Transcript Book 6 Chapter 21 Measures of Dispersion

21
Measures of Dispersion
Case Study
21.1 Range and Inter-quartile Range
21.2 Box-and-whisker Diagrams
21.3 Standard Deviation
21.4 Applications of Standard Deviation
21.5 Effects on the Dispersion with a Change in Data
Chapter Summary
Case Study
Both divers get the same total mark.
But it seems that A’s performance
is less consistent. Is that true?
You need to know the meaning
of dispersion first.
Although both drivers got the same total mark, it does not mean that
both of them had a consistent performance for all ten dives.
If we plot a broken line graph for both
divers, we find that the marks of diver A
fluctuate more than those of diver B.
In the above example, we consider the
spread of the data.
In this chapter, we will learn how to
represent this by statistical methods.
P. 2
21.1 Range and Inter-quartile Range
In junior forms, we learnt three measures of central tendency
of a set of data, namely mean, median and mode. However,
these measures tell us only limited information about the data.
Consider two boxes of apples A and B.
The weight (in g) of each apple in each box is given below:
Box A: 100, 100, 100, 104, 110, 110
Box B: 85, 95, 100, 104, 116, 124
The mean and the median of the weights for both boxes of apples is 104 g
and 102 g respectively, and the weights of the apples in Box B are more
widely spread than those in Box A.
The spread or variability of data is called the dispersion of the data.
In this chapter, we are going to learn the following measures of dispersion:
1. Range
2. Inter-quartile range
3. Standard deviation
P. 3
21.1 Range and Inter-quartile Range
A. Range
The range is a simple measure of the dispersion of a set of
data.
1. For ungrouped data, the range is the difference between the
largest value and the smallest value in the set of data.
Range  Largest value – Smallest value
2.
For grouped data, the range is the difference between the highest
class boundary and the lowest class boundary
Range  Highest class boundary – Lowest class boundary
P. 4
21.1 Range and Inter-quartile Range
A. Range
Example 21.1T
The weights (in g) of eight pieces of meat are given below:
210, 230, 245, 180, 220, 240, 175, 195
(a) Find the range of the weights.
(b) If the meat is sold at $3 per 100 g, find the range of the prices of
the meat.
Solution:
(a) Range  (245  175) g
 70 g
70
100
 $2.1
(b) Range of the prices  $3 
P. 5
21.1 Range and Inter-quartile Range
A. Range
Example 21.2T
The following table shows the weights of the boys in S6A.
Weight (kg)
50 – 54
55 – 59
60 – 64
65 – 69
70 – 74
Frequency
3
6
8
5
2
(a) Write down the upper class boundary of the class 70 kg – 74 kg.
(b) Write down the lower class boundary of the class 50 kg – 54 kg.
(c) Hence find the range of the weights.
Solution:
(a) 74.5 kg
(b) 49.5 kg
(b) Range  (74.5  49.5) kg
 25 kg
P. 6
21.1 Range and Inter-quartile Range
B. Inter-quartile Range
When a set of data is arranged in ascending order of magnitude,
the quartiles divide the data into four equal parts.
Full set of data arranged in order of magnitude
25% of data
25% of data
Q1
25% of data
Q2
25% of data
Q3
Q1 : lower quartile  25% of data less than it
Q2 : median  50% of data less than it
Q3 : upper quartile  75% of data less than it
Q1, Q2 and Q3 are also called
the first, the second and the
third quartiles respectively.
The inter-quartile range is defined as the difference between the
upper quartile and the lower quartile of the set of data.
Inter-quartile range  Q3 – Q1
P. 7
21.1 Range and Inter-quartile Range
B. Inter-quartile Range
Example 21.3T
The marks of 13 boys in a Chinese test are recorded below:
72 78 80 65 62 62 78
81 84 70 68 69 60
(a) Arrange the marks in ascending order.
(b) Find the median mark.
(c) Find the range and the inter-quartile range.
