WISKUNDIGE GELETTERDHEID
Graad 12
NOG OEFENINGE
Statistiek : antwoorde.
  
MATHEMATICAL LITERACY
Grade 12
MORE EXERCISES
Statistics : answers
  
  
  
Antwoorde / Answers  1
  
     
      1.1  Rangskik : 16 ; 17 ; 18 ; 19 ; 19 ;
                        20 ; 21 ; 22 ; 24 ; 26 ; 28
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{230}{11}\ =\ 20,909 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 20\ ┃\ waarde\ presies\ in } $$ $$ \hspace*{41 mm}\mathrm{die\ middel } $$

$$ \hspace*{13 mm}\mathrm{Modus\ =\ 19\ ┃\ kom\ die\ meeste } $$ $$ \hspace*{38 mm}\mathrm{kere\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ =\ omvang } $$ $$ \hspace*{13 mm}\mathrm{Omvang\ =\ maksimum\ ━\ minimum} $$
$$ \hspace*{28 mm}=\ 28\ ━\ 16\ = 12 $$

[ Vraag 1. ]
  
  
     
      1.2  Rangskik : 11 ; 12 ; 14 ; 16 ; 17 ;
                               17 ; 17 ; 18 ; 21 ; 22
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{165}{10}\ =\ 16,5 } $$

$$ \hspace*{13 mm}\mathrm{As\ daar\ 'n\ aantal\ ewe\ waardes } $$ $$ \hspace*{13 mm}\mathrm{voorkom\ is\ die\ mediaan\ die } $$ $$ \hspace*{13 mm}\mathrm{gemiddelde\ van\ die\ twee } $$ $$ \hspace*{13 mm}\mathrm{waardes\ naaste\ aan\ die\ middel } $$


$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{17 + 17}{2}\ = 17 } $$

$$ \hspace*{13 mm}\mathrm{Modus\ =\ 17\ ┃\ kom\ 3\ keer\ voor } $$ $$ \hspace*{38 mm}\mathrm{voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 22\ ━\ 11\ = 11 } $$

[ Vraag 1. ]
  
  
     
      1.3  Rangskik : 38 ; 57 ; 58 ; 59 ; 60 ;
                               60 ; 63
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{395}{7}\ =\ 56,429 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 59\ middelste\ waarde } $$

$$ \hspace*{13 mm}\mathrm{Modus\ =\ 60\ ┃\ kom\ 2\ keer\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 63\ ━\ 38\ = 25 } $$

[ V 1. ]
  
  
     
      1.4  Rangskik : 49 ; 50 ; 50 ; 51 ; 51 ;
                               52 ; 53 ; 58 ; 80
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{494}{9}\ =\ 54,889 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 51\ middelste\ waarde } $$

$$ \hspace*{13 mm}\mathrm{Modus\ =\ 50\ en\ 51\ ┃\ albei\ kom\ } $$ $$ \hspace*{48 mm}\mathrm{2\ keer\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 80\ ━\ 49\ = 31 } $$

[ V 1. ]
  
  
     
      1.5  Rangskik : 14 ; 18 ; 23 ; 27 ; 30 ;
                               38
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{150}{6}\ =\ 25 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{23 + 27}{2}\kern4mm=\kern2mm25 } $$

$$ \hspace*{13 mm}\mathrm{Geen\ modus\ ┃\ alle\ waardes\ } $$ $$ \hspace*{38 mm}\mathrm{1\ keer\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 38\ ━\ 14\ = 24 } $$

[ V 1. ]
  
  
     
      1.6  Rangskik : 13 ; 19 ; 19 ; 19 ; 19 ;
                               21 ; 23
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{133}{7}\ =\ 19 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 19\kern2mmmiddelste\ waarde } $$ $$ \hspace*{38 mm}\mathrm{\ waarde } $$


$$ \hspace*{13 mm}\mathrm{Modus\ == 19\ ┃\ kom\ 4\ keer } $$ $$ \hspace*{38 mm}\mathrm{\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 23\ ━\ 13\ = 10 } $$

[ V 1. ]
  
  
     
      1.7  Rangskik : 4 ; 7 ; 8 ; 11 ; 11 ;
                               15 ; 15 ; 15 ; 17 ; 25
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{128}{10}\ =\ 12,8 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{11 + 15}{2}\kern2mm=\ 13 } $$

$$ \hspace*{13 mm}\mathrm{Modus\ == 15\ ┃\ kom\ 3\ keer } $$ $$ \hspace*{38 mm}\mathrm{\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 25\ ━\ 4\ = 21 } $$

[ V 1. ]
  
  
     
      1.8  Rangskik : 30 ; 31 ; 31 ; 32 ; 33 ;
                               34 ; 37 ; 37 ; 37
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{302}{9}\ =\ 33,556 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 33 } $$

$$ \hspace*{13 mm}\mathrm{Modus\ == 37\ ┃\ kom\ 3\ keer } $$ $$ \hspace*{38 mm}\mathrm{\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 37\ ━\ 30\ = 7 } $$

[ V 1. ]
  
  
     
      1.1  Arrange : 16 ; 17 ; 18 ; 19 ; 19 ;
                        20 ; 21 ; 22 ; 24 ; 26 ; 28
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{230}{11}\ =\ 20,909 } $$

$$ \hspace*{13 mm}\mathrm{Median\ =\ 20\ ┃\ value\ exactly\ in } $$ $$ \hspace*{39 mm}\mathrm{the\ middle } $$

$$ \hspace*{13 mm}\mathrm{Mode\ =\ 19\ ┃\ appears\ most\ often } $$

$$ \hspace*{13 mm}\mathrm{Range\ =\ maximum\ ━\ minimum} $$ $$ \hspace*{25 mm}=\ 28\ ━\ 16\ = 12 $$



[ Question 1. ]
  
  
     
      1.2  Arrange : 11 ; 12 ; 14 ; 16 ; 17 ;
                             17 ; 17 ; 18 ; 21 ; 22
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{165}{10}\ =\ 16,5 } $$

$$ \hspace*{13 mm}\mathrm{If\ there\ is\ an\ even\ number\ of } $$ $$ \hspace*{13 mm}\mathrm{values\ the\ median\ is\ the } $$ $$ \hspace*{13 mm}\mathrm{mean\ of\ the\ two\ midmost\ values } $$ $$ \hspace*{13 mm}\mathrm{(those\ nearest\ to\ the\ centre) } $$


$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{17 + 17}{2}\ = 17 } $$

$$ \hspace*{13 mm}\mathrm{Mode\ =\ 17\ ┃\ appears\ 3\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 22\ ━\ 11\ = 11 } $$

[ Question 1. ]
  
  
     
      1.3  Arrange : 38 ; 57 ; 58 ; 59 ; 60 ;
                             60 ; 63
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{395}{7}\ =\ 56,429 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = 59\ midmost\ value } $$

$$ \hspace*{13 mm}\mathrm{Mode\ =\ 60\ ┃\ appears\ 2\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 63\ ━\ 38\ = 25 } $$

