Chapter 2 Descriptive Statistics
1
Constructing a Frequency Distribution The following sample data set lists the number of minutes 50 Internet subscribers spent on the Internet during their most recent session. Construct a frequency distribution that has seven classes. 50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86 41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20 18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44
2
Constructing a Frequency Distribution 1. Decide on the number of classes. Usually between 5 and 20; otherwise, it may be difficult to detect any patterns. 2. Find the class width. Determine the range of the data. Divide the range by the number of classes. Round up to the next convenient number. Max value – min value / no. of classes
3
Constructing a Frequency Distribution 3. Find the class limits. You can use the minimum data entry as the lower limit of the first class. Find the remaining lower limits (add the class width to the lower limit of the preceding class). Find the upper limit of the first class. Remember that classes cannot overlap. Find the remaining upper class limits.
4
Constructing a Frequency Distribution 4. Make a tally mark for each data entry in the row of the appropriate class. 5. Count the tally marks to find the total frequency f for each class.
5
Frequency Distribution Frequency Distribution Class Frequency, f Class width • A table that shows 1–5 5 classes or intervals of 6 – 1 = 5 6 – 10 8 data with a count of the 11 – 15 6 number of entries in each 16 – 20 8 class. 21 – 25 5 • The frequency, f, of a class is the number of 26 – 30 4 data entries in the class. Lower class Upper class limits
limits 6
Solution: Constructing a Frequency Distribution 50 40 41 17 11 7 22 44 28 21 19 23 37 51 54 42 86 41 78 56 72 56 17 7 69 30 80 56 29 33 46 31 39 20 18 29 34 59 73 77 36 39 30 62 54 67 39 31 53 44
1. Number of classes = 7 (given) 2. Find the class width max min 86 7 11.29 #classes 7
Round up to 12
7
Solution: Constructing a Frequency Distribution 3. Use 7 (minimum value) as first lower limit. Add the class width of 12 to get the lower limit of the next class. 7 + 12 = 19 Find the remaining lower limits.
Class width = 12
Lower limit 7
Upper limit
19 31 43 55 67 79 8
Solution: Constructing a Frequency Distribution The upper limit of the first class is 18 (one less than the lower limit of the second class). Add the class width of 12 to get the upper limit of the next class. 18 + 12 = 30 Find the remaining upper limits.
Lower limit 7 19 31 43 55 67 79
Upper limit
18 30 42 54 66
Class width = 12
78 90 9
Solution: Constructing a Frequency Distribution 4. Make a tally mark for each data entry in the row of the appropriate class. 5. Count the tally marks to find the total frequency f for each class. Class
Tally
Frequency,
f 7 – 18
IIII I
6
19 – 30
IIII IIII
10
31 – 42
IIII IIII III
13
43 – 54
IIII III
8
55 – 66
IIII
5
67 – 78
IIII I
6
79 – 90
II
Σf 2= 50
10
Determining the Midpoint Midpoint of a class (Lower class limit) (Upper class limit) 2 Class 7 – 18
Midpoint 7 18 12.5 2
19 – 30
19 30 24.5 2
31 – 42
31 42 36.5 2
Frequency, f 6
Class width = 12 10 13
11
Determining the Relative Frequency Relative Frequency of a class • Portion or percentage of the data that falls in a particular class. class frequency f • relative frequency Sample size n Class
Frequency, f
7 – 18
6
19 – 30
10
31 – 42
13
Relative Frequency 6 0.12 50 10 0.20 50 13 0.26 50
12
Determining the Cumulative Frequency Cumulative frequency of a class • The sum of the frequency for that class and all previous classes. Class
Frequency, f
Cumulative frequency
7 – 18
6
6
19 – 30
+ 10
16
31 – 42
+ 13
29
13
Expanded Frequency Distribution
Class
Frequency, f
Midpoint
Relative frequency
7 – 18
6
12.5
0.12
6
19 – 30
10
24.5
0.20
16
31 – 42
13
36.5
0.26
29
43 – 54
8
48.5
0.16
37
55 – 66
5
60.5
0.10
42
67 – 78
6
72.5
0.12
48
79 – 90
2
84.5
0.04 f 1 n
50
Σf = 50
Cumulative frequency
14
Graphs of Frequency Distributions
frequency
Frequency Histogram • A bar graph that represents the frequency distribution. • The horizontal scale is quantitative and measures the data values. • The vertical scale measures the frequencies of the classes. • Consecutive bars must touch.
data values 15
Class Boundaries Class boundaries • The numbers that separate classes without forming gaps between them.