Solution:
(a) Arrange the marks in ascending order:
60, 62, 62, 65, 68, 69, 70, 72, 78, 78, 80, 81, 84
(b) Median  70
62  65
78  80
(c) Range  84 60
Q3 
Q1 
 24
2
 63.5
2
 79
 Inter-quartile range  79 63.5
 15.5
P. 8
21.1 Range and Inter-quartile Range
B. Inter-quartile Range
Example 21.4T
The cumulative frequency polygon shows the
lifetimes (in hours) of 80 bulbs.
(a) Find the range of the lifetimes.
(b) Find the median lifetime.
(c) Find the inter-quartile range.
Solution:
(a) Range  (780  140) hours
 640 hours
(b) From the graph,
Median  420 hours
(c) Q1  240 hours, Q3  600 hours
 Inter-quartile range  (600  240) hours
 360 hours
P. 9
21.1 Range and Inter-quartile Range
B. Inter-quartile Range
Example 21.5T
Consider the ages of passengers in two mini-buses.
Mini-bus A: 18, 24, 25, 19, 12, 10, 34, 39, 45, 23, 34, 40, 24, 28
Mini-bus B: 23, 26, 28, 32, 38, 34, 19, 26, 29, 32, 35, 30, 29, 22
By comparing the ranges and the inter-quartile ranges of the ages,
determine which group of passengers has a larger dispersion of ages.
Solution:
For Mini-bus A,
For Mini-bus B,
the range  45  10  35
the range  38  19  19
For Mini-bus A,
For Mini-bus B,
the inter-quartile range  34  19
the inter-quartile range  32  26
 15
6
Since the range and the inter-quartile range of the ages of passengers
of mini-bus A are larger, the passengers on mini-bus A have a larger
dispersion.
P. 10
21.2 Box-and-whisker Diagrams
A box-and-whisker diagram is a statistical diagram that
provides a graphical summary of the set of data by showing the
quartiles and the extreme values of the data.
The difference between the
two end-points of the line is
the range. The length of the
box is the inter-quartile range.
A box-and-whisker diagram shows the greatest value, the least value, the
median, the lower quartile and the upper quartile of a set of data.
P. 11
21.2 Box-and-whisker Diagrams
Example 21.6T
The following box-and-whisker diagram shows the number of family
members of a class of students.
(a) Find the median and the range of the number of family members.
(b) Find the inter-quartile range.
Solution:
(a)
Median  2.5
Maximum value  6
and minimum value  1
Range  61
5
(b) Q1  2 and Q3 
4
Inter-quartile range  4  2
2
P. 12
21.2 Box-and-whisker Diagrams
Example 21.7T
The following shows the measurements of the waists (in inches) of
the students in a class.
Girls: 27 26 25 23 26 28 32 24 25 29 30 28 22 25 26
Boys: 32 32 30 29 28 25 28 26 29 31 32 34 28 27 28
(a) Find the median, the lower quartile and the upper quartile of the
waists for both boys and girls.
(b) Draw box-and-whisker diagrams of their waist measurements
on the same graph paper.
Solution:
(a) Arrange the measurements in ascending order:
Girls:22, 23, 24, 25, 25, 25, 26, 26, 26, 27, 28, 28, 29, 30, 32
Boys: 25, 26, 27, 28, 28, 28, 28, 29, 29, 30, 31, 32, 32, 32, 34
For girls, median  26 inches, Q1  25 inches and Q3  28 inches
For boys, median  29 inches, Q1  28 inches and Q3  32 inches
P. 13
21.2 Box-and-whisker Diagrams
Example 21.7T
The following shows the measurements of the waists (in inches) of
the students in a class.
Girls: 27 26 25 23 26 28 32 24 25 29 30 28 22 25 26
Boys: 32 32 30 29 28 25 28 26 29 31 32 34 28 27 28
(a) Find the median, the lower quartile and the upper quartile of the
waists for both boys and girls.
