# CURRICULUM INSPIRATIONS: INNOVATIVE CURRICULUM ONLINE EXPERIENCES: TANTON TIDBITS:

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1 CURRICULUM INSPIRATIONS: wwwmaaorg/ci MATH FOR AMERICA_DC: wwwmathforamericaorg/dc INNOVATIVE CURRICULUM ONLINE EXPERIENCES: wwwgdaymathcom TANTON TIDBITS: wwwjamestatocom TANTON S TAKE ON MEAN ad VARIATION JULY 01 I our early grades we lear that the average of a collectio of data measuremets represets, i some way, a typical or middle value for the data For example, the average of the umbers 1,, 5, is: = 5 Geometrically, the average is the level of a sad-box after we smooth out colums of sad of heights give by the data: I a statistics class the average value of a collectio of data values is called the mea of the data (The word still meas average ) Oe deotes the mea of the data by puttig a bar over whichever letter is beig uses to deote the data For example, the mea of a1, a, ad a is: a + a + a a= 1 ad the mea of x1, x, K, x is: x= x + x + L+ x 1 If the data set is extraordiarily large ad oe does t have ay hope of determiig the mea of the full data set, the that true, but ukow, mea is usually deoted with the Greek letter µ For example, we have o hope of kowig the average height of wwwjamestatocom ad wwwgdaymathcom

3 The media of the data set,, 5, 6, 7, 16, 16, 19, 7 is 7 The media of,,, 5, 8, 8, 10, is = 6 5 The media is a value that divides the data set ito two equally sized groups The midrage of a data set is the average of the smallest ad largest values The midrage of the data set 5, 6, 9, 9 is = 7 The midrage provides a quick estimate to a cetral value It is easy to compute, but is highly affected by extremely low or high values i the data set Exercise: a) Fid FIVE data values with: Media = 10 Mode = 10 Mea = 1000 b) Now fid five data values with media = 10, mode = 1000 ad mea = 10 c) Ca you fid five data values with media = 1000, mode = 10, mea = 10? COOL Exercise: Repeat the previous exercise but this time for SIX data values DEVIATION FROM THE MEAN The data set 1,,5, has mea 5 So too does the data set: 01, 0, 1, 110 These are two very differet data sets, with the secod beig much more spread out tha the first We ca measure the degree of spread by calculatig the average deviatio from the mea for each DATA SET 1,,5, : Deviatios: 1 5 = 15 5 = = 5 5 = 05 Average deviatio: = 15 DATA SET 01, 0, 1, 110: Deviatios: = = = = 1075 Average deviatio: = 865 The umbers 15 ad 865, the average deviatios from the mea, do give a quatitative measure of the amout of spread of each data set THE POINT OF THIS ESSAY Usig the absolute value, the distace of a particular data value from the mea value of the data, is the atural ad appropriate way to measure data variatio But statisticias DON T use absolute values i their work! This is very strage ad cofusig for studets (There is also a secod piece of cofusio, which we shall leave to later i this essay) Here are two ratioales for the switch away from absolute values: wwwjamestatocom ad wwwgdaymathcom

4 RATIONALE ONE: Workig with absolute values is hard Ca we avoid them? Ideed, workig with absolute values i mathematical equatios is really tough! Optioal Exercises: a) Sketch the curve x + y = b) Fid all values of w which satisfy: w w 5 w = 7 c) (From last moth s essay) Three data poits A= (, ), B= (5,8) ad C = (7,5) are plotted o a graph A horizotal lie y = k will be draw but a value k eeds to be chose so that the sum of the three vertical deviatios from the horizotal lie is at a miimum (NOTE: We ve draw the horizotal lie so that A lies below it ad B ad C above it This eed ot be the case) O a calculator, type i a fuctio that represets the sum of these three deviatios ad graph it Which value of k seems to give a miimum value for this sum of three deviatios? But we still eed a measure, a positive umber that represets the deviatio of each data value from the mea If we wat to avoid absolute value, how else ca we obtai positive values? Aswer: Square the values! Let s square all the deviatios ad take the average of those squared deviatios: DATA SET 1,,5, : Deviatios squared: 1 5 = 5 5 = = 65 5 = 05 Average squared deviatio: = 5 DATA SET 01, 0, 1, 110: Deviatios squared: = = = = Average squared deviatio: = 5575 These average squared deviatios still give a good sese of the differet spreads the two data sets possess Oe subtle poit: Data ofte comes from physical measuremets the height of a perso, the speed of a car o a highway, ad so o ad so has uits associated with them If x 1, x, K, x are i uits of iches, say, x1 + x + L+ x the the mea x= also has uits of iches, but the average squared deviatio: wwwjamestatocom ad wwwgdaymathcom

9 where the sum is over each ad every data poit i the set This simplifies to: 1 x y N ( + +L ) which is ideed µ! For the average value of the variaces, we eed to work with: + L+ ( x 1 +! ( N )! 1 + L This is equivalet to: ( ) ( N ) ( x ' 1 x ') + L+ ( x ' x ') + L+ ( x + ( x ' 1 x ') + L+ ( x ' x ') 1!! + L = 1!! ( ) ( N ) 1 1 x1 + L+ x) + L+ x + L+ x) x' 1 ( x ' 1+ L+ x ' ) + L+ x' ( x ' 1+ L+ x' ) + L = 1! N! (( 1) x1 x L + ( x1 + ( 1) x L + L (( 1 ) x ' 1 x ' x ' ) ( x ' 1 ( 1 ) x ' x' ) L + + L + L + L By expadig terms ad coutig how may times a particular data poit squared x appears ad how may times the pair 1 x x appears (ad these couts are the 1 same for all data poits), oe ca show that this expressio does ideed equal: ( µ ) + L+ ( µ ) y1 y N N 1 the variace over all the data poits We ll leave the details to the truly gug-ho reader! Exercise: To get a (maageable) feel for the algebra, do work through the details for the case of N = data poits: x1, x, x, x Write dow ad simplify the formulas for the variaces of each of the subsets,,,, x, x, x, { x1 x x },{ x1 x x },{ 1 } {,, } x x x ad a expressio for the average of these four values Show this average equals: x1 + x + x + x x1 x + x + x + x + x 1 x + x + x + x + x x + x + x + x + x 1 1 1, 01 James Tato wwwjamestatocom ad wwwgdaymathcom

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