WISKUNDE
GRAAD 11
NOG OEFENINGE
Statistiek : antwoorde.
MATHEMATICS
GRAAD 11
NOG OEFENINGE
Statistics : answers.
1.1
|
Punte
p
Marks |
Klas middelpunt
(x)
Class midpoint |
Frekwensie
f
Frequency |
Kum. Frekw.
cf
Cum. Freq. |
f . x |
0 < p ≤ 5 |
2,5 |
4 |
4 |
10 |
5 < p ≤ 10 |
7,5 |
9 |
13 |
67,5 |
10 < p ≤ 15 |
12,5 |
16 |
29 |
200 |
15 < p ≤ 20 |
17,5 |
7 |
36 |
122,5 |
20 < p ≤ 25 |
22,5 |
4 |
40 |
90 |
|
Σ 40
490
1.2
Σf.x
490
Gemiddelde / mean = ──── = ──── = 12,25
n
40
1.3
modale interval / modal interval is 10 < p ≤ 15 {f = 16}
n
40
1.4.1
Posisie van / Position of Q1 = ───
= ────
= 10
4
4
Q1 is die tiende getal en dus lê Q1 in die interval 5 < p ≤ 10
Q1 is the tenth number and thus Q1 lies in die interval 5 < p ≤ 10
2n
2 × 40
1.4.2
Posisie van / Position of Q2 = ───
= ──────
= 20
4
4
Q2 is die 20ste getal en dus lê Q2 in die interval 10 < p ≤ 15
Q2 is the 20th number and thus Q2 lies in die interval 10 < p ≤ 15
3n
3 × 40
1.4.3
Posisie van / Position of Q3 = ───
= ──────
= 30
4
4
Q3 is die 30ste getal en dus lê Q3 in die interval 15 < p ≤ 20
Q3 is the 30th number and thus Q3 lies in die interval 15 < p ≤ 20
90n
90 × 40
1.4.4
Posisie van / Position of P90 = ───
= ──────
= 36
100
100
P90 is die 36ste getal en dus lê P90 in die interval 15 < p ≤ 20
P90 is the 36th number and thus P90 lies in die interval 15 < p ≤ 20
1.5
1.6.1
Q1 = 12; Q1 = 12; Q2 = 12; Q3 = 12
1.6.2
IQR = Q3 − Q1 = 15 − 12 = 3
1.7
minimum = 6; Q1 = 12;
Q2 = 12; Q3 = 12;
maksimum / maximum = 24
1.8
1.9
Nee want die gemiddelde is ongeveer gelyk aan die mediaan. /
No because the mean is approximately equal to the median.
1.10
Minimum = 18 [P90 = 18 ]
2.1
|
|
Klas middelpunt
(x)
Class midpoint |
Frekwensie f
f
Frequency |
Kum. Frekw.
cf
Cum. Freq. |
f . x |
40 < m ≤ 50 |
45 |
2 |
2 |
90 |
50 < m ≤ 60 |
55 |
4 |
6 |
220 |
60 < m ≤ 70 |
65 |
7 |
13 |
455 |
70 < m ≤ 80 |
75 |
5 |
18 |
375 |
80 < m ≤ 90 |
85 |
2 |
20 |
70 |
|
Σ 20
1 310
2.2
Σf.x
1 310
Gemiddelde / mean = ──── = ──── = 65,5
n
20
2.3
modale interval / modal interval is 60 < m ≤ 70 {f = 7}
n
20
2.4.1
Posisie van / Position of Q1 = ───
= ────
= 5
4
4
Q1 is die vyfde getal en dus lê Q1 in die interval 50 < m ≤ 60
Q1 is the fifth number and thus Q1 lies in die interval 5 < m ≤ 10
2n
2 × 20
2.4.2
Posisie van / Position of Q2 = ───
= ──────
= 10
4
4
Q2 is die 10de getal en dus lê Q2 in die interval 60 < m ≤ 70
Q2 is the 10th number and thus Q2 lies in die interval 60 < m ≤ 70
3n
3 × 20
2.4.3
Posisie van / Position of Q3 = ───
= ──────
= 15
4
4
Q3 is die 15de getal en dus lê Q3 in die interval 70 < m ≤ 80
Q3 is the 15th number and thus Q3 lies in die interval 70 < m ≤ 80
30n
30 × 20
2.4.4
Posisie van / Position of P30 = ───
= ──────
= 6
100
100
P30 is die 6de getal en dus lê P30 in die interval 50 < m ≤ 60
P30 is the 6th number and thus P30 lies in die interval 50 < m ≤ 60
8n
8 × 20
2.4.5
Posisie van / Position of D8 = ───
= ──────
= 16
10
10
D8 is die 16de getal en dus lê D8 in die interval 70 < m ≤ 80
D8 is the 16th number and thus D8 lies in die interval 70 < m ≤ 80
2.5
2.6.1
Q1 = 58;
Q2 = 65;
Q3 = 73
2.6.2
IQR = Q3 − Q1
= 73 − 58
= 15
2.7
minimum = 47; Q1 = 58;
Q2 = 65; Q3 = 73;
maksimum / maximum = 83
2.8
2.9
Nee want die gemiddelde is ongeveer gelyk aan die mediaan. /
No because the mean is approximately equal to the median.