(b) Draw box-and-whisker diagrams of their waist measurements
on the same graph paper.
Solution:
(b) For boys, minimum  25 inches
maximum  34 inches
For girls, minimum  22 inches
maximum  32 inches
Refer to the figure on the right.
P. 14
21.3 Standard Deviation
A. Standard Deviation for Ungrouped Data
Standard deviation describes how the spread out of the data
are around the mean. It is usually denoted by .
Consider a set of ungrouped data x1, x2, …, xn.
Standard deviation,  
( x1  x ) 2  ( x2  x ) 2  ... ( xn  x ) 2
n
n

 ( xi  x)2 ,
i 1
n
where x is the mean and n is the total number of data.
Notes:
The quantity  2 is called the variance of the data.
_
(xi – x) is the deviation of the
ith data from the mean.
P. 15
21.3 Standard Deviation
A. Standard Deviation for Ungrouped Data
Example 21.8T
Six students joined the inter-school cross-country race. The times taken
(in min) to complete the race are recorded below:
45, 46, 49, 50, 52, y
If the mean time is 49.5 min, find
(a) the value of y and
(b) the standard deviation of the times taken.
(Give the answer correct to 3 significant figures.)
Solution:
45  46  49  50  52  y
6
y  55
(a) 49.5 
6
(b)
( xi  x)2  (45  49.5)2 2(46  49.5)2 2(49  49.5)2
 (50  49.5)  (52  49.5)  (55  49.5)2
 20.2512.25 0.25 0.25 6.25 30.25
 69.5
69.5
Standard deviation 
min  3.40 min (cor. to 3 sig. fig.)
6
i 1
P. 16
21.3 Standard Deviation
A. Standard Deviation for Ungrouped Data
Example 21.9T
The following table shows the marks of Eric in five tests of two subjects.
Test 1 Test 2 Test 3 Test 4 Test 5
Chinese 65
70
76
68
78
English
72
81
85
90
80
(a) Find the standard deviations of the marks of each subject.
(Give the answers correct to 3 significant figures.)
(b) In which subject is his performance more consistent?
Solution:
(a) For Chinese, mean 
65  70  76  68  78
 71.4
5
Standard deviation
(65 71.4)2  (70  71.4)2  (76  71.4)2  (68 71.4)2  (78 71.4)2

5
 4.88 (cor. to 3 sig. fig.)
P. 17
21.3 Standard Deviation
A. Standard Deviation for Ungrouped Data
Example 21.9T
The following table shows the marks of Eric in five tests of two subjects.
Test 1 Test 2 Test 3 Test 4 Test 5
Chinese 65
70
76
68
78
English
72
81
85
90
80
(a) Find the standard deviations of the marks of each subject.
(Give the answers correct to 3 significant figures.)
(b) In which subject is his performance more consistent?
Solution:
For English, mean 
72  81 85  90  80
 81.6
5
Standard deviation
(72  81.6)2  (81 81.6)2  (85 81.6)2  (90  81.6)2  (80  81.6)2

5
 5.95 (cor. to 3 sig. fig.)
P. 18
21.3 Standard Deviation
A. Standard Deviation for Ungrouped Data
Example 21.9T
The following table shows the marks of Eric in five tests of two subjects.
Test 1 Test 2 Test 3 Test 4 Test 5
Chinese 65
70
76
68
78
English
72
81
85
90
80
(a) Find the standard deviations of the marks of each subject.
(Give the answers correct to 3 significant figures.)
(b) In which subject is his performance more consistent?
Solution:
(b) For Chinese, standard deviation  4.88
For English, standard deviation  5.95
The standard deviation of Chinese is smaller than that of English,
so Eric’s performance in Chinese is more consistent.
P. 19
21.3 Standard Deviation
A. Standard Deviation for Ungrouped Data
Example 21.10T
The numbers of students in five classes are given as:
y – 15, y + 6, y + 9, y – 20, y + 15
(a) Find the mean and the standard deviation.