• The distance from the upper limit of the first class to the lower limit of the second class is 19 – 18 = 1. • Half this distance is 0.5.
Class
Class Frequency, Boundaries f
7 – 18
6.5 – 18.5
6
19 – 30
10
31 – 42
13
• First class lower boundary = 7 – 0.5 = 6.5 • First class upper boundary = 18 + 0.5 = 18.5 16
Class Boundaries Class 7 – 18 19 – 30 31 – 42 43 – 54 55 – 66 67 – 78 79 – 90
Class boundaries 6.5 – 18.5 18.5 – 30.5 30.5 – 42.5 42.5 – 54.5 54.5 – 66.5 66.5 – 78.5 78.5 – 90.5
Frequency,
f 6 10 13 8 5 6 2
17
Example: Frequency Histogram Construct a frequency histogram for the Internet usage frequency distribution. Class
Class boundaries
Frequency, Midpoint
f
7 – 18
6.5 – 18.5
12.5
6
19 – 30
18.5 – 30.5
24.5
10
31 – 42
30.5 – 42.5
36.5
13
43 – 54
42.5 – 54.5
48.5
8
55 – 66
54.5 – 66.5
60.5
5
67 – 78
66.5 – 78.5
72.5
6
79 – 90
78.5 – 90.5
84.5
2 18
Solution: Frequency Histogram (using Midpoints)
19
Solution: Frequency Histogram (using class boundaries)
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
You can see that more than half of the subscribers spent between 19 and 54 minutes on the Internet during their most recent session. 20
Graphs of Frequency Distributions
frequency
Frequency Polygon • A line graph that emphasizes the continuous change in frequencies.
data values
21
Example: Frequency Polygon Construct a frequency polygon for the Internet usage frequency distribution. Class
Midpoint
Frequency, f
7 – 18
12.5
6
19 – 30
24.5
10
31 – 42
36.5
13
43 – 54
48.5
8
55 – 66
60.5
5
67 – 78
72.5
6
79 – 90
84.5
2
22
Solution: Frequency Polygon Internet Usage
Frequency
The graph should begin and end on the horizontal axis, so extend the left side to one class width before the first class midpoint and extend the right side to one class width after the last class midpoint.
14 12 10 8 6 4 2 0 0.5
12.5
24.5
36.5
48.5
60.5
72.5
84.5
96.5
Time online (in minutes)
You can see that the frequency of subscribers increases up to 36.5 minutes and then decreases. 23
Graphs of Frequency Distributions
relative frequency
Relative Frequency Histogram • Has the same shape and the same horizontal scale as the corresponding frequency histogram. • The vertical scale measures the relative frequencies, not frequencies.
data values 24
Example: Relative Frequency Histogram Construct a relative frequency histogram for the Internet usage frequency distribution. Frequency,
Class
Class boundaries
f
Relative frequency
7 – 18
6.5 – 18.5
6
0.12
19 – 30
18.5 – 30.5
10
0.20
31 – 42
30.5 – 42.5
13
0.26
43 – 54
42.5 – 54.5
8
0.16
55 – 66
54.5 – 66.5
5
0.10
67 – 78
66.5 – 78.5
6
0.12
79 – 90
78.5 – 90.5
2
0.04 25
Solution: Relative Frequency Histogram
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
From this graph you can see that 20% of Internet subscribers spent between 18.5 minutes and 30.5 minutes online. 26
Graphs of Frequency Distributions
cumulative frequency
Cumulative Frequency Graph or Ogive • A line graph that displays the cumulative frequency of each class at its upper class boundary. • The upper boundaries are marked on the horizontal axis. • The cumulative frequencies are marked on the vertical axis.
data values 27
Constructing an Ogive 1. Construct a