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# statistics week 2 of notes Math-K310

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This 9 page Class Notes was uploaded by Katelyn Scott on Thursday October 1, 2015. The Class Notes belongs to Math-K310 at Indiana University taught by Linda Krause in Summer 2015. Since its upload, it has received 49 views. For similar materials see Statistical Techniques in Mathematics (M) at Indiana University.

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Date Created: 10/01/15

Chapter 2 Characteristics of Data 2.1 1. Center- measures of center; a value that indicates where the middle of the data is 2. Variation- a measure of the amount that the data values vary 3. Distribution- the shape of the spread of the data over the range of values 4. Outliers- a data value that lies far away from the majority of other values EXAMPLE- Constructa groupedfrequencydistributionand cumulativefrequencydistributionwith5 classes. Test Data: 85 79 82 73 79 96 62 88 99 78 84 91 *Find the range: 99-62=37 *Find the class width: 37÷5= 7.4 (remember to round up to the nearest whole number for class widths) - class width = 8 Begin with the smallest number (62) Class Class boundaries tally frequency Cumulativefreq. < Class 62-69 61.5-69.5 l 1 1 69.5 70-77 69.5-77.5 l 1 2 77.5 78-85 77.5-85.5 l/ll l 6 8 85.5 86-93 85.5-93.5 ll 2 10 93.5 94-101 93.5-101.5 ll 2 12 101.5 • To find the class width: o Find the range = greatest value – least value o Divide by the number of classes desired o Always round up for class width, even if it is a whole number • Rule of thumb rounding rule o Round to 1 more decimal place than the given data 2.4 • Examples of bad graphs • Graph B appears double B A 90 90 A B 0 o Starting at 0 shows that this is not the case B A o Area of A = 1x1=1 o Area of B= 2x2=4 When doubling only double one dimension • Normal distribution o Approximately symmetric o Frequency starts and ends at low with highest point in the middle o Gaps in a distribution (frequency=0) may indicate data are from more than one population • A histogram is a bar graph. Each and every class is labeled across the bottom. The frequency is represented by vertical height. Vertical heights can be related to frequency • EXAMPLE: Using data from EX 1 construct a histogram using class boundaries and frequencies and then using class boundaries and relative frequencies. 7 7/12 6 6/12 5 5/12 4/12 4 3 3/12 2 2/12 1 1/12 0 0 61.5 69.5 77.5 88.5 93.5 101.5 61.5 69.5 77.5 88.5 93.5 101.5 o Class boundaries are in the x axis and the frequencies are in the y axis respectively • Types of graphs to learn o Histograms o Frequency polygon- x=midpoint, y= frequency/relative frequency o Ogive- x= class boundaries, y= cumulative frequency/ cumulative relative frequency o Dot plots- stack of x values o Stem leaf plot- tens l units--- like 70 = 7 l 0 • EXAMPLE: Graph a frequency polygon and Ogive using data from Ex 1. 7 Frequency Polygon Ogive 6 14 5 12 4 10 8 3 6 Frequency 2 4 Cumulative Frequency 1 0 0 61.5 69.5 77.5 85.5 93.5 101.5 65.5 73.5 81.5 89.5 97.5 Class Boundaries Midpoint • EXAMPLE:A test was administered to a group of thirty-six students.Their scores were as follows: 40 26 38 18 42 17 25 43 46 46 19 26 35 34 15 44 40 35 63 25 35 3 33 29 34 41 49 52 35 48 22 32 43 51 27 14 Construct a stemplot and a dot plot. 0 3 456789 1 2556679 2 Stem Plot 3 234455558 4 00123346689 5 12 6 3 0 10 20 30 40 50 60 • Do not use 2 or 3 dimensionaldata to represent 1 dimensional data • Do not start a graph at a nonzero number • Do not graph flawed data 3.2 • EXAMPLE: find the mean, median, midrange and mode of the data set. 17 12 13 11 14 18 17 18 16 15 Order the data points- 11 12 13 14 15 16 17 17 18 18 Mean= ∑data÷ 10 data values = 15.1 Median= (15+16)÷ 2 = 15.5 Mode=17, 18 Midrange= (11+18)÷ 2 = 14.5 • Measuring of Center o Mode- most often used number, there can be no mode or many modes o Mean- total of all values added together divided by the number of values Mean: x̄= (1 + 2 + …nx ) ÷ n o Median- the middle most value If the number of values is even, add the middle two numbers and divide then by 2 o Midrange- the average of the highest and the lowest values (high+low)÷2 = • Do this problem on your own before continuing. An answer key will be providedon the next page EXAMPLE: Find the mean of the data in the frequency distribution. 1 by hand , 2 by calculator. For 108 randomly selected college students, this exam score frequency distribution was obtained. For calculator TI 84you’ll need to go to stat > 1.Edit > insert midpoints into L1 > insert frequencies into L2 > stat > calc > 1 1-var stats > list: 2L1 > freqlist: 2 L2 > calculate > mean = x ̄ If this were not a frequency distribution you would leave freqlist blank Class limits Frequency midpoint midpointfrequency 90 – 98 6 99 – 107 22 108 – 116 43 117 – 125 28 126 – 134 9 • Answer Find the mean of the data in the frequency distribution. 1 by hand, 2 by calculator For 108 randomly selected college students, this exam score frequency distribution was obtained. Class limits Frequency midpoint midpoint x frequency 90 – 98 6 94 864 99 – 107 22 103 2266 108 – 116 43 112 4816 117 – 125 28 121 3388 126 – 134 9 130 1170 108 12204 Mean = 12204/108= 113.0 • Given sample data the mean is represented by x̄ o x̄= ∑x ÷ n • In a population it is represented by µ o µ = ∑x ÷ n • to find the median (n+1)÷ 2 is the location of it

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