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by: Sophie Levy

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# Week 2 MATH 1140

Sophie Levy
Tulane

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CH 2 HANDOUT
COURSE
PROF.
Robert Herbert,
TYPE
Class Notes
PAGES
12
WORDS
CONCEPTS
Statistics
KARMA
25 ?

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This 12 page Class Notes was uploaded by Sophie Levy on Sunday September 11, 2016. The Class Notes belongs to MATH 1140 at Tulane University taught by Robert Herbert, in Fall 2016. Since its upload, it has received 7 views. For similar materials see Statistics For Business in Math at Tulane University.

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Date Created: 09/11/16
Ch#2#(Handout/#Week#1) Wednesday,)August)31,)2016 2:15)PM 2.1\$Qualitative\$Data\$ Example:)An)instructor)has)given)an)exam)to)30)students.)The)letter)grades)of)the)exams)are) recorded)below.) X)=)grade=)A,)B,)A,)B,)C,)C,)C,)C,)D,)F,)D,)F,)A,)A,)B,)D,)A,)C,)C,)C,)B,)B,)B,)B,)D,)C,)C,)C,)C,)F)Give) the) Frequency)distribution:)simply)count)the)occurrence)of)each)value) ! A:)5 ! B:)7 ! C:)11 ! D:)4) ! F:)3) Relative)Frequencyf :requency)divide)by)the)number)of)values;)these)are)often)better)than) the)frequency)distribution)because,)if)these)are)only)samples,)the)percentages)can)easily)be) applied)to)the)population)for)a)reliable)estimate ! A:)5/30=0.1667) ! B:)7/30=0.2333) ! C:)11/30=0.3667) ! D:)4/30=0.1333) ! F:)3/30)=)0.1) Give)bar)graph)for)the)grades ! Credo)diagram:)put)the)most)important)values)first)(would)be)C,)B,)A,)D,)then)F) ! Pareto)diagram:)natural) Give)a)pie)chart)for)the)grades.) ! Angle)will)be)Relative)Frequency)x)360) Give)a)Pareto)diagram)for)the)grades.) 2.2.2.5\$Quantitative\$Data\$ Data)Set)A:)x)= 3,4,5,5,6,7,7,8,8,8,9,9,9,10,11,11,12,13,13,13,14,15,17,19,20,25,29,34,17,37,18,41) Classes:)intervals)that)must)be)set)up)so)that)every)element)of)the)data)lies)in)one) ! Classes)of)numbers))that)are)"A<))but))≤B") ! Class)width)could)be)BYA ! You)don't)want)to)make)the)classes)too)narrow)(too)many)classes)and)too)much)info))or) too)broad)(all)data)falls)within)one)class)and)that)doesn't)tell)you)anything)) ! LLCL:)Least)lower)class)limit:)left)endpoint)of)the)leftmost)class)aka)starting)point;)must) be)smaller)than)the)smallest)element)in)the)data) ! Example: Relative)Frequency ! Example: Relative)Frequency ))2<))))≤9 13 .41 ))9<))))≤16 9 .28 16<))))≤23 5 .16 23<))))≤30 2 .06 30<))))≤37 2 .06 37<))))≤44 1 .03 ○ Choice)LLCL:)2 ○ Choice)class)width:)7) What)do)we)learn?)Data)is)clumped)up)between)2)and)16) ! Histogram:s )ame)as)bar)graph)but)does)it)with)numerical)data,)and)all)the)rectangles) touch)each)other ○ In)this)example)it)is) right9skewed:)a)few)highly)unusual)elements)in)the)data;) "globbed)towards)the)left";)because)most)elements)are)below)the)value)of)16 ! Bellshapeddata)("normal)data"):)the)values)are)clumped)towards)the)middle;)we) LOVE)it)in)stats) ! You)can)relativize)the)frequency)tables)r fquency(/(n(where)n is)the)number)of)data)) ○ Take)all)of)the)numbers)and)divide)by)32) ○ They)will)add)up)to)100%)/)1.0 Summarize)data)set)A)with)a)stem)and)leaf)display.) ! Stems)=)10s) ! Leaves)=)1s ! Our)stems:)0,)1,)2,)3,)4 Stem))))Leaves 0 3,)4,)5,)6,)7,)7,)8,)8,)8,)9,)9,)9 1 0,)1,)1,)2,)3,)3,)3,)4,)5,)7,)7,)8),9 2 0,)5,)9 3 4,)7 4 1 Give)a)frequency)distribution)for)the)data)set)(occurrence)of)each)value)) 2 0,)5,)9 3 4,)7 4 1 Give)a)frequency)distribution)for)the)data)set)(occurrence)of)each)value)) ! 