(Give the answers correct to 3 significant figures if necessary.)
(b) Find the range and the median if the mean is 34.
Solution:
( y 15)  ( y  6)  ( y  9)  ( y  20)  ( y  15)
 y 1
5
(a) Mean 
Standard deviation
( y 15  y  1) 2  ( y  6  y  1) 2  ( y  9  y  1) 2
 ( y  20  y  1) 2  ( y  15  y  1) 2

5
962
(b) ∵ y  1  34

∴
y  35
5
The five numbers are 20, 41, 44, 15 and 50.
 13.9
(cor. to 3 sig. fig.)
∴ Range  50 15 and Median  41
P. 20
 35
21.3 Standard Deviation
B. Standard Deviation for Grouped Data
For a set of grouped data, we have to consider the frequency
of each group.
Standard deviation,  =
f1 ( x1  x ) 2  f 2 ( x2  x ) 2  ...  f n ( xn  x ) 2
f1  f 2  ...  f n
n
 fi ( xi  x )2
=
i 1
n
 fi
,
i 1
where fi and xi are the frequency and the class mark of the ith class
interval respectively, x is the mean and n is the total number of class
marks.
P. 21
21.3 Standard Deviation
B. Standard Deviation for Grouped Data
Example 21.11T
The following table shows the ages of 50 workers in a company.
Age
11 – 20 21 – 30 31 – 40 41 – 50 51 – 60
Number of workers
4
24
12
8
2
(a) Find the mean age of the workers.
(b) Find the standard deviation of the ages.
(Give the answer correct to 3 significant figures.)
Solution:
(a) Class mark 15.5 25.5 35.5 45.5 55.5
Frequency 4
24 12
8
2
4(15.5)  24(25.5)  12(35.5)  8(45.5)  2(55.5)
 31.5
50
5
2
2
2
(b)  f i ( xi  x)  4(15.5 31.5)  24(25.5 31.5)
i 1
12(35.5 31.5)2 8(45.5 31.5)2  2(55.5 31.5)2
 4800
4800
Standard deviation 
 9.80 (cor. to 3 sig. fig.)
50
P. 22
Mean age 
21.3 Standard Deviation
C. Finding Standard Deviation by a Calculator
In actual practice, it is quite difficult to calculate the standard
deviation if the amount of data is very large.
In such circumstances, a calculator can help us to find the standard
deviation.
In order to use a calculator, we have to set the function mode of the
calculator to standard deviation ‘SD’. We also have to clear all the
previous data in the ‘SD’ mode.
For both ungrouped and grouped data, we can
use a calculator to find the mean and the standard
deviation.
For grouped data, use the
class mark to represent the
entire group.
P. 23
21.3 Standard Deviation
The following table summarizes the advantages and
disadvantages of the three different measures of dispersion.
Measure of dispersion
1. Range
Advantage
Only two data are
involved, so it is the
easiest one to calculate.
Disadvantage
Only extreme values
are considered which
may give a misleading
impression.
2. Inter-quartile range It only focuses on the
Cannot show the
middle 50% of data, thus dispersion of the
avoiding the influence of whole group of data.
extreme values.
3. Standard deviation
It takes all the data into
account that can show
the dispersion of the
whole group of data.
Difficult to compute
without using a
calculator.
P. 24
21.4 Applications of Standard Deviation
A. Standard Scores
Standard score is used to compare data in relation with the
mean and the standard deviation .
The standard score z of a given value x from a set of data
with mean x and standard deviation  is defined as:
z
xx

Notes:
The standard score may be positive, negative or zero.
A positive standard score means the given value is z times the standard
deviation above the mean while a negative standard score means the
given value is z times the standard deviation below the mean.
P. 25
21.4 Applications of Standard Deviation
A. Standard Scores
Example 21.12T
Ryan sat for a mathematics examination which consisted of two papers.