[ Q 1. ]
  
  
     
      1.4  Arrange : 49 ; 50 ; 50 ; 51 ; 51 ;
                             52 ; 53 ; 58 ; 80
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{494}{9}\ =\ 54,889 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = 51\ midmost\ value } $$

$$ \hspace*{13 mm}\mathrm{Mode\ =\ 50\ and\ 60\ ┃\ both\ appear } $$ $$ \hspace*{48 mm}\mathrm{2\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 80\ ━\ 49\ = 31 } $$

[ Q 1. ]
  
  
     
      1.5  Arrange : 14 ; 18 ; 23 ; 27 ; 30 ;
                             38
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{150}{6}\ =\ 25 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = \frac{23 + 27}{2}\kern4mm=\kern2mm25 } $$

$$ \hspace*{13 mm}\mathrm{No\ mode\ ┃\ all\ values\ differ } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 38\ ━\ 14\ = 24 } $$

[ Q 1. ]
  
  
     
      1.6  Arrange : 13 ; 19 ; 19 ; 19 ; 19 ;
                             21 ; 23
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{133}{7}\ =\ 19 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = 19\kern2mmmidmost\ value } $$


$$ \hspace*{13 mm}\mathrm{Mode\ = 19 ┃\ appears\ 4\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 23\ ━\ 13\ = 10 } $$

[ Q 1. ]
  
  
     
      1.7  Arrange : 4 ; 7 ; 8 ; 11 ; 11 ;
                             15 ; 15 ; 15 ; 17 ; 23
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{128}{10}\ =\ 12,8 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = \frac{11 + 15}{2}\kern2mm=\kern2mm13 } $$

$$ \hspace*{13 mm}\mathrm{Mode\ = 19 ┃\ appears\ 3\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 25\ ━\ 4\ = 21 } $$

[ Q 1. ]
  
  
     
      1.8  Arrange : 30 ; 31 ; 31 ; 32 ; 33 ;
                             34 ; 37 ; 37 ; 37
                               34 ; 37 ; 37 ; 37
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{302}{9}\ =\ 33,556 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = 33 } $$

$$ \hspace*{13 mm}\mathrm{Mode\ = 19 ┃\ appears\ 3\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 37\ ━\ 30\ = 7 } $$

[ Q 1. ]
  
  
  
  
Antwoorde / Answers  2
  
     
      2.1  Rangskik : 31 ; 32 ; 34 ; 34 ; 34 ;
                               36 ; 37
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{238}{7}\ =\ 34 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 34\kern2mm┃\ middelste } $$ $$ \hspace*{40 mm}\mathrm{\kern2mmwaarde } $$

$$ \hspace*{13 mm}\mathrm{Modus\ == 34\ ┃\ kom\ 3\ keer } $$ $$ \hspace*{38 mm}\mathrm{\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 37\ ━\ 31\ = 6 } $$

             Die datawaardes is nie te
             verspreid nie, omvang = 6 en
             daarom sal al die waardes,
             gemiddelde, mediaan en modus,
             die data goed beskryf.
[ V 2. ]
  
  
     
      2.2  Rangskik : 6 ; 6 ; 7 ; 7 ; 7 ;
                               8 ; 8 ; 9 ; 31
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{89}{9}\ =\ 9,889 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 7\kern2mm┃\ middelste } $$ $$ \hspace*{40 mm}\mathrm{\kern2mmwaarde } $$

$$ \hspace*{13 mm}\mathrm{Modus\ == 7\ ┃\ kom\ 3\ keer } $$ $$ \hspace*{38 mm}\mathrm{\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 31\ ━\ 6\ = 31 } $$

             Die uitskieter, 31, veroorsaak
             dat die gemiddelde nie 'n
             betroubare beskrywing van die
             data is nie, terwyl die mediaan
             en modus wel goeie beskrywings is.
[ V 2. ]
  
  
     
      2.3  Rangskik : 3 ; 11 ; 15 ; 15 ; 16 ;
                               16
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{76}{6}\ =\ 12,667 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{15 + 15}{2}\kern2mm=\ 15 } $$

$$ \hspace*{13 mm}\mathrm{Modus\ == 15\ en\ 16 ┃\ albei\ kom } $$ $$ \hspace*{44 mm}\mathrm{\ 2\ keer\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 16\ ━\ 3\ = 13 } $$

             Die uitskieter, 3, veroorsaak
             dat die gemiddelde nie 'n
             betroubare beskrywing van die
             data is nie. Die mediaan
             en modus is beter beskrywings.
[ V 2. ]
  
  
     
      2.4  Rangskik : 17 ; 18 ; 20 ; 20 ; 20 ;
                               23
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{118}{6}\ =\ 19,667 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{20 + 20}{2}\kern2mm=\ 20 } $$

$$ \hspace*{13 mm}\mathrm{Modus\ == 20\ ┃\ kom\ 3 } $$ $$ \hspace*{38 mm}\mathrm{\ keer\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 23\ ━\ 17\ = 6 } $$

             Die data is effens skeef na
             links getrek en die
             gemiddelde word daardeur
             beïnvloed. Die mediaan en
             modus is goeie beskrywings.
[ V 2. ]
  
  
     
      2.5  Rangskik : 21 ; 22 ; 22 ; 22 ; 24 ;
                               25 ; 26 ; 62
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{224}{8}\ =\ 28 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{22 + 24}{2}\kern2mm=\ 23 } $$

$$ \hspace*{13 mm}\mathrm{Modus\ == 22\ ┃\ kom\ 3 } $$ $$ \hspace*{38 mm}\mathrm{\ keer\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 62\ ━\ 22\ = 40 } $$

             Die uitskieter 62 het 'n groot
             invloed op die gemiddelde.
             Die gemiddelde is dus nie 'n
             goeie beskrywing van die
             data nie. Die modus is
             effens te klein. Die mediaan
             is die beste beskrywing.
[ V 2. ]
  
  
     
      2.6  Rangskik : 13 ; 48 ; 51 ; 52 ; 53 ;
                               54 ; 57 ; 92
$$ \hspace*{13 mm}\mathrm{Gemiddelde\ =\ \frac{420}{8}\ =\ 52,5 } $$

$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{52 + 53}{2}\kern2mm=\ 52,5 } $$

$$ \hspace*{13 mm}\mathrm{Geen\ modus\ want\ al\ die\ waardes } $$ $$ \hspace*{13 mm}\mathrm{\ kom\ net\ een\ keer\ voor } $$

$$ \hspace*{13 mm}\mathrm{Variasiewydte\ = 62\ ━\ 22\ = 40 } $$

             Die uitskieters, 13 en 92,
             balanseer" mekaar sodat die
             gemiddelde 'n goeie beskrywing
             van die data is. Die mediaan is
             ook 'n goeie beskrywing van
             die data. Die modus bestaan nie
             en kan dus nie gebruik word nie.