3:)1 ! 14:)1 ! 4:)1 ! 15:)1 ! 5:)1 ! 17:)2 ! 6:)1 ! 18:)1 ! 7:)2 ! 19:)1 ! 8:)3 ! 20:)1 ! 9:)3 ! 25:)1 ! 10:)1 ! 29:)1 ! 11:)2 ! 34:)1 ! 12:)1 ! 37:)1 ! 13:)3 ! 41:)1 Give)a)histogram)for)the)data Give)a)relative)frequency)distribution)for)the)data)(frequency)divided)by)number)of)values) ! 3:)1/32=0.0313 ! 14:)1/32=0.0313 Give)a)relative)frequency)distribution)for)the)data)(frequency)divided)by)number)of)values) ! 3:)1/32=0.0313 ! 14:)1/32=0.0313 ! 4:)1/32=0.0313 ! 15:)1/32=0.0313 ! 17:)2/32=0.0625 ! 5:)1/32=0.0313 ! 6:)1/32=0.0313 ! 18:)1/32=0.0313 ! 7:)2/32=0.0625 ! 19:)1/32=0.0313 ! 8:)3/32=0.0938 ! 20:)1/32=0.0313 ! 9:)3/32=0.0938 ! 25:)1/32=0.0313 ! 29:)1/32=0.0313 ! 10:)1/32=0.0313 ! 11:)2/32=0.0625 ! 34:)1/32=0.0313 ! 12:)1/32=0.0313 ! 37:)1/32=0.0313 ! 13:)3/32=0.0938 ! 41:)1/32=0.0313 Give)a)relative)histogram)for)the)data: Measures)of)Central)Tendency: ! Mode:)most)frequently)occurring)element)of)the)data)set,)if)there)is)such;)there)can) Measures)of)Central)Tendency: ! Mode:)most)frequently)occurring)element)of)the)data)set,)if)there)is)such;)there)can) only)be)one)mode ○ Corresponds)to)the)highest)point)in)the)histogram) ! Average/Mean:)find)the)sum)of)all)numbers)/)amount)of)numbers ! Median:)known)as)the)50th)percentile)or)as)the)2nd)quartile)(Q )2 ○ Where)at)least)50)percent)of)the)elements)ar≤ e)the)value)and)50)percent)are)≥ the)value ○ Odd)number)of)elements=)a)single)median)value Even)number)of)elements=)must)find)average)between)the)two)middle)numbers) ○ ! Good)way)to)calculate)skew:) ○ If)the)mean)value)is)larger)than)median=)it)is)right)skewed ○ If)the)mean)value)is)less)than)the)median=)it)is)left)skewed ○ If)the)mean)value)is)equal)to)the)median=)it)is)symmetrical) Does)Data)Set)A)have)a)mode?) ! Our)data)has)no)mode)because)there)is)not)a)single)value)which)occurs)the)most Find)the)mean)value)of)Data)Set)A.) ! 14.28 Find)the)median)of)data)Set)A.) ! If)there)are)32)elements,)you)want)16)to)be) ≤value)and)16)to)be) ≥value ! In)this)example,)any)number)between)11)and)12)will)satisfy)the)median) ! Q 2)ll)be)the)average)of)11)and)12) ! Q 2))11.5 Does)the)median)or)the)mean)better)describe)the)“)middle”)of)the)data?) ! The)mean)value)is)deceptive)because)it)is)righ Ytkewed)so)strongly ! The)median)offers)a)better)description)of)the)"middle" Is)Data)Set)A)skewed?) ! The)data)is)rightYskewed) ! The)mean)>)median) Measures)of)Consistency ! Parameters:) μ,\$σ2 2 ! Statisticsx,)s ! You)can)measure)byR ) ange=)the)largest)element)in)the)dat–a)mallest)element) ! Ex:)41–3=)a)range)of)38 ! Or)by)Population)Variance=)σ ,)average)square)deviation)from)the)mean)value ! Population)average)parameter= μ ! All)elements)in)population)added/number)of)elements) ! Estimated)by x,)which)is)the)mean) 2 2 ! σ =)(Σx–iμ) /)N ! Where)the)sum)is)taken)over)all)values) ! N=)number)of)elements)in)the)population) ! Population)standard)deviation:)a)measure)that)is)used)to)quantify)the)amount)of) variation)or)dispersion)of)a)set)of)data)values. ! Ex:)population)N=5 ! 