3.1
|
|
Frekwensie f
f
Frequency |
Kum. Frekw.
cf
Cum. Freq. |
Klas middelpunt
(x)
Class midpoint |
f . x |
140 < h ≤ 150 |
3 |
3 |
145 |
435 |
150 < h ≤ 160 |
10 |
13 |
155 |
1 550 |
160 < h ≤ 170 |
14 |
27 |
165 |
2 310 |
170 < h ≤ 180 |
12 |
39 |
175 |
2 100 |
180 < h ≤ 190 |
1 |
40 |
185 |
185 |
|
Σ 40
6 580
3.2
Σf.x
6 580
Gemiddelde / mean = ──── = ───── = 164,5
n
40
3.3
3.4
Op grafiek getoon. / Shown on the graph.
3.5
IQR = Q3 − Q1 = 172 − 157 = 15
3.6
Minimum lengte / height = 165 cm [ Q2 = 165 en stel 50% voor. / and represents 50% ]
4.1
|
|
Frekwensie f
f
Frequency |
Kum. Frekw.
cf
Cum. Freq. |
Klas middelpunt
(x)
Class midpoint |
f . x |
5 < t ≤ 10 |
4 |
4 |
7,5 |
30 |
10 < t ≤ 15 |
8 |
12 |
12,5 |
100 |
15 < t ≤ 20 |
6 |
18 |
17,5 |
105 |
20 < t ≤ 25 |
2 |
20 |
22,5 |
45 |
|
Σ 40
280
4.2
Σf.x
280
Gemiddelde / mean = ──── = ───── = 14
n
20
4.3
modale interval / modal interval is 10 < t ≤ 15 {f = 8}
4.4
4.5
Op grafiek getoon. / Shown on the graph.
4.6
IQR = Q3 − Q1 = 17 − 11 = 6
4.7
Minimum tyd / time = 17 [ Q3 = 17 en stel boonste 25% voor. / and represents top 25% ]
5.1
|
|
Frekwensie f
f
Frequency |
Kum. Frekw.
cf
Cum. Freq. |
Klas middelpunt
(x)
Class midpoint |
f . x |
10 < a ≤ 20 |
5 |
5 |
15 |
75 |
20 < a ≤ 30 |
19 |
24 |
25 |
475 |
30 < a ≤ 40 |
25 |
49 |
35 |
875 |
40 < a ≤ 50 |
11 |
60 |
45 |
495 |
|
Σ 40
1 920
5.2
Σf.x
1 920
Gemiddelde / mean = ──── = ───── = 32
n
40
5.3
5.4
Op grafiek getoon. / Shown on the graph.
5.5
IQR = Q3 − Q1 = 38 − 27 = 11
5.6
Die kleinste ouderdom van die oudste 60% van die kopers is 30 jaar. [ D4 = 30 en
stel 40% voor ] /
The youngest age of the elder 60% of the shoppers is 30 years. [ D4 = 30 and
represents 40% ]
6.1
|
|
Frekwensie f
f
Frequency |
Kum. Frekw.
cf
Cum. Freq. |
Klas middelpunt
(x)
Class midpoint |
f . x |
0 < p ≤ 5 |
3 |
3 |
2,5 |
7,5 |
5 < p ≤ 10 |
7 |
10 |
7,5 |
52,5 |
10 < p ≤ 15 |
10 |
20 |
12,5 |
125 |
15 < p ≤ 20 |
12 |
32 |
17,5 |
210 |
20 < p ≤ 25 |
8 |
40 |
22,5 |
180 |
|
Σ 40
575
6.2
Σf.x
575
Gemiddelde / mean = ──── = ───── = 14,375
n
40
6.3
10 leerlinge / 10 pupils.
40 × 25
6.4
Slaagsyfer / Pass mark = 40%, d.i. / i.e. punt / mark = ─────── = 10
100
30 leerlinge het meer as 10 punte. / 30 pupils have more than 10 marks.