[ V 2. ]
  
  
     
      2.1  Arrange : 31 ; 32 ; 34 ; 34 ; 34 ;
                             36 ; 37
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{238}{7}\ =\ 34 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = 34\ ┃\ midmost\ value } $$

$$ \hspace*{13 mm}\mathrm{Mode\ = 34 ┃\ appears\ 3\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 37\ ━\ 31\ = 7 } $$

             The data values are not too spread
             out, range = 6 and therefore all three
             the measures, average/mean,
             median and mode, describe
             the data well.
[ Q 2. ]
  
  
     
      2.2  Arrange : 6 ; 6 ; 7 ; 7 ; 7 ;
                             8 ; 8 ; 9 ; 31
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{89}{9}\ =\ 9,889 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = 7\ ┃\ midmost\ value } $$

$$ \hspace*{13 mm}\mathrm{Mode\ = 7 ┃\ appears\ 3\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 31\ ━\ 6\ = 31 } $$

             The outlier, 31, affects the mean so
             that the mean is not a reliable
             description of the data. The median
             and mode are better descriptions.

[ Q 2. ]
  
  
     
      2.3  Arrange : 3 ; 11 ; 15 ; 15 ; 16 ;
                             16
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{76}{6}\ =\ 12,667 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = \frac{15 + 15}{2}\ =\ 15 } $$

$$ \hspace*{13 mm}\mathrm{Mode\ = 14\ and\ 16\ ┃\ both\ appear } $$ $$ \hspace*{46 mm}\mathrm{\ 2\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 16\ ━\ 3\ = 13 } $$

             The outlier, 3, affects the mean so
             that the mean is not a reliable
             description of the data. The median
             and mode are better descriptions.

[ Q 2. ]
  
  
     
      2.4  Arrange : 17 ; 18 ; 205 ; 20 ; 20 ;
                             23
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{118}{6}\ =\ 19,667 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = \frac{20 + 20}{2}\ =\ 20 } $$

$$ \hspace*{13 mm}\mathrm{Mode\ = 20\ ┃\ appears\ 3\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 23\ ━\ 17\ = 6 } $$

             The mean is influenced by the
             values that are slightly slanted
             to the left.The median and
             mode are better descriptions.

[ Q 2. ]
  
  
     
      2.5  Arrange : 21 ; 22 ; 22 ; 22 ; 24 ;
                             25 ; 26 ; 62
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{224}{8}\ =\ 28 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = \frac{22 + 24}{2}\ =\ 23 } $$

$$ \hspace*{13 mm}\mathrm{Mode\ = 22\ ┃\ appears\ 3\ times } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 62\ ━\ 22\ = 40 } $$

             The outlier 62 has a great
             influence on the mean.
             The mean is therefore not a
             reliable description of the data.
             The mode is slightly too small.
             The median is the best description.

[ Q 2. ]
  
  
     
      2.6  Arrange : 13 ; 48 ; 51 ; 52 ; 53 ;
                             54 ; 57 ; 92
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{420}{8}\ =\ 52,5 } $$

$$ \hspace*{13 mm}\mathrm{Median\ = \frac{52 + 53}{2}\ =\ 52,5 } $$

$$ \hspace*{16 mm}\mathrm{\ No\ mode\ as\ all\ the\ values } $$ $$ \hspace*{16 mm}\mathrm{\ appear\ once\ only } $$

$$ \hspace*{13 mm}\mathrm{Range\ = 92\ ━\ 13\ = 79 } $$

             The outliers, 13 and 92, "balance"
             one another so that the mean is a
             good description of the data.
             The median is also a good
             description of the data.
             The mode does not exsist and
             cannot be used to describe the data.

[ Q 2. ]
  
  
  
  
Antwoorde / Answers  3
  
             Gebruik onderstaande formules
             om die posisie van die kwartiele
             en persentiele re bereken :
             n = aantal waardes in die datastel
             p = die nommer van die kwartiel of
                   persentiel
$$ \hspace*{13 mm}\mathrm{Q_{p}\ =\ \frac{p(n + 1)}{4}\ ;\ P_{p}\ =\ \frac{p(n + 1)}{100} } $$

        Benader die getal tot die naaste
        heelgetal.

     
      3.1  Rangskik : 3 ; 8 ; 15 ; 16 ; 18 ; 21 ; 23 ;
                               24 ; 33 ; 35 ; 36 ; 37 ; 38
$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 23\ ┃\ middelste\ waarde } $$

             Die posisie van die eerste kwartiel
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ =\ \frac{1(13 + 1)}{4}\ =\ 3,5 } $$


             Die eerste kwartiel is dus presies
             halfpad tussen die 3de en 4de
             data waardes.
$$ \hspace*{13 mm}\mathrm{Waarde\ van\ Q_{1}\ =\ \frac{15 + 16}{2} } $$
$$ \hspace*{39 mm}\mathrm{=\ 15,5 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ =\ \frac{2(13 + 1)}{4}\ =\ 7,0 } $$

$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ =\ \frac{3(13 + 1)}{4}\ =\ 10,5 } $$


             Die tweede kwartiel is dus
             die 7de waarde en die
             3de kwartiel is presies
             halfpad tussen die 10de en
             die 11de data waardes.
             Die waarde van Q2 = 23 en
$$ \hspace*{13 mm}\mathrm{Q_{2}\ =\ \frac{35 + 36}{2}\ =\ 35,5 } $$

$$ \hspace*{13 mm}\mathrm{Interkwartielvariasiewydte\ =\ IKV } $$ $$ \hspace*{31 mm}\mathrm{IKV\kern2mm=\kern2mmQ_{3}\ ─\ Q_{1} } $$ $$ \hspace*{40 mm}\mathrm{=\kern2mm35,5\ ─\ 15,5 } $$ $$ \hspace*{40 mm}\mathrm{=\kern2mm20 } $$



$$ \hspace*{13 mm}\mathrm{p(P_{20})\ =\ \frac{20(13 + 1)}{100}\ =\ 2,8 } $$

$$ \hspace*{13 mm}\mathrm{p(P_{25})\ =\ \frac{25(13 + 1)}{100}\ =\ 3,5 } $$

             P20 is dus die derde waarde en
             P25 is presies halfpad tussen
             die derde en 4de data waardes.
             Die waarde van P20 = 15 en die
$$ \hspace*{13 mm}\mathrm{van\ P_{25}\ =\ \frac{15 + 16}{2}\kern3mm=\kern2mm15,5 } $$

$$ \hspace*{13 mm}\mathrm{p(P_{75})\ =\ \frac{75(13 + 1)}{100}\ =\ 10,5 } $$

$$ \hspace*{13 mm}\mathrm{p(P_{80})\ =\ \frac{80(13 + 1)}{100}\ =\ 11,2 } $$

             P75 is presies halfpad tussen
             die 10de en 11de data waardes.
             en P80 is die 11de data waarde.
$$ \hspace*{13 mm}\mathrm{Waarde\ van\ P_{75}\ =\ \frac{35 + 36}{2} } $$
$$ \hspace*{41 mm}\mathrm{=\kern2mm35,5 } $$
             en die waarde van P80 = 36
             Die middelste 50% van die
             waardes is tussen Q1 en Q3,
             d.i. tussen 15,5 en 35,5
             Die maksimum van die onderste
             25% van die waardes is 15.
             Die minimum van die boonste
             80% van die waardes = P80 = 36.
[ V 3. ]
  
