In hierdie les hersien ons slegs die graad 11 variansie en ... - AdMaths

GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Page 1 of 11
In this lesson we only revise the grade 11 variance and standard deviation. The data
is ungrouped and no frequency tables are used.
You already know that variance and standard deviation (s.d.) are measures of
dispersion / spread.
For ungrouped data without frequency tables
Mean (Average) :
Variance :
Population :
Sample :
x
2
2
s
Standard deviation :
s.
d.
x
n
2
( x – x )
n
2
( x – x )
n – 1
( x – x )
n
2
Another form that is sometimes used
x
n
xi
i 1
n
are x1 , x2 , x3 , … , xn .
2
i
n
( x
1
i
n
– x )
, where the n data values
2
This formula is given on the NSC formula sheet
* See note further on
s.
d.
( x
– x )
Standard deviation = var iance
s.d. =
The standard deviation is therefore the square root of the average squared
deviations from their mean ( x ).
i
n
1
i
n
2
2

GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Page 2 of 11
Remember : x – x (or xi – x ) represents the distance that a data value is away
from the mean ( x ). If the data value is smaller than the average, the
difference is negative and the other way around.
The more widely the data values are spread out, the bigger the standard deviation.
The closer the data values are together (bundled), the smaller the standard
deviation.
So remember that the standard deviation tells us how the data is spread out around
the mean.
The standard deviation uses all the data values and is therefore a better measure of
dispersion than the IQR which only uses 2 data values.
The standard deviation is one of the most common, popular and most noted
measures of dispersion around the mean.
Conclusions and predictions can be made from the standard deviation and variance.
Outliers do have an effect on the standard deviation.
Outliers will increase the standard deviation (data is more spread out).
Example 1
The monthly rainfall (in mm) for Bellville during 2009 was recorded as follows :
20 12 50 125 65 48 35 27 53 84 16 5
Determine, correct to 2 decimals, the
a) mean
b) variance
c) standard deviation of this information.

Solution : ( using TABLES )
x
a) Mean ( x ) = ( OR
n
GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
=
20
540
=
12
Page 3 of 11
12
= 45 mm
x
50
n
xi
i 1
125
n
65
)
48 35
12
b) Variance : Remember that x = 45 mm from (a)
Rainfall (x) in mm x – x ( x – x ) 2
20 -25 625
12 -33 1089
50 5 25
125 80 6400
65 20 400
48 3 9
35 -10 100
27 -18 324
53 8 64
84 39 1521
16 -29 841
5 -40 1600
Som 12998
Variance ( 2 ) =
2
( x – x )
n
12998
=
12
27
53
84
= 1083,17 mm 2
16
5

GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Page 4 of 11
You could also have called the data values x1, x2, x3, …, xi and then have
used
n
2
( xi
– x )
2 i 1
.
n
c) Standard deviation (s.d.) =
=
=
2
2
( x – x )
n
= 1083 , 17
= 32,91 mm
OR
n
2
( xi
– x )
i 1
n
The square root has to be taken, because the unit for variance is mm 2 and not mm.
[ Also notice that you have used ( x – x ) 2 in the table and formula and then
taken the square root afterwards.
Why is this necessary? Why did you not just use x – x ?
That would have given you the correct unit namely mm immediately …
Reason : Add column 2’s values …
I.o.w. ( x – x )
= -25 – 33 + 5 + 80 + 20 + 3 – 10 – 18 + 8 + 39 – 29 – 40
= 0 !!!
And such an answer for spread is meaningless! ]

Solution : ( using CALCULATOR )
GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Page 5 of 11
Mean and standard deviation can be determined much quicker by using statistical
functions on a calculator.
Follow the instructions below for your specific calculator and test your answer.
x = 45 mm and s.
d.
32,
91 mm
Raw data (Ungrouped)
In stat mode
Enter data values
Mean
Standard deviation
(population)
Standard deviation
(sample)
Clear memory
Out of stat mode
x
SHARP EL – 531 WH CASIO FX – 82ES
mode
0
Enter all data values (x1 etc.)
followed by each time.
M+
RCL
Use standard deviation (population)!
1 0
X
RCL X
RCL
SX
0
2nd F CA AC
mode 2 1
Enter all data values followed by
AC
mode 0 mode 1
and when finished.
SHIFT STAT 5 2 =
SHIFT STAT 5 3 =
=