The following table shows his marks as well as the means and the
standard deviations of the marks for the whole class in these papers.
(a) Find his standard scores in the two papers.
(Give the answers correct to 3 significant figures.)
Paper I Paper II
(b) In which paper did he perform
Marks
66
71
better?
Mean
60.8
62.4
Solution:
Standard deviation
4.2
6.4
66 60.8
(a) Paper I: z 
4.2
 1.24 (cor. to 3 sig. fig.)
71 62.4
Paper II: z 
6.4
 1.34 (cor. to 3 sig. fig.)
(b) Since 1.34  1.24, Ryan performed better in Paper II than in Paper I.
P. 26
21.4 Applications of Standard Deviation
A. Standard Scores
Example 21.13T
Given that the standard scores of Doris’s marks in Art and Music are
–2.3 and 1.4 respectively, find
(a) her mark in Art if the mean and the standard deviation of the marks
are 30 and 2 respectively;
(b) the mean mark of Music if Doris got 41.5 marks and the standard
deviation of the marks is 3.5.
Solution:
(a) Doris’s mark in Art  (2.3)  2 + 30
 25.4
(b) The mean mark of Music  41.5  1.4  3.5
 36.6
P. 27
21.4 Applications of Standard Deviation
B. Normal Distribution
For a large number of the frequency distributions we meet in
our daily life, their frequency curves have the shape of a bell:
The bell can have different shapes. This bell-shaped
frequency curve is called the normal curve and the
corresponding frequency distribution is called the
normal distribution.
For a normal distribution, the mean, median and the
mode of the data lie at the centre of the distribution.
Therefore, the normal curve is symmetrical about
the mean, i.e., the axis of symmetry for the normal
curve is x  x.
In a normal distribution,
mean  median  mode.
P. 28
21.4 Applications of Standard Deviation
B. Normal Distribution
In addition, we can tell the percentage of the data lie within
a number of standard deviations from the mean:
1. About 68% of the data lie within
one standard deviation from the
mean, i.e., x –  and x + .
2. About 95% of the data lie within
two standard deviations from the
mean, i.e., x – 2 and x + 2.
3. About 99.7% of the data lie within
three standard deviations from the
mean, i.e., x – 3 and x + 3.
P. 29
21.4 Applications of Standard Deviation
B. Normal Distribution
Example 21.14T
The heights of some soccer players are normally distributed with a mean of
180 cm and a standard deviation of 8 cm. Find the percentage of players
(a) whose heights are between 172 cm and 188 cm,
(b) whose heights are greater than 188 cm.
Solution:
Given x  180 and   8.
(a) 172  180  8  x 
 34% of the players’ heights lie between ( x  ) cm and x cm.
188  180 + 8  x  
 34% of the players’ heights lie between x cm and ( x   ) cm.
Percentage of players whose heights are between 172 cm and 188 cm
 34% + 34%  68%
(b) 188  180 + 8  x  
 Percentage of players whose heights are greater than 188 cm
 50%  34%  16%
P. 30
21.4 Applications of Standard Deviation
B. Normal Distribution
Example 21.15T
The weights of 2000 children are normally distributed with a mean of
56.4 kg and a standard deviation of 6.2 kg.
(a) How many children have weights between 50.2 kg and 68.8 kg?
(b) How many children are heavier than 50.2 kg?
Solution:
Given x  56.4 and   6.2.
(a) 50.2  56.4  6.2  x 
68.8  56.4 + 2  6.2  x  2
81.5% of children have weights between ( x  ) kg and ( x  2 ) kg.
 Number of children  2000  81.5%
 1630
(b) 50.2  56.4  6.2  x 
Percentage of children who are heavier than 50.2 kg  50% + 34%
 84%
 Number of children  2000  84%
 1680
P. 31
21.5 Effects on the Dispersion with a
Change in Data
A. Removal of the Largest or Smallest item
from the Data
In junior forms, we learnt that if we remove a datum greater
than the mean of the data set, then the mean will decrease.
