12.25 (Used excel values) (f) Data Set | | | | | | | | | 11 | 12 | 14 | 15 | 16 | 18 | 24 | 25 | 28 | 82 | Min | Quartile 1 | | Median | | Quartile 3 | Max | Table [ 2 ]. Ages of 10 randomly selected students in a Judo school in Brisbane (Corrected data set) Mean: 11+12+14+15+16+18+24+25+28+8210=24.5 Median: 16+182 =17 Range: 82 – 11 = 71 IQR: 12.25 ...
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...AS PER APPENDIX ‘B’ TO DPM 2009 STATEMENT OF CASE FOR DEVELOPMENT OF MECHATRONIC CONTROL SYSTEM AT AFTC, JALAHALLI (W), BANGALORE |1. Unit |Air Force Technical College, Jalahalli West | |2. Name of the item being produced |Development of mechatronic control based security gate at AFTC | |3. Justification of the procurement |The Aim of the project is to make a Mechatronic Control system which would| | |enable automatic operations of Main Security gate barrier and act against | | |any security threat effectively. As in examples in some Asian countries, | | |there is possibility of a terror attack in India when some terrorists | | |might try to infiltrate inside a high security bases in a speeding | | |vehicle. The project demonstrates some concepts which help in quick | | |automated reaction of the system which will prevent such types of attacks.| | | ...
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...I. NORMALITY IN THE TESTPAPER II. EQUAL VARIANCES | Ho: | The two data have equal variances. | | | | | | | | | Ha: | The two data did not have equal variances. | | | | | | | | | | | | | | | | | | | | | Since the p value (0.006035) is lesser than alpha (0.05), then Ho is rejected. | | | | | | | | | | | | | | | | | | | Therefore, the two data did not have two equal variances. Meaning the two data comes from the different population. | Ho: | The two data have equal variances. | | | | | | Ha: | The two data did not have equal variances. | | | | | | | | | | | | | | | Since the p value (0.428217) is greater than alpha (0.05), then Ho is accepted. | | | | | | | | | | | | | Therefore, the two data have equal variances. Meaning the two data comes from the same population. | III. SAMPLE WITH NORMAL DISTRIBUTION 1-2 IN THE NOTEBOOK IV. 1. Ho: The data is equal to the population average. Ha: The data is not equal to the population average. Since the p value of normality test (0.010) is lesser than alpha (0.05), then data did not follows a normal distribution and use 1- sample wilcoxon. Since the p value of 1 sample wilcoxon (0.183) is greater than alpha (0.05) then Ho is accepted.Therefore, the data is equal to the population average. 2. Ho: the data is equal to the population average Ha: the data is not equal than the population average. Since the p value of normality...
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