1.1 Arrange
: 16 ; 17 ; 18 ; 19 ; 19 ;
20 ; 21 ; 22 ; 24 ; 26 ; 28
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{230}{11}\ =\ 20,909 } $$
$$ \hspace*{13 mm}\mathrm{Median\ =\ 20\ ┃\ value\ exactly\ in } $$
$$ \hspace*{39 mm}\mathrm{the\ middle } $$
$$ \hspace*{13 mm}\mathrm{Mode\ =\ 19\ ┃\ appears\ most\ often } $$
$$ \hspace*{13 mm}\mathrm{Range\ =\ maximum\ ━\ minimum} $$
$$ \hspace*{25 mm}=\ 28\ ━\ 16\ = 12 $$
1.2 Arrange
: 11 ; 12 ; 14 ; 16 ; 17 ;
17 ; 17 ; 18 ; 21 ; 22
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{165}{10}\ =\ 16,5 } $$
$$ \hspace*{13 mm}\mathrm{If\ there\ is\ an\ even\ number\ of } $$
$$ \hspace*{13 mm}\mathrm{values\ the\ median\ is\ the } $$
$$ \hspace*{13 mm}\mathrm{mean\ of\ the\ two\ midmost\ values } $$
$$ \hspace*{13 mm}\mathrm{(those\ nearest\ to\ the\ centre) } $$
$$ \hspace*{13 mm}\mathrm{Mediaan\ =\ \frac{17 + 17}{2}\ = 17 } $$
$$ \hspace*{13 mm}\mathrm{Mode\ =\ 17\ ┃\ appears\ 3\ times } $$
$$ \hspace*{13 mm}\mathrm{Range\ = 22\ ━\ 11\ = 11 } $$
1.3 Arrange
: 38 ; 57 ; 58 ; 59 ; 60 ;
60 ; 63
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{395}{7}\ =\ 56,429 } $$
$$ \hspace*{13 mm}\mathrm{Median\ = 59\ midmost\ value } $$
$$ \hspace*{13 mm}\mathrm{Mode\ =\ 60\ ┃\ appears\ 2\ times } $$
$$ \hspace*{13 mm}\mathrm{Range\ = 63\ ━\ 38\ = 25 } $$
1.4 Arrange
: 49 ; 50 ; 50 ; 51 ; 51 ;
52 ; 53 ; 58 ; 80
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{494}{9}\ =\ 54,889 } $$
$$ \hspace*{13 mm}\mathrm{Median\ = 51\ midmost\ value } $$
$$ \hspace*{13 mm}\mathrm{Mode\ =\ 50\ and\ 60\ ┃\ both\ appear } $$
$$ \hspace*{48 mm}\mathrm{2\ times } $$
$$ \hspace*{13 mm}\mathrm{Range\ = 80\ ━\ 49\ = 31 } $$
1.5 Arrange
: 14 ; 18 ; 23 ; 27 ; 30 ;
38
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{150}{6}\ =\ 25 } $$
$$ \hspace*{13 mm}\mathrm{Median\ = \frac{23 + 27}{2}\kern4mm=\kern2mm25 } $$
$$ \hspace*{13 mm}\mathrm{No\ mode\ ┃\ all\ values\ differ } $$
$$ \hspace*{13 mm}\mathrm{Range\ = 38\ ━\ 14\ = 24 } $$
1.6 Arrange
: 13 ; 19 ; 19 ; 19 ; 19 ;
21 ; 23
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{133}{7}\ =\ 19 } $$
$$ \hspace*{13 mm}\mathrm{Median\ = 19\kern2mmmidmost\ value } $$
$$ \hspace*{13 mm}\mathrm{Mode\ = 19 ┃\ appears\ 4\ times } $$
$$ \hspace*{13 mm}\mathrm{Range\ = 23\ ━\ 13\ = 10 } $$
1.7 Arrange
: 4 ; 7 ; 8 ; 11 ; 11 ;
15 ; 15 ; 15 ; 17 ; 23
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{128}{10}\ =\ 12,8 } $$
$$ \hspace*{13 mm}\mathrm{Median\ = \frac{11 + 15}{2}\kern2mm=\kern2mm13 } $$
$$ \hspace*{13 mm}\mathrm{Mode\ = 19 ┃\ appears\ 3\ times } $$
$$ \hspace*{13 mm}\mathrm{Range\ = 25\ ━\ 4\ = 21 } $$
1.8 Arrange
: 30 ; 31 ; 31 ; 32 ; 33 ;
34 ; 37 ; 37 ; 37
34 ; 37 ; 37 ; 37
$$ \hspace*{13 mm}\mathrm{Mean\ =\ \frac{302}{9}\ =\ 33,556 } $$
$$ \hspace*{13 mm}\mathrm{Median\ = 33 } $$
$$ \hspace*{13 mm}\mathrm{Mode\ = 19 ┃\ appears\ 3\ times } $$
$$ \hspace*{13 mm}\mathrm{Range\ = 37\ ━\ 30\ = 7 } $$
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.