Similarly, if we remove a datum less than the mean of the data set, then
the mean will increase.
We can deduce that:
If the greatest or the least value (assuming the removed datum is unique)
in a data set is removed, then
1. the range will decrease;
2. the inter-quartile range may increase, decrease or remain unchanged;
3. the standard deviation may increase or decrease.
P. 32
21.5 Effects on the Dispersion with a
Change in Data
B. Adding a Common Constant to the Whole
Set of Data
We have the following conclusion:
If a constant k is added to each datum in a set of data, then
the following measures of dispersion will not change:
1. the range,
2. the inter-quartile range and
If a constant k is added to
each datum, then the mean,
3. the standard deviation
median and the mode will
also increase by k.
P. 33
21.5 Effects on the Dispersion with a
Change in Data
C. Multiplying the Whole Set of Data by a
Common Constant
We have the following conclusion:
The range, the inter-quartile range and the standard deviation
will be k times the original values if each datum in a set of data
is multiplied by a constant k.
Notes:
If the quartiles are not members of the data set, the conclusion will be
the same.
P. 34
21.5 Effects on the Dispersion with a
Change in Data
C. Multiplying the Whole Set of Data by a
Common Constant
Example 21.16T
Consider the nine different numbers:
20, 45, 25, 30, 32, 28, 35, 51, 40
(a) Find the range, the inter-quartile range and the standard deviation.
(b) Find the new range, the new inter-quartile range and the new standard
deviation of the new set of data if
(i) 10 is subtracted from each data;
(ii) each number is halved;
(iii) the datum 45 is removed.
(Give the answers correct to 3 significant figures if necessary.)
Solution:
(a) Range  51 20  31
25  28
40  45
 26.5 and Q3 
 42.5
2
2
Inter-quartile range  42.5  26.5  16
Standard deviation  9.31 (cor. to 3 sig. fig.)
Q1 
P. 35
21.5 Effects on the Dispersion with a
Change in Data
C. Multiplying the Whole Set of Data by a
Common Constant
Example 21.16T
Consider the nine different numbers:
20, 45, 25, 30, 32, 28, 35, 51, 40
(a) Find the range, the inter-quartile range and the standard deviation.
(b) Find the new range, the new inter-quartile range and the new standard
deviation of the new set of data if
(i) 10 is subtracted from each data;
(ii) each number is halved;
(iii) the datum 45 is removed.
(Give the answers correct to 3 significant figures if necessary.)
Solution:
(b) (i) If 10 is subtracted from each datum, the range, the inter-quartile
range and the standard deviation of the new data remain unchanged.
Range  31
Inter-quartile range 16
Standard deviation  9.31 (cor. to 3 sig. fig.)
P. 36
21.5 Effects on the Dispersion with a
Change in Data
C. Multiplying the Whole Set of Data by a
Common Constant
Example 21.16T
Consider the nine different numbers:
20, 45, 25, 30, 32, 28, 35, 51, 40
(a) Find the range, the inter-quartile range and the standard deviation.
(b) Find the new range, the new inter-quartile range and the new standard
deviation of the new set of data if
(i) 10 is subtracted from each data;
(ii) each number is halved;
(iii) the datum 45 is removed.
(Give the answers correct to 3 significant figures if necessary.)
Solution:
(b) (ii) If each datum is halved, the range, the inter-quartile range and the
standard deviation of the new data are multiplied by 0.5.
Range  310.5 15.5
Inter-quartile range 160.5  8
Standard deviation  9.30950.5  4.65 (cor. to 3 sig. fig.)
P. 37
21.5 Effects on the Dispersion with a
Change in Data
C. Multiplying the Whole Set of Data by a
Common Constant
Example 21.16T
Consider the nine different numbers:
20, 45, 25, 30, 32, 28, 35, 51, 40
(a) Find the range, the inter-quartile range and the standard deviation.