1,)2,)6,)10,)11 ! Σx i)N=) μ ! =(1)+)2)+)6)+)10)+)11))/)5 ! =6)=μ ))))))X )))))μxY (xYμ ) 1 Y5 ))))))))25 2 Y4 ))))))))16 6 0 0 10 4 16 11 5 25 =Sum)of)Deviations=)82 ! Average)square)deviations)=)82)/)5)=)16.4) =Sum)of)Deviations=)82 ! Average)square)deviations)=)82)/)5)=)16.4) ! Ex)2:)Sample)of)elements)of)1,)6,)7,)8,)9,)10,)15 ! n=)7 ! 2) An)estimate)of) σ is) □ x̅)=)1)+)6)+)7)+)8)+)9)+)10)+)15)/)7)=)8 ! Sigma)(xYxbar)squared)=)49/108 ! BOARD Find)the)sample)variance of)the)data:)measures)the)consistency)of)the)variability)of)data;)the) sum)of)the)squared)deviations)from)the)mean)divided)by)(n Y1).)The)symbol) S is)used)to) represent)this) 1. Calculate)average)of)data 2. Calculate)variation)of)each)element)of)data)away)from)the)mean)value) 3. N=)number)of)elements)in)a)population:)32) 4. S =)92.46673 5. S=)9.61516 Find)the)sample)standard)deviationof)the)data:)s,)is)defined)by)the)positive)square)root)of) the)sample)variance,)s2) Ch  2  (Handout/  Week  2)   Monday,  September   5,  2016 11:55  AM 2.8   Percentiles(p):  a  p percentile  is  a  value  so  that  at  least  p%  of  the  values  in  the  data   are  ≤the  value  and  at  least  100–p%  are  ≥the  value.   -­ 0  <  p  ≤ 100 -­ The  percentile  always  exists,  but  need  not  be  unique -­ Ex:  if  you  choose  the  60th  percentile  of  a  value  set  of  1,  3,  4,  6,  10,  12… ○ At  least  60%  have  to  be   ≤the  value ○ At  least  40%  have  to  be   ≥the  value  (it  ends  up  being  50%)   ○ 6  is  the  60th  percentile -­ Ex2:  if  you  choose  the  50th  percentile  of  a  value  set  of  1,  2,  6,  8,  9,  12… ○ Any  number  between  6  and  8  works   -­ In  excel:  percentile.exc ○ Arrange  the  data  in  ascending  order ○ Calculate  (n+1)p § This  formula  gives  you  the  POSITION  of  the  value  of  the  percentile § P=  written  decimal,  rounded  up  and  down  to  give  you  approximate   position  of  the  percentile  in  the  ascending  array   ○ Interpolate  between  the  values  by  the  fractional  part   § The  decimal  point  will  give  you  the  percentage  between  the  two   elements   ○ Ex:  80th  percentile  of  1,  3,  4,  6,  10,  12 § Estimate=  (n+1)p  =  (7)(.8)=    5.6  with  the  .6  going  up  and  down   because  there  is  no  element  at  5.6! § 12-­‐10=2 § 2(.6)=1.2 § 10  +  1.2  =  11.2   -­ Q i1  the  1st  quartile,  the  25th  percentile   -­ Q 2s  the  2nd  quartile,  median,  and  50th  percentile -­ Q 3   the  3rd  quartile,  the  75th  percentile   Handout  cntd…   Data  Set  A:  x  = 2 -­ Q 3   the  3rd  quartile,  the  75th  percentile   Handout  cntd…   Data  Set  A:  x  = 3,4,5,5,6,7,7,8,8,8,9,9,9,10,11,11,12,13,13,13,14,15,17,19,20,25,29,34,17,37,18,41   Find  the  thpercentile  of  the  data.  We  will  use  Excel  percentiles   -­ Q =  8 1   -­ Work: ○ (n+1)(.25)  =  (33)(.25)=8.25   ○ It's  25%  between  elements  8  and  9 ○ Which  is  between  8  and  8   Find  the  thpercentile  of  the  data.  Find  the  interquartile  range  of  the  data.   -­ Q 3  17.75 -­ Work:   ○ (n+1)(.75)  =  (33)(.75)  =  24.75 ○ Look  75%  of  the  way  between  elements  24  and  25   ○ Which  is  between  17  and  18 Z  score:  to  measure  distances  in  units  of  standard  deviations -­ Population  z  score:   z  =  (x  – μ)  /  σ -­ Sample  z  score:  z  = ̅  /  s   -­ Ex:  population  with  aμ   of  10  and  a  σ of  3,  what  is  the  distance  from  16  to  the   mean  value? ○ z  =  (x  –μ)  /  σ ○ Z  =  (16  -­‐10)  /  3 ○ Z=  2  standard  deviations Give  the  five  number  summary  of  the  data.   How  many  standard  deviations  is  the  value  29  from  the  mean  value.   -­ x̅:  14.28 -­ So  z=(29  – 14.28)/  9.615962 -­ 1.54  standard  deviations Find  the  interquartile  range   Interquartile  range  (IQR)=  Q 3­‐Q =1  7.75–   8  =  9.75  9.75 Give  the  value  of  the  upper  fence   Find  the  interquartile  range   Interquartile  range  (IQR)=  Q 3­‐Q =1  7.75–   8  =  9.75  9.75 Give  the  value  of  the  upper  fence   Upper  fence:  upper  quartile  (Q3)  plus  1.5  interquartile  ranges   -­ Q +3  1.5  IQR   -­ 17.75  +  (1.5)(9.75) -­ =32.375 Give  the  value  of  the  lower  fence   -­ Q –11.5  IQR -­ 8  –1.5(9.75) -­ =  -­‐6.625 Give  the  value  which  corresponds  to  the  upper  whisker   Upper  whisker:  largest  element  in  the  data  that  is  less  than  or  equal  to  the  upper   fence   -­ In  the  data  set,  it's  29 Give  the  value  which  corresponds  to  the  lower  whisker   Lower  whisker :  smallest  element  in  the  data  that  is  greater  than  or  equal  to  the  lower   fence -­ In  the  data  set,  it's  3   Are  there  any  outliers  in  the  data?   Outliers:  elements  that  are  outside  the  fences   -­ In  the  data  set,  34,  37,  41 -­ If  (x ̅  /  s  >  3,  then  x  is  an  outlier Bell-­‐shaped(mound-­‐shaped)  data:  no  skew;  perfectly  symmetrical Empirical  Rule:s  uppose  data  is  bell  shaped… -­ 67-­‐68%  of  the  values  will  lie  within  1  standard  deviation  of  the  mean  value Empirical  Rule:s  uppose  data  is  bell  shaped… -­ 67-­‐68%  of  the  values  will  lie  within  1  standard  deviation  of  the  mean  value -­ 95%  of  the  values  will  fall  within  2  standard  deviations  of  the  mean  value   -­ 99.7%  of  the  values  will  fall  within  3  standard  deviations  of  the  mean  value -­ Example:  Suppose  population  has μ  =  20   =  2 ○ What  fraction  lies  within  16  and  24? ○ Should  describe  interval  using  standard  deviations ○ Z=  (24  -­‐ 20)  /  2   § Z=  2   Z  =  (16  -­‐20)  /  2   ○ § Z=  -­‐2   ○ Thus,  [16,24]  is  the  interval  of  the  value  which  lies  within  2  sd  of  the   mean;  so  about  95% Cheyhyshev's  Theorem   -­ Suppose  a  population  has  a  mean  value   μand  standard  deviationσ,  then  at   least  (1   1/k )  x  100%  of  the  values  in  the  populationk standard  hin   deviations  of  the  mean  value -­ Ex:  μ=15  andσ=2…  What  is  the  minimal  fraction  of  the  data  which  lie  within  the   range  11  to  19    NOT  assuming  that  the  population-llhaped   ○ (19-­‐15)/2  =  2   ○ (11-­‐15)/2=-­‐2   ○ So  K  is  2   2 ○ At  least  (1  -­‐1/2 )  x  100%  =  75%  of  the  values  lie  within11  to  19

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