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.
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.
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.
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.
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.
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
3
rd and 4
th 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 7
th value and
the 3
rd quartile
is exactly halfway between the
10
th and 11
th data values.
The value of Q
2 = 23 and
$$ \hspace*{13 mm}\mathrm{Q_{3}\ = \frac{35 + 36}{2}\ = 35,5 } $$
$$ \hspace*{13 mm}\mathrm{Interquartile range (IQR)\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{32 mm}\mathrm{IQR\ = Q_{3}\ ─ Q_{1} } $$
$$ \hspace*{40 mm}\mathrm{=\ 35,5\ ─ 15,5 } $$
$$ \hspace*{40 mm}\mathrm{\ =\ 20 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{20})\ = \frac{20(13 + 1)}{100}\ = 2,8 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{25})\ = \frac{25(13 + 1)}{100}\ = 3,5 } $$
P
20 is therefore the 3
rd value
and P
25 is exactly halfway between
the 3
rd and 4
th data values.
The value of P
20 = 15 and that
$$ \hspace*{13 mm}\mathrm{of\ P_{25}\ = \frac{15 + 16)}{2}\ = 15,5 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{75})\ = \frac{75(13 + 1)}{100}\ = 10,5 } $$
$$ \hspace*{13 mm}\mathrm{p(P_{80})\ = \frac{80(13 + 1)}{100}\ = 11,2 } $$
P
75 is exactly halfway between the
10
th and 11
th data values
and P
80 is the 11
th data value.
$$ \hspace*{13 mm}\mathrm{Value\ of\ P_{75}\ = \frac{35 + 36)}{2}\ = 35,5 } $$
and the value of P
80 is 36
The middle 50% of the values
lie between Q
1 and Q
3,
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 = P
80 = 36.
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 Q
1 and Q
3,
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 = P
80 = 92.
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 Q
1 and Q
3,
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 = P
80 = 34.
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 Q
1 and Q
3,
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 = P
80 = 48.
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
$$ \hspace*{29 mm}\mathrm{a\ = e\ ─\ range } $$
$$ \hspace*{33 mm}\mathrm{= 28\ ─\ 10\kern{2mm}=\kern{2mm}18 } $$
$$ \hspace*{13 mm}\mathrm{Values\ :\ 18\ ;\ b\ :\ 24\ :\ d\ :\ 28 } $$
4.2 There is no mode because
all the values differ.
4.3 There values : 18 ; b ; 24 ; d ; 28
There are 2 values smaller than
the median and 2 greater than
the median.
4.4 There is only one value. d. between
the median, 24, and the
greatest values, 24.
4.5 Approximate sum =
mean X number of values
sum = 23,2 X 5
= 160,0
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
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.
5.3 Approximate sum =
= mean X number of data values
$$ \hspace*{29 mm}\mathrm{som\ = 17,143\ X\ 7 } $$
$$ \hspace*{38 mm}\mathrm{= 120 } $$
5.4 There are two values
between 14 and 17.
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.
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
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.
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.
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
6.5 Approximate sum =
mean X number of data values
$$ \hspace*{32 mm}\mathrm{sum\ = 18\ X\ 9 } $$
$$ \hspace*{41 mm}\mathrm{= 162 } $$
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 ]
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 5
th and 6
th 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 ]
8.1 Totale aantal lopies
=
gemiddelde X aantal wedstryde
= 23,1 X 7
= 161,7 = 162
Lopies in 7
de 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 7
th 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
9.1 A
= kleinste waarde, minimum
B
= eerste kwartiel Q
1
C
= tweede kwartiel Q
2, mediaan
D
= derde kwartiel Q
3
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
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 = Q
3 ━ Q
1
E
= 82 | maksimum / maximum
Vr. / Qu. 11