(b) Find the new range, the new inter-quartile range and the new standard
deviation of the new set of data if
(i) 10 is subtracted from each data;
(ii) each number is halved;
(iii) the datum 45 is removed.
(Give the answers correct to 3 significant figures if necessary.)
Solution:
(b) (iii)The remaining data are 20, 25, 28, 30, 32, 35, 40, 51.
Range  51 20  31
35 40 25 28

Inter-quartile range 
2
2
11
Standard deviation  8.97 (cor. to 3 sig. fig.)
P. 38
21.5 Effects on the Dispersion with a
Change in Data
D. Insertion of Zero in the Data Set
We have the following conclusion:
If a zero value is inserted in a non-negative data set, then
1. the range may increase or remain unchanged;
2. the inter-quartile range may increase, decrease or remain unchanged;
3. the standard deviation may increase or decrease.
P. 39
21.5 Effects on the Dispersion with a
Change in Data
D. Insertion of Zero in the Data Set
Example 21.17T
The daily income ($) of a hawker during the last two weeks was:
200, 220, 230, 240, 250, 320, 340, 360, 380, 400, 450, 580, 650
(a) Find the inter-quartile range and the standard deviation.
(b) There was a thunderstorm last Monday and the income on that day
was zero. If the income from last Monday is also considered, find the
new inter-quartile range and the new standard deviation.
(Give the answers correct to 1 decimal place if necessary.)
Solution:
230  240
 $235
2
400  450
Q3  $
 $425
2
Inter-quartile range  $(425  235)
 $190
(a) Q1  $
The standard deviation
 $133.9
(cor. to 1 d. p.)
(b) Q1  $230
Q3  $400
Inter-quartile range
 $(400  230)
 $170
The standard deviation
 $158.2 (cor. to 1 d. p.)
P. 40
Chapter Summary
21.1 Range and Inter-quartile Range
1. The range is the difference between the largest value (highest
class boundary) and the smallest value (lowest class boundary)
in a set of ungrouped (grouped) data.
2. The inter-quartile range is the difference between the upper
quartile Q3 and the lower quartile Q1 of a set of data.
P. 41
Chapter Summary
21.2 Box-and-whisker Diagrams
A box-and-whisker diagram illustrates the spread of a set of data.
It shows the greatest value, the least value, the median, the lower
quartile and the upper quartile of the data.
P. 42
Chapter Summary
21.3 Standard Deviation
Standard deviation  is the measure of dispersion that
describes how spread out a set of data is around the mean value.
n
2
(
x

x
)
 i
i 1
1. For ungrouped data:  
n
n
2. For grouped data:  
 fi ( xi  x )2
i 1
n
 fi
i 1
Larger values for the range, the inter-quartile range and the standard
deviation of the data indicate a larger dispersion and vice versa.
P. 43
Chapter Summary
21.4 Applications of Standard Deviation
1. The standard score z is the number of standard deviations that
a given value is above or below the mean, and is given by
z
xx

.
2. The curve of a normal distribution is bell-shaped and is called
the normal curve.
In the normal distribution, different percentages of data lie within
different standard deviations from the mean.
P. 44
Chapter Summary
21.5 Effects on the Dispersion with a Change
in Data
1. If the greatest or the least value (assuming both are unique) in a data
set is removed, then the range will decrease. However, the interquartile range may increase, decrease or remain unchanged and the
standard deviation may increase or decrease.
2. If a constant k is added to each datum in a set of data, then the range,
the inter-quartile range and the standard deviation will not change.
3. If each item in the data is multiplied by a positive constant k, then
the range, the inter-quartile range and the standard deviation will be
k times their original values.
4. If a zero value is inserted in a non-negative data set, then the range
may increase or remain unchanged, the inter-quartile range may
increase, decrease or remain unchanged and the standard deviation
may increase or decrease.
P. 45