     
      3.2  Rangskik : 31 ; 48 ; 49 ; 51 ; 51 ; 53 ; 54 ;
                               55 ; 57 ; 58 ; 63 ; 67 ; 71
                               72 ; 73 ; 75 ; 92 ; 93 ; 95
                               96
$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{58 + 63}{2}\ ┃\ ewe\ waardes } $$
$$ \hspace*{29 mm}\mathrm{=\ 60,5 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ = \frac{1(20 + 1)}{4}\ = 5,25 } $$
$$ \hspace*{13 mm}\mathrm{5^{de}\ waarde\ en\ dus\ Q_{1}\ =\ 51 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{2(20 + 1)}{4}\ = 10,5 } $$

$$ \hspace*{13 mm}\mathrm{tussen\ 10^{de}\ en\ 11^{de}\ waardes } $$
$$ \hspace*{13 mm}\mathrm{Q_{2}\ = \frac{58 + 63}{2}\ = 60,5 } $$

$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ = \frac{3(20 + 1)}{4}\ = 15,75 } $$
$$ \hspace*{13 mm}\mathrm{16^{de}\ waarde\ en\ dus\ Q_{3}\ =\ 75 } $$
$$ \hspace*{13 mm}\mathrm{IKV\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{22 mm}\mathrm{= 75\ ─ 51\kern{2mm}= 24 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(20 + 1)}{100}\ = 4,2 } $$

$$ \hspace*{13 mm}\mathrm{4^{de}\ waarde\ en\ dus\ P_{20}\ =\ 51 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(20 + 1)}{100}\ = 5,25 } $$

$$ \hspace*{13 mm}\mathrm{5^{de}\ waarde\ en\ dus\ P_{25}\ =\ 51 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(20 + 1)}{100}\ = 15,75 } $$

$$ \hspace*{13 mm}\mathrm{16^{de}\ waarde\ en\ dus\ P_{75}\ =\ 75 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(20 + 1)}{100}\ = 16,8 } $$

$$ \hspace*{13 mm}\mathrm{17^{de}\ waarde\ en\ dus\ P_{80}\ =\ 92 } $$

             Die middelste 50% van die
             waardes is tussen Q1 en Q3,
             d.i. tussen 51 en 75
             Die maksimum van die onderste
             25% van die waardes is 51.
             Die minimum van die boonste
             80% van die waardes = P80 = 36.
[ V 3. ]
  
     
      3.3  Rangskik : 2 ; 3 ; 4 ; 5 ; 7 ; 9 ; 11 ;
                               12 ; 13 ; 13 ; 13 ; 14 ; 15
                               16 ; 16 ; 18 ; 23 ; 25 ; 31
                               23 ; 25 ; 41 ; 42 ; 43 ; 44
$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 14\ ┃\ middelste\ waarde } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ = \frac{1(23 + 1)}{4}\ = 6 } $$
$$ \hspace*{13 mm}\mathrm{6^{de}\ waarde\ en\ dus\ Q_{1}\ =\ 9 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{2(23 + 1)}{4}\ = 12 } $$

$$ \hspace*{13 mm}\mathrm{12^{de}\ waarde\ en\ dus\ Q_{2}\ =\ 14 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ = \frac{3(23 + 1)}{4}\ = 18 } $$
$$ \hspace*{13 mm}\mathrm{18^{de}\ waarde\ en\ dus\ Q_{3}\ =\ 31 } $$
$$ \hspace*{13 mm}\mathrm{IKV\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{22 mm}\mathrm{= 31\ ─ 9\kern{2mm}= 22 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(23 + 1)}{100}\ = 4,8 } $$

$$ \hspace*{13 mm}\mathrm{5^{de}\ waarde\ en\ dus\ P_{20}\ =\ 7 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(23 + 1)}{100}\ = 6 } $$

$$ \hspace*{13 mm}\mathrm{6^{de}\ waarde\ en\ dus\ P_{25}\ =\ 9 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(23 + 1)}{100}\ = 18 } $$

$$ \hspace*{13 mm}\mathrm{18^{de}\ waarde\ en\ dus\ P_{75}\ =\ 31 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(23 + 1)}{100}\ = 19,2 } $$

$$ \hspace*{13 mm}\mathrm{19^{de}\ waarde\ en\ dus\ P_{80}\ =\ 34 } $$

             Die middelste 50% van die
             waardes is tussen Q1 en Q3,
             d.i. tussen 9 en 31
             Die maksimum van die onderste
             25% van die waardes is 9.
             Die minimum van die boonste
             80% van die waardes = P80 = 34.
[ V 3. ]
  
     
      3.43  Rangskik : 31 ; 32 ; 34 ; 35 ; 36 ; 37 ; 11 ;
                               40 ; 41 ; 41 ; 41 ; 42 ; 44
                               48 ; 49 ; 52
$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ 40,5 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ = \frac{1(16 + 1)}{4}\ = 4,25 } $$
$$ \hspace*{13 mm}\mathrm{4^{de}\ waarde\ en\ dus\ Q_{1}\ =\ 35 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{2(16 + 1)}{4}\ = 8,5 } $$

$$ \hspace*{13 mm}\mathrm{tussen\ 8^{ste}\ en\ 9^{de}\ waarde } $$
$$ \hspace*{13 mm}\mathrm{Q_{2}\ = \frac{40 + 41}{2}\ = 40,5 } $$

$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ = \frac{3(16 + 1)}{4}\ = 12,75 } $$
$$ \hspace*{13 mm}\mathrm{13^{de}\ waarde\ en\ dus\ Q_{3}\ =\ 44 } $$
$$ \hspace*{13 mm}\mathrm{IKV\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{22 mm}\mathrm{= 44\ ─ 35\kern{2mm}= 9 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(16 + 1)}{100}\ = 3,4 } $$

$$ \hspace*{13 mm}\mathrm{3^{de}\ waarde\ en\ dus\ P_{20}\ =\ 34 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(16 + 1)}{100}\ = 4,25 } $$

$$ \hspace*{13 mm}\mathrm{4^{de}\ waarde\ en\ dus\ P_{25}\ =\ 35 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(16 + 1)}{100}\ = 12,75 } $$

$$ \hspace*{13 mm}\mathrm{13^{de}\ waarde\ en\ dus\ P_{75}\ =\ 44 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(16 + 1)}{100}\ = 13,6 } $$

$$ \hspace*{13 mm}\mathrm{14^{de}\ waarde\ en\ dus\ P_{80}\ =\ 48 } $$

             Die middelste 50% van die
             waardes is tussen Q1 en Q3,
             d.i. tussen 35 en 44
             Die maksimum van die onderste
             25% van die waardes is 35.
             Die minimum van die boonste
             80% van die waardes = P80 = 48.
[ V 3. ]
  
  
             Use the formulae below to calculate
             the position of the quartiles and
             percentiles :
             n = number of values in the data set
             p = the number of the quartile or
                   percentile
$$ \hspace*{13 mm}\mathrm{Q_{p}\ =\ \frac{p(n + 1)}{4}\ ;\ P_{p}\ =\ \frac{p(n + 1)}{100} } $$

        Round off this number to the nearest
        integer.