30 × 100
Persentasie geslaag / Percentage passed = ─────── = 75%
40
80 × 25
6.5
80% van die punte / of the marks = ─────── = 20
100
8 leerlinge het 80% of meer. / 8 pupils have 80% or more.
6.6
Q2 deel data in twee gelyke dele. Stel dus 50% voor. /
Q2 divides data into two equal parts. Thus, represents 50%.
2n
2 × 40
Q2 = ─── = ───── = 20
4
4
Q2 = 15 punte / marks. Maksimum is dus 15 punte. / Maximum is thus 15 marks.
6.6
Q3 deel die boonste 50% van die data in twee dele. Stel dus laagste 75% en hoogste 25% voor. /
Q3 divides the top 50% of the data into two parts. Thus, represents 75% and top 25%.
3n
3 × 40
Posisie van / Position of Q3 = ─── = ───── = 30
4
4
Trek Q3 op ogief om punt te kry. Q3 = 19 punte. Laagste punt van top 25% is dus 19 punte.
Draw Q3 on the ogive to obtain mark. Q3 = 19 marks. Lowest mark of top 25% is thus 19 marks.
7.1
|
|
Frekwensie f
f
Frequency |
Kum. Frekw.
cf
Cum. Freq. |
Klas middelpunt
(x)
Class midpoint |
f . x |
40 < m ≤ 50 |
2 |
2 |
45 |
90 |
50 < m ≤ 60 |
13 |
15 |
55 |
715 |
60 < m ≤ 70 |
30 |
45 |
65 |
1 950 |
70 < m ≤ 80 |
13 |
58 |
75 |
975 |
80 < m ≤ 90 |
2 |
60 |
85 |
170 |
|
Σ 60
3900
7.2
Σf.x
3 900
Gemiddelde / mean = ──── = ───── = 65
n
60
n
60
7.3
Posisie van / Position of Q2 = ─── = ───── = 15
4
4
Trek Q1 op ogief en lees waarde.
Draw Q1 on ogive and read the value.
Waarde van Q1 = 60.
Value of Q1 = 60.
Maksimum massa van ligste 25% is 60 kg.
Maximum mass of lightest 25% is 60 kg.
7.4
Trek vertikale lyn vanaf (65 ; 0) om ogief te sny
Draw a vertical line from (65 ; 0) to intersect the
en lees die waarde van die kumulatiewe
ogive and read the value of the cumulative
frekwensie op die y-as af.
frequency on the y-axis.
Kumulatiewe frekwensie = 30.
Cumulative frequance = 30.
Daar is dus (60 − 30), d.w.s. 30 leerlinge met
Thus there are (60 − 30), i.e. 30 pupils that
'n massa groter as 65 kg.
have a mass greater than 65 kg.
8.1
Q1 = R7 000 en Q1 stel die maksimum
8.1
Q1 = R7 000 and Q1 represents the maximum
van die onderste 25% voor.
of the lower 25% of the data.
n + 1
27 + 1
Posisie van / Position of Q1 = ──── = ───── = 7
4
4
Die 7de werknemer verdien R7 000 sodat
The 7th employee earns R7 000 so that
6 werknemers minder as R7 000 verdien.
6 employees earn less than R7 000.
8.2
Q2 stel die onderste grens van die boonste
8.2
Q2 represents the minimum of the top 50%.
50% voor.
Dus, minimum = R12 000 en
Thus, minimum =R12 000 and
maksimum = R34 000.
maximum = R34 000.
8.3
IQR = Q3 − Q1 = 18 000 − 7 000 = 11 000
Grense om uitskieters te identifiseer is
Boundaries to identify outliers are given by
Onderste grens = Q1 − 1,5 × IQR
Lower boundary = Q1 − 1,5 × IQR
= 7 000 − 1,5 × 11 000
= 7 000 − 1,5 × 11 000
= −9 500 < 0
= −9 500 < 0
Boonste grens = Q3 + 1,5 × IQR
Top boundary = Q3 + 1,5 × IQR
= 18 000 + 1,5 × 11 000
= 18 000 + 1,5 × 11 000
= 34 500
= 34 500
Elke waarneming lê tussen die grense en
Every observation lies between the boundaries
dus is daar geen uitskieters nie.
and thus there are no outliers.
9.1
Ja, dit is skeef na links of negatief skeef
9.1
Yes, it is skewed to the left or negatively skewed
want die data is meer verspreid links
because the data is more spread out to
van die mediaan.
the left of the median.