NOTE : WATCH OUT!
GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Page 6 of 11
On the formula sheet of the National Senior Certificate (NSC)
they give the
s.
d.
n
2
( xi
– x )
i 1
n
Here xi does NOT represent the midpoint value as usual, but xi represents
the raw ungrouped data from x1 to xn!!
Therefore it is the same formula used for raw ungrouped data without
using a frequency table!
Example 2 ( Here we compare the standard deviation of 2 groups of data. )
Take the reaction time in a test of 2 groups of learners from 2 different classes.
Gr. 12A
Time (sec)
Gr. 12B
Time (sec)
Determine the
a) mean
b) variance
70 21 60 84 63 55
58 59 45 56 69 66
c) standard deviation of the reaction time of 2 classes.

Solution : ( using TABLES )
a) Gr. 12A
x
70
Gr. 12B
x
21
353
6
58,
8 sec
58
59
353
6
58,
8 sec
GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
60 84
6
45 56
6
63
69
Page 7 of 11
55
66
Coincidently the same mean implies that we can compare them easier.
b) Gr. 12A
Data (x) x – x (x – x ) 2
70 11,2 125,44
21 -37,8 1 428,84
60 1,2 1,44
84 25,2 635,04
63 4,2 17,64
55 -3,8 14,44
Sum 2 222,84

Variance ( 2 ) =
GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
=
Page 8 of 11
2
( x – x )
n
2222,
84
6
= 370,47 sec 2
This causes a problem, because the unit for reaction time is now sec 2 !
Gr. 12B
Data (x) x – x (x – x ) 2
58 -0,8 0,64
59 0,2 0,04
45 -13,8 190,44
56 -2,8 7,84
69 10,2 104,04
66 7,2 51,84
Sum 354,84
Variance ( 2 ) =
=
2
( x – x )
n
354,
84
6
= 59,14 sec 2

c) Gr. 12A
GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Standard deviation = var iance
Gr12B
Page 9 of 11
2
( x x)
n
= 370 , 47
= 19,25
Standard deviation = var iance
2
( x x)
n
= 59 , 14
= 7,69
Solution : ( using CALCULATOR )
Redo the problem and use your calculator this time. Follow the steps on the
diagram. Check your answers.
Gr. 12A Gr. 12B
Mean 58,8 sec 58,8 sec
Standard deviation 19,25 sec 7,69 sec

EXERCISE 3
GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Page 10 of 11
1. The total daily muffin sales at the school’s tuck shop for the past 3 weeks
was as follows :
15 27 25 45 73 18 17 34 57 61 21 19 43 56 89
1.1 Show all calculations and determine the following correct to 2 decimals :
( Note : No statistical functions on your calculator may be used.)
1.1.1 mean
1.1.2 standard deviation
1.2 Now test your answers by using the statistical functions on your calculator.
2. The number of oranges in 2 kg bags was
15 ; 10 ; 12 ; 17 ; 9 ; 13 ; 14 ; 11 ; 15 ; 16 ; 17 ; 14
Determine :
2.1 the mean number of oranges in the bags
2.2 the standard deviation of the number of oranges in the bags.
3. The netball coach must decide which of her goal shooters must play the goal
shooter at the next match. Here follow the mean number of goals per match
for three goal shooters.
Lizelle : 5 8 10 3 7
Suzi : 9 7 4 3 2 1 11
Marthie : 13 4 8 11
3.1 Determine the mean for all three goal shooters.
3.2 Determine the standard deviation for all three goal shooters.
3.3 Who must she choose?

GR. 12 THIRD PAPER
DATA HANDLING (STATISTICS)
VARIANCE AND STANDARD DEVIATION I
LESSON 3
Page 11 of 11
4. Five numbers 12 ; 15 ; 7 ; x and y has a mean of 13 and a standard
deviation of 13. Determine the values of x and y correct to 2 decimals if x
is positive.
5. Gr 12! Advanced! Try this sum with grouped data and frequency tables.
(Lesson 5)
Marilu is a human resource manager. She made a summary of the daily
wages of the workers in a company.
The results were as follows:
Daily wages
(in Rands)
Number of women Number of men
70 x < 75 5 3
75 x < 80 15 5
80 x < 85 25 10
85 x < 90 30 15
90 x < 95 16 20
95 x < 100 14 18
100 x < 105 10 16
105 x < 110 8 20
5.1 Determine the mean daily wages of both groups.
5.2 Determine the standard deviation in daily wages for both groups.
5.3 Is there a big difference in their wages? Discuss.