     
      3.1  Arrange : 3 ; 8 ; 15 ; 16 ; 18 ; 21 ; 23 ;
                             24 ; 33 ; 35 ; 36 ; 37 ; 38
$$ \hspace*{13 mm}\mathrm{Median\ = 23\ ┃\ midmost\ value } $$
$$ \hspace*{13 mm}\mathrm{Position\ of\ first\ quartile : } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ = \frac{1(13 + 1)}{4}\ =\ 3,5 } $$


             The first quartile is therefore
             exactly halfway between the
             3rd and 4th data values.
$$ \hspace*{13 mm}\mathrm{Value\ of\ Q_{1}\ = \frac{15 + 16}{2}\ = 15,5 } $$


$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{2(13 + 1)}{4}\ = 7,0 } $$

$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ = \frac{3(13 + 1)}{4}\ = 10,5 } $$


             The second quartile is thus
             the 7th value and
             the 3rd quartile
             is exactly halfway between the
             10th and 11th data values.
             The value of Q2 = 23 and
$$ \hspace*{13 mm}\mathrm{Q_{3}\ = \frac{35 + 36}{2}\ = 35,5 } $$

$$ \hspace*{13 mm}\mathrm{Interquartile range (IQR)\ = Q_{3}\ ─ Q_{1} } $$ $$ \hspace*{32 mm}\mathrm{IQR\ = Q_{3}\ ─ Q_{1} } $$ $$ \hspace*{40 mm}\mathrm{=\ 35,5\ ─ 15,5 } $$ $$ \hspace*{40 mm}\mathrm{\ =\ 20 } $$



$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(13 + 1)}{100}\ = 2,8 } $$

$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(13 + 1)}{100}\ = 3,5 } $$

             P20 is therefore the 3rd value
             and P25 is exactly halfway between
             the 3rd and 4th data values.
             The value of P20 = 15 and that
$$ \hspace*{13 mm}\mathrm{of\ P_{25}\ = \frac{15 + 16)}{2}\ = 15,5 } $$

$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(13 + 1)}{100}\ = 10,5 } $$

$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(13 + 1)}{100}\ = 11,2 } $$

             P75 is exactly halfway between the
             10th and 11th data values
             and P80 is the 11th data value.
$$ \hspace*{13 mm}\mathrm{Value\ of\ P_{75}\ = \frac{35 + 36)}{2}\ = 35,5 } $$

             and the value of P80 is 36

             The middle 50% of the values
             lie between Q1 and Q3,
             i.e. between 15,5 and 35,5
             The maximum of the bottom
             25% of the values is 15.
             The minimum of the top
             80% of the values = P80 = 36.
[ Q 3. ]
  
     
      3.2  Arrange : 31 ; 48 ; 49 ; 51 ; 51 ; 53 ; 54 ;
                             55 ; 57 ; 58 ; 63 ; 67 ; 71
                             72 ; 73 ; 75 ; 92 ; 93 ; 95
                             96
$$ \hspace*{13 mm}\mathrm{Median\ = \frac{58 + 63}{2}\ ┃\ even\ number } $$
$$ \hspace*{27 mm}\mathrm{= 60,5\ \ of\ values } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ = \frac{1(20 + 1)}{4}\ = 5,25 } $$
$$ \hspace*{13 mm}\mathrm{5^{th}\ value\ and\ thus\ Q_{1}\ =\ 51 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{2(20 + 1)}{4}\ = 10,5 } $$

$$ \hspace*{13 mm}\mathrm{between\ 10^{th}\ and\ 11^{th}\ values } $$
$$ \hspace*{13 mm}\mathrm{Q_{2}\ = \frac{58 + 63}{2}\ = 60,5 } $$

$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ = \frac{3(20 + 1)}{4}\ = 15,75 } $$
$$ \hspace*{13 mm}\mathrm{16^{th}\ value\ and\ thus\ Q_{3}\ =\ 75 } $$
$$ \hspace*{13 mm}\mathrm{IQR\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{22 mm}\mathrm{= 75\ ─ 51\kern{2mm}= 24 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(20 + 1)}{100}\ = 4,2 } $$

$$ \hspace*{13 mm}\mathrm{4^{th}\ value\ and\ thus\ P_{20}\ =\ 51 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(20 + 1)}{100}\ = 5,25 } $$

$$ \hspace*{13 mm}\mathrm{5^{th}\ value\ and\ thus\ P_{25}\ =\ 51 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(20 + 1)}{100}\ = 15,75 } $$

$$ \hspace*{13 mm}\mathrm{16^{th}\ value\ and\ thus\ P_{75}\ =\ 75 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(20 + 1)}{100}\ = 16,8 } $$

$$ \hspace*{13 mm}\mathrm{17^{th}\ value\ and\ thus\ P_{80}\ =\ 92 } $$

             The middle 50% of the values
             lie between Q1 and Q3,
             i.e. between 51 and 75
             The maximum of the bottom
             25% of the values is 51.
             The minimum of the top
             80% of the values = P80 = 92.
[ Q 3. ]
  
