Mathematics
NQF Level 4
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Mathematics NQF Level 4
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Maths Level 4 Module 12
Level 4 Mathematics study guide, 21 pages. Read online or get the free PDF download.
Subject: MathematicsMultiple grades21 pages
Mathematics. Study guide, 21 pages. Read online or download the PDF.
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- Mathematics
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- Multiple grades
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- TVET
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Data handling and
probability models
Topic 4
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Mathematics NQF Level 4
Use variance and regression
analysis to interpolate and
extrapolate bivariate data
Model 12
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Mathematics NQF Level 4
Learning activity 12.3
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Mathematics NQF Level 4
Scatter plot vid 21
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Learning activity 12.3
Learning activity 12.5
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Mathematics NQF Level 4
Least squares progression vid 22
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Learning activity 12.5
1) The numbers of cars passing an observation
point every 5 minutes during an hour are as
follows:
120; 200; 320; 350; 220; 200; 120; 100; 80;
60; 50; 40
Calculate the following:
1.1) Mean
40 + 50 + 60 + 80 + 100 + 120 + 120 + 200 + 200 + 220 + 320 + 350
x=
12
1860
=
12
x = 155
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Summative assessment – Module 12
1.2) Standard deviation
Mark (x) Deviation from mean or Mark (x) x2
(x − x ) ( x − x )2
40 40 − 155 = −115 13 225 40 1 600
50 50 − 155 = −105 11 025 50 2 500
60 60 − 155 = −95 9 025 60 3 600
80 80 − 166 = −75 5 625 80 6 400
100 100 − 155 = −55 3 025 100 10 000
120 120 − 155 = −35 1 225 120 14 400
120 120 − 155 = −35 1 225 120 14 400
200 200 − 155 = 45 2 025 200 40 000
200 200 − 155 = 45 2 025 200 40 000
220 220 − 155 = 65 4 225 220 48 400
320 320 − 155 = 165 27 225 320 102 400
350 350 − 155 = 195 38 025 350 122 500
0 117 900 1 860 ∑ x 2 =
∑x= 406 200
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Summative assessment – Module 12
1.2) Continued
Σ( x − x ) 2 1
Variance = σ
= nΣ x 2 − ( Σ x ) 2
N n
OR 1
117
= 900 12(406 200) − (1 860) 2
= 12
12 1
=
= 9 825 4 874 400 − 3 459 000
12
σ = 9825 = 99,12
σ = 99,12
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Summative assessment – Module 12
2) Peter scored the following marks out of 10
for ten tests in Mathematics:
4; 8; 3; 7; 5; 4; 6; 2; 9; 4
Calculate the following:
2.1) mean
2+3+ 4+ 4+ 4+5+ 6+ 7 +8+9
x=
10
52
=
10
= 5, 2
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Summative assessment – Module 12
2.2) Standard deviation
Mark (x) Deviation from mean or Mark (x) x2
(x − x ) ( x − x )2
2 2 − 5,2 = −3,2 10,24 2 4
3 3 − 5,2 = −2,2 4,84 3 9
4 4 − 5,2 = −1,2 1,44 4 16
4 4 − 5,2 = −1,2 1,44 4 16
4 4 − 5,2 = −1,2 1,44 4 16
5 5 − 5,2 = −0,2 0,04 5 25
6 6 − 5,2 = 0,8 0,64 6 36
7 7 − 5,2 = 1,8 3,24 7 49
8 8 − 5,2 = 2,8 7,84 8 64
9 9 − 5,2 = 3,8 14,44 9 81
0 45,6 ∑x=52 ∑ x2 =
316
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Summative assessment – Module 12
2.2) Continued
Σ( x − x ) 2 1
Variance = OR σ
= n Σ x 2
− ( Σ x ) 2
n n
45, 6 1
= = 10(316) − (52) 2
10 10
= 4,56 1
= 3160 − 2704
σ = 4.56 10
σ = 2,135 = 2,135
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Summative assessment – Module 12
3) The relationship between the height of a
sample of male students and their shoe size
is given below. Draw a scatter plot of the
data.
Height in 1,70 1,75 1,77 1,80 1,80 1,81 1,85 1,89 1,90 1,92
metres
Shoe size 7 8 9 8 10 9 11 10 11 12
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Summative assessment – Module 12
3) Continued
Figure 12.18
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Summative assessment – Module 12
4) Draw the intuitive line of best fit for the
following data:
x 1 2 3 4 5 6 7 8 9
y 9 8,1 7 5,8 4,9 6 6,9 2,2 1
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Figure 12.19
Summative assessment – Module 12
5) The following table shows the braking
distance and speed in an experiment:
Speed km/h (x) 60 80 100 120 140
Braking distance m (y) 32 33 49 64 72
5.1 Determine the equation of the least squares
regression line.
Speed Braking
x − x y − y ( x − x )( y − y ) ( x − x ) 2
km/h distance m (y)
60 32 −40 −20,2 808 1 600
80 44 −20 −8,2 164 400
100 49 0 −3,2 0 0
120 64 20 11,8 236 400
140 72 40 19,8 792 1 600
Sum 500 261 0 0 2 000 4 000
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Summative assessment – Module 12
5.1 Continued
Σ( x − x )( y − y ) y= a + bx
b=
Σ( x − x ) 2
52, 2= a + 0,5(100)
2 000
=
4 000 =a 52, 2 − 50
= 0,5 = −2, 2
yˆ =
−2, 2 + 0,5 x
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Summative assessment – Module 12
5.2) Represent the data in a scatter plot.
Figure 12.20
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Summative assessment – Module 12
5.3) Use the regression line to predict the
braking distance in metres when the
speed is 160 km/h.
82,2 km from A
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Summative assessment – Module 12
Mathematics NQF Level 2
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Mathematics NQF Level 4
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