9.2
IVK / IQR = Q3 − Q1 = 82 − 62 = 20
Onderste grens / Bottom boundary = Q1 − 1,5 × IQR = 62 − 1,5 × 20 = 62 − 30 = 32
Boonste grens / Top boundary = Q3 + 1,5 × IQR = 82 + 1,5 × 20 = 82 + 30 = 112
Al die waarnemings lê binne die grense en dus is daar geen uitskieters nie.
All the data items lie within the boundaries and thus there are no outliers.
10.1
Ja, dit is skeef na regs of positief
10.1
Yes, it is skewed to the right or positively
skeef want die data is meer verspreid
skewed because the data is more spread
regs van die mediaan.
out to the right of the median.
10.2
IVK / IQR = Q3 − Q1 = 83 − 30 = 53
Onderste grens / Bottom boundary = Q1 − 1,5 × IQR = 30 − 1,5 × 53 = 30 − 79,5 = −49,5
Boonste grens / Top boundary = Q3 + 1,5 × IQR = 83 + 1,5 × 53 = 83 + 79,5 = 162,5
Al die waarnemings lê binne die grense en dus is daar geen uitskieters nie.
All the data items lie within the boundaries and thus there are no outliers.
Σx
484
11.1.1
Gemiddelde / Average = ──── = ───── = 44
n
11
11.1.2
Rangskik in stygende volgorde / Arrange in ascending order :
22; 29; 33; 35; 36; 42; 43; 50; 52; 66; 76
Q1
Q2
Q3
Q1 = 33; Q2 = 42; Q3 = 52
11.1.3
IVK / IQR = Q3 − Q1 = 52 − 33 = 19
11.2
Nee, die data is byna simmetries gerangskik om die mediaan. /
No, the data is arranged very nearly symmetrically about the median.
11.3
Onderste grens / Lower boundary = Q1 − 1,5 × IQR
= 33 − 1,5 × 19 = 33 − 28,5 = 4,5
Boonste grens / Top boundary = Q3 + 1,5 × IQR
= 52 + 1,5 × 19 = 52 + 28,5 = 80,5
Al die data items is binne die grense en dus is daar geen uitskieters nie. /
All data items are within the boundaries and therefore there are no outliers.
12.1
Gemiddelde / Average = 38,35 sakrekenaar / calculator
12.2
Standaardafwyking / Standard deviation = 18,53 sakrekenaar / calculator
12.3
Rangskik in stygende volgorde / Arrange in ascending order :
14; 16; 17; 20; 22; 24; 25; 32; 33; 35; 38; 38; 40; 43; 48; 52; 53; 61; 72; 84
Onderste grens / Lower boundary = gemid. / mean − σ
= 38,35 − 18,53 = 19,82
Boonste grens / Top boundary = gem. / mean + σ
= 38,35 + 18,53 = 56,88
Ouderdomme buite grense / Ages outside the boundaries : 14; 16; 17; 61; 72; 84
Ses mense se ouderdomme is buite een standaardafwyking van die gemiddelde. /
Six people have an age outside the one standard deviation of the mean.
13.1
Gemiddelde / Average = 21 sakrekenaar / calculator
13.2
Standaardafwyking / Standard deviation = 6,83 sakrekenaar / calculator
13.3
Rangskik in stygende volgorde / Arrange in ascending order :
10; 15; 15; 18; 19; 20; 24; 26; 30; 33
Onderste grens / Lower boundary = gemid. / mean − σ
= 21 − 6,83 = 14,17
Boonste grens / Top boundary = gem. / mean + σ
= 21 + 6,83 = 27,83
Afwesighede binne grense / Abscences within the boundaries : 15; 15; 18; 19; 20; 24; 26
Sewe weke se afwesighede is binne een standaardafwyking van die gemiddelde. /
Seven weeks abscences are inside one standard deviation of the mean.
14.1
Gemiddelde / Average = 21,4 sakrekenaar / calculator
14.2
Standaardafwyking / Standard deviation = 3,38 sakrekenaar / calculator
14.3
Rangskik in stygende volgorde / Arrange in ascending order :
15; 18; 19; 20; 21; 22; 23; 24; 25; 27
Onderste grens / Lower boundary = gemid. / mean − σ
= 21,4 − 3,38 = 18,02
Boonste grens / Top boundary = gem. / mean + σ
= 21 + 3,38 = 24,78
Tyd binne grense / Time within the boundaries : 19; 20; 21; 22; 23; 24
Ses hardlopers se tyd is binne een standaardafwyking van die gemiddelde. /
Six runners' time are inside one standard deviation of the mean.