     
      3.3  Arrange : 2 ; 3 ; 4 ; 5 ; 7 ; 9 ; 11 ;
                             12 ; 13 ; 13 ; 13 ; 14 ; 15
                             16 ; 16 ; 18 ; 23 ; 25 ; 31
                             23 ; 25 ; 41 ; 42 ; 43 ; 44
$$ \hspace*{13 mm}\mathrm{Median\ = 14 ┃\ midmost\ value } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ = \frac{1(23 + 1)}{4}\ = 6 } $$
$$ \hspace*{13 mm}\mathrm{6^{th}\ value\ and\ thus\ Q_{1}\ =\ 9 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{2(23 + 1)}{4}\ = 12 } $$

$$ \hspace*{13 mm}\mathrm{12^{th}\ value\ and\ thus\ Q_{2}\ =\ 14 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ = \frac{3(23 + 1)}{4}\ = 18 } $$
$$ \hspace*{13 mm}\mathrm{18^{th}\ value\ and\ thus\ Q_{3}\ =\ 31 } $$
$$ \hspace*{13 mm}\mathrm{IQR\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{22 mm}\mathrm{= 31\ ─ 9\kern{2mm}= 22 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(23 + 1)}{100}\ = 4,8 } $$

$$ \hspace*{13 mm}\mathrm{5^{th}\ value\ and\ thus\ P_{20}\ =\ 7 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(23 + 1)}{100}\ = 6 } $$

$$ \hspace*{13 mm}\mathrm{6^{th}\ value\ and\ thus\ P_{25}\ =\ 9 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(23 + 1)}{100}\ = 18 } $$

$$ \hspace*{13 mm}\mathrm{18^{th}\ value\ and\ thus\ P_{75}\ =\ 31 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(23 + 1)}{100}\ = 19,2 } $$

$$ \hspace*{13 mm}\mathrm{19^{th}\ value\ and\ thus\ P_{80}\ =\ 34 } $$

             The middle 50% of the values
             lie between Q1 and Q3,
             i.e. between 9 and 31
             The maximum of the bottom
             25% of the values is 9.
             The minimum of the top
             80% of the values = P80 = 34.
[ Q 3. ]
  
     
      3.4  Arrange : 31 ; 32 ; 34 ; 35 ; 35 ; 36 ; 37 ;
                             40 ; 41 ; 41 ; 41 ; 42 ; 44
                             48 ; 49 ; 52
$$ \hspace*{13 mm}\mathrm{Median\ = 40,5 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{1})\ = \frac{1(16 + 1)}{4}\ = 4,25 } $$
$$ \hspace*{13 mm}\mathrm{4^{th}\ value\ and\ thus\ Q_{1}\ =\ 35 } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{2(16 + 1)}{4}\ = 8,5 } $$

$$ \hspace*{13 mm}\mathrm{between\ 8^{th}\ and\ 9^{th}\ values } $$
$$ \hspace*{13 mm}\mathrm{p(Q_{2})\ = \frac{40 + 41}{2}\ = 40,5 } $$

$$ \hspace*{13 mm}\mathrm{p(Q_{3})\ = \frac{3(16 + 1)}{4}\ = 12,75 } $$
$$ \hspace*{13 mm}\mathrm{13^{th}\ value\ and\ thus\ Q_{3}\ =\ 44 } $$
$$ \hspace*{13 mm}\mathrm{IQR\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{22 mm}\mathrm{= 44\ ─ 35\kern{2mm}= 9 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(16 + 1)}{100}\ = 3,4 } $$

$$ \hspace*{13 mm}\mathrm{3^{th}\ value\ and\ thus\ P_{20}\ =\ 34 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(16 + 1)}{100}\ = 4,25 } $$

$$ \hspace*{13 mm}\mathrm{4^{th}\ value\ and\ thus\ P_{25}\ =\ 35 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(16 + 1)}{100}\ = 12,75 } $$

$$ \hspace*{13 mm}\mathrm{13^{th}\ value\ and\ thus\ P_{75}\ =\ 44 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(16 + 1)}{100}\ = 13,6 } $$

$$ \hspace*{13 mm}\mathrm{14^{th}\ value\ and\ thus\ P_{80}\ =\ 48 } $$

             The middle 50% of the values
             lie between Q1 and Q3,
             i.e. between 35 and 44
             The maximum of the bottom
             25% of the values is 35.
             The minimum of the top
             80% of the values = P80 = 48.
[ Q 3. ]
  
  
  
Antwoorde / Answers  4
  
             Dalk mag die volgende nuttig wees :
             Daar is 5 verskillende waardes en
             dus kan ons skryf : a ; b ; c ; d ; e
             Ons weet ook dat die grootste
             waarde 28 is, die mediaan is
             24 en die omvang is 10.
             Grootste waarde is 28, sodat e = 28.
             Die mediaan is presies in die middel
             en daarom c = 24. Ons kan die
             waardes neerskryf :
                                    a ; b ; 24 ; d ; 28
  
     
      4.1  Die kleinste waarde =
                         = grootste waarde ─ omvang
$$ \hspace*{29 mm}\mathrm{a\ = e\ ─\ omvang } $$ $$ \hspace*{33 mm}\mathrm{= 28\ ─\ 10\kern{2mm}=\kern{2mm}18 } $$ $$ \hspace*{13 mm}\mathrm{Waardes\ :\ 18\ ;\ b\ :\ 24\ :\ d\ :\ 28 } $$


[ V 4. ]
  
     
      4.2  Daar is geen modus nie
             want al die waardes verskil.
[ V 4. ]
  
     
      4.3  Die waardes : 18 ; b ; 24 ; d ; 28
             Daar is 2 waardes kleiner as die
             mediaan en 2 groter as die mediaan.
[ V 4. ]
  

     
      4.4  Daar is slegs een waarde, d,
             tussen die mediaan, 24, en
             die grootste waarde, 28.
[ V 4. ]
  
     
      4.5  Benaderde som =
             gemiddelde X aantal waardes.
             som = 23,2 X 5
                     = 160,0
[ V 4. ]
  
             Perhaps the following may help :
             There are 5 different values and thus
             we can write : a ; b ; c ; d ; e
             We also know that the greatest
             value is 28, the median is 24
             and the range is 10. The
             greatest value is 28, thus e = 28.
             The median is exactly in the middle
             and thus c = 24. We can write the
             values : a ; b ; 24 ; d ; 28
  
  
     
      4.1  The smallest value =
                          greatest value ━ range
$$ \hspace*{29 mm}\mathrm{a\ = e\ ─\ range } $$ $$ \hspace*{33 mm}\mathrm{= 28\ ─\ 10\kern{2mm}=\kern{2mm}18 } $$ $$ \hspace*{13 mm}\mathrm{Values\ :\ 18\ ;\ b\ :\ 24\ :\ d\ :\ 28 } $$


[ Q 4. ]
  
     
      4.2  There is no mode because
             all the values differ.
[ Q 4. ]
  
     
      4.3  There values : 18 ; b ; 24 ; d ; 28
             There are 2 values smaller than
             the median and 2 greater than
             the median.
[ Q 4. ]
  
     
      4.4  There is only one value. d. between
             the median, 24, and the
             greatest values, 24.
[ Q 4. ]
  
     
      4.5  Approximate sum =
             mean X number of values
             sum = 23,2 X 5
                     = 160,0
[ Q 4. ]
  
  
  
Antwoorde / Answers  5
  
             Hulplys  : a ; b ; c ; d ; e ; f ; g
             Daar is 7 waardes en ons weet
             dat die mediaan 17 is, die kleinste
             waarde is 14, die modus is 19 en
             die omvang(variasiewydte) is 6
             Kleinste waarde is 14, sodat a = 14.
             Daar is 7 waardes en die mediaan
             is presies in die middel
             sodat f = 17.
  
  
     
      5.1  Die omvang =
              = (grootste ─ kleinste) waarde
                      6 = g ─ a
                      g = 6 + a
                      g = 6 + 14
                      g = 20
             Waardes :
                            14 ; b ; c ; 17 ; e ; f ; 20
[ V 5. ]
  
     
      5.2  Die mediaan is presies in die middel
             sodat daar 3 waardes kleiner as die
             mediaan en 3 waardes groter as
             die mediaan is.
[ V 5. ]
  
     
      5.3  Benaderde som =
             = gemiddelde X aantal datawaardes
$$ \hspace*{29 mm}\mathrm{som\ = 17,143\ X\ 7 } $$ $$ \hspace*{38 mm}\mathrm{= 120 } $$

[ V 5. ]
  
     
      5.4  Daar is twee waardes
             tussen 14 en 17.
[ V 5. ]
  
     
      5.5  Daar is drie waardes groter as 17
             en 20 is die grootste waarde.
             Daarom is daar twee waardes
             tussen 17 en 20. Die modus is 19
             en dit is die waarde met die
             hoogste frekwensie (kom
             die meeste voor)
             Daar is 2 onbekende waardes en
             daarom moet hulle 19 en 19 wees.
[ V 5. ]
  
             List to assist  : a ; b ; c ; d ; e ; f ; g
             There are 7 values and we know
             that the median is 17, the
             smallest value is 14, the mode is
             19 and the range is 6 The smallest
             value is 14 and thus a = 14.
             There are 7 valueses and the
             median is exactly in the middle
             so that f = 17.
  

     
      5.1  The range =
                      greatest value ━ smallest value
                      6 = g ━ a
                      g = 6 + a
                      g = 6 + 14
                      g = 20
                      Values :
                      14 ; b ; c ; 17 ; e ; f ; 20
[ V 5. ]
  
     
      5.2  The median is exactly in the middle
             so that there are 3 values smaller
             than the median and 3 values
             greater than the median.
[ V 5. ]
  
     
      5.3  Approximate sum =
             = mean X number of data values
$$ \hspace*{29 mm}\mathrm{som\ = 17,143\ X\ 7 } $$ $$ \hspace*{38 mm}\mathrm{= 120 } $$

[ V 5. ]
  
     
      5.4  There are two values
             between 14 and 17.
[ V 5. ]
  
     
      5.5  There are three values greater than
             17 and 20 is the greatest value.
             Thus there are two values
             between 17 and 20. The modus is
             19 and it is the value that has the
             highest frequency (appears most)
             There are 2 unknown values and
             therefore they have to be 19 and 19.
[ V 5. ]
  
  
  
Antwoorde / Answers  6
  
             Hulplys  : a ; b ; c ; d ; e ; f ; g ; h ; i
             Daar is 9 waardes en ons
             weet dat die mediaan 17 is, die
             kleinste waarde is 7, die modus is
             27 en die omvang (variasiewydte)
             is 26 . Kleinste waarde is 7, sodat
             a = 7. Daar is 9 waardes en die
             mediaan is presies in die middel
             sodat e = 17.
  
  
     
      6.1  Die omvang =
             (grootste ─ kleinste) waarde
                       26 = i ─ a
                          i = 7 + 26
                            = 33
             Waardes :
                     7 ; b ; c ; d ; 17 ; f ; g ; h ; 20
[ V 6. ]
  
     
      6.2  Die mediaan is presies in die
             middel sodat daar 4 waardes
             kleiner as die mediaan en 4
             groter as die mediaan is.
[ V 6. ]
  
     
      6.3  Die gemiddelde is 18 en al die
             waardes is heelgetalle sodat die
             gemiddelde, 18, dus die eerste
             waarde na die mediaan, 17, is.
             Die waardes is :
             7 ; b ; c ; d ; 17 ; 18 ; g ; h ; 33
             Daar is dus 3 waardes groter
             as die gemiddelde.

[ V 6. ]
     
      6.4  Al die waardes is heelgetalle sodat
             die mediaan, 17, die
             eerste getal groter as 16 is.
             Die modus is 27. Daar is
             twee getalle tussen 18 en
             33 sodat g = h = 27.
             Die getalle groter as 16 is
             17 ; 18 ; 27 ; 27 ; 33
[ V 6. ]
  
     
      6.5  Die benaderde som =
             gemiddelde X aantal datawaardes
$$ \hspace*{32 mm}\mathrm{som\ = 18\ X\ 9 } $$ $$ \hspace*{41 mm}\mathrm{= 162 } $$

[ V 6. ]
  
     
      6.6  Die som van die waardes in
             6.4 is 122.
[ V 6. ]
  
     
      6.7  Som = (som in 6.5) ━ (som in 6.6)
                     = 162 ━ 122.
                     = 40
[ V 6. ]
  
             List  : a ; b ; c ; d ; e ; f ; g ; h ; i
             There are 9 values and we know
             that the median is 17, the
             smallest value is 7, the mode is
             27 and the range is 26.
             Smallest value is 7, so that a = 7.
             There are 9 values and the median
             is exactly in the middle
             so that e = 17.
  
  
     
      6.1  The range =
                  greatest value ━ smallest value
                      26 = i ━ a
                      i = 7 + 26
                      i = 33
                      Values :
                7 ; b ; c ; d ; 17 ; f ; g ; h ; 20
[ Q 6. ]
  
     
      6.2  The median is exactly in the middle
             so that there are 4 values smaller
             than the median and 4 values
             greater than the median.
[ Q 6. ]
  
     
      6.3  The mean is 18 and all the
             values are integers so that the
             mean, 18, is thus the
             first value after the median, 17.
             The values are :
             7 ; b ; c ; d ; 17 ; 18 ; g ; h ; 33
             There are therefore 3 values
             greater than the mean.
[ Q 6. ]
  
     
      6.4  All the values are integers so
             that the median, 17, is the
             fist number greater than 16.
             The mode is 27. There are
             two values between 18 and
             33 so that g = h = 27
             The numbers greater than 16 are
             17 ; 18 ; 27 ; 27 ; 33
[ Q 6. ]
  
     
      6.5  Approximate sum =
             mean X number of data values
$$ \hspace*{32 mm}\mathrm{sum\ = 18\ X\ 9 } $$ $$ \hspace*{41 mm}\mathrm{= 162 } $$

[ Q 6. ]
  

       6.6  The sum of the valuesin 6.4 is 122
                                                 [ Q. 6 ]
  
  

       6.7  Sum = (sum in 6.5) - (sum in 6.6)
                     = 162 - 122
                     = 40

                                                 [ Q. 6 ]
  
  


  
  
Antwoorde / Answers  7
  
             Hulplys  : 
                     a ; b ; c ; d ; e ; f ; g ; h ; i ; j
             Daar is 10 waardes en ons
             weet dat die mediaan 9 is, die
             kleinste waarde is 2, die modus is
             8 en die omvang (variasiewydte) is 19
             Kleinste waarde is 2, sodat a = 2.
             Daar is 10 waardes en die mediaan
             is presies in die middel
             sodat die mediaan, 9, tussen die
             5de en 6de waardes is.
             Waardes  :
              a ; b ; c ; d ; e ; (9) ; f ; g ; h ; i ; j
  
             9 is die mediaan
  
  
       7.1   Die omvang =
                 (grootste ━ kleinste) waarde
                      19  =  j  ━  a
                      j  =  2  +  19
                        i  =  21
              Waardes :
              2 ; b ; c ; d ; e ; (9) ; f ; g ; h ; i ; 21
                                                       [ V 7 ]
  
  
       7.2  Die mediaan is presies in die middel
             sodat daar 5 waardes kleiner as die
             mediaan is en 5 waardes is groter
             as die mediaan.
                                                       [ V 7 ]
  
  
       7.3  Ja. Die mediaan is die middelste
              waarde (midmost) en dus is die
              helfte of 50% van die waardes
              kleiner as die mediaan.
                                                      [ V 7 ]
  
  
       7.4  Die gemiddelde = 10,5.
              As f = 10, dan is 4 waardes groter
              as die gemiddelde. As f = 11 is
              daar 5 waardes groter as
              die gemiddelde.
                                                       [ V 7 ]
  
  
       7.5  Benaderde som  =
               gemiddelde X aantal datawaardes
               som  =  10,5  X  10
                       =  105
                                                       [ V 7 ]
  
  
      
7.6  Die modus het 'n waarde van 8
              met 'n frekwensie van 2.
              Volgens ons tabel is daar 5
              waardes kleiner as 9.
              Twee van die waardes is 8.
              Daar is dus 3 waardes kleiner .
              as 8. Drie waardes is kleiner
              as die modus.
                                                       [ V 7 ]
  
  
        List  : a ; b ; c ; d ; e ; f ; g ; h ; i ; j
        There are 10 values and we know
        that the median is 9, the smallest
        value is 2, the mode is 8 and the
        range is 19.
        Smallest value is 2, so that a = 2.
        There are 10 values and the median
        is exactly in die middle so
        that the median, 9, lies between
        the 5th and 6th values.
        Values  :
         a ; b ; c ; d ; e ; (9) ; f ; g ; h ; i ; j
        9 is the median
  
  


       7.1   The range =
                greatest value ━ smallest value
                      19  =  j  ━  a
                      j  =  2  +  19
                        i  =  21
              Values :
              2 ; b ; c ; d ; e ; (9) ; f ; g ; h ; i ; 21
  
                                                       [ Q 7 ]
  
       7.2  The median is the midmost value
             so that there are 5 values smaller
             than the median and 5 values greater
             than the median.
                                                       [ Q 7 ]
  
  
       7.3  Yes. The median is the midmost
               value and thus half or 50% of
               the values are smaller than
               the median.
                                                       [ Q 7 ]
  
  
       7.4  The mean = 10,5.
              If f = 10, then 4 values are greater
              than the mean. If f = 11 then 5
              values are greater than
              yhe mean.
                                                       [ Q 7 ]
  
  
       7.5  Approximate sum  =
               mean X number of data values
               sum  =  10,5  X  10
                       =  105
                                                       [ Q 7 ]
  
  
       7.6  The mode has a value of 8 with
              a frequency of 2.
              According to our list 5 values
              are smaller than 9.
              Two of these values are 8.
              Thus 3 values are smaller
              than 8. Thee values are smaller
              than the mode.
                                                       [ Q 7 ]
  
  
  
Antwoorde  / Answers  8  
  
       8.1   Totale aantal lopies =
  
                    gemiddelde X aantal wedstryde
  
               = 23,1 X 7
  
               = 161,7  = 162
  
               Lopies in 7de wedstryd = 162 ━ 131
  
                                                  = 31
  
                                                       Vr. / Qu. 8
  
  
       8.2  Nee. Vier tellings onder 10 en
  
             5 tellings is onder die gemiddelde.
  
                                                       Vr. / Qu. 8
  
  
       8.3  Gemiddelde = 138  ÷  7
  
                                = 19,7
  
                                                       Vr. / Qu. 8
  
  
       8.4  Die tweede kolwer.
  
              Sy tellings was gereeld
  
              naby aan sy gemiddelde.
  
                                                       Vr. / Qu. 8
  
  
       8.1   Total runs = mean X matches
  
                              = 23,1 X 7
  
                              = 161,7   = 162
  
               Runs in 7th match = 162 ━ 131
  
                                          = 31
  
  
                                                       Vr. / Qu. 8
  
  
       8.2  No. Four scores were below 10 and
  
             5 scores are below the mean.
  
                                                       Vr. / Qu. 8
  
  
       8.3  Mean = 138  ÷  7
  
                       = 19,7
  
                                                       Vr. / Qu. 8
  
  
       8.4  The second batnman.
  
              His scores were consustently
  
              close to his average.
  
                                                       Vr. / Qu. 8
  
  


  
Antwoorde  / Answers  9  
  
       9.1  A = kleinste waarde, minimum
  
               B = eerste kwartiel  Q1
  
               C = tweede kwartiel  Q2, mediaan
  
               D = derde kwartiel  Q3
  
               E = grootste waarde, maksimum
  
                                                       Vr. / Qu. 9
  
  
       9.2.1  Omvang = E ━ A
  
                                                       Vr. / Qu. 9
  
  
       9.2.2 
  
               Interkwartielomvang = D ━ B
  
                                                       Vr. / Qu. 9
  
  
       9.1  A = smallest value, minimum
  
               B = first quartile  Q1
  
               C = second quartile  Q2, median
  
               D = third quartile  Q3
  
               E = greatest value, maximum
  
                                                       Vr. / Qu. 9
  
  
       9.2.1  Range = E ━ A
  
                                                       Vr. / Qu. 9
  
  
       9.2.2 
  
               Inter-quartile range = D ━ B
  
                                                       Vr. / Qu. 9
  
  


  
Antwoorde  / Answers  10  
  
                A = 3        |  minimum
  
                B = 16      |  eerste kwartiel / first quartile
  
                A = 26      |  tweede kwartiel, mediaan / second quartile, median
  
                A = 36      |  derde kwartiel / third quartile
  
                A = 38      |  maksimum / maximum
                                                                                                                    Vr. / Qu. 10
  
  
  
Antwoorde  / Answers  11  
  
                A = 13        |  E - A = 69 (omvang = grootste - kleinste) / range = maximum - minimum
  
                B = 37      |  eerste kwartiel - maksimum onderste 25% / 
                                    first quartile - maximum lowest 25%
  
                C = 55      |  tweede kwartiel, mediaan / second quartile, median
  
                D = 68      |  D - B = 31 /  IKO / IQR  =  Q3 ━ Q1
  
                E = 82      |  maksimum / maximum
                                                                                                                    Vr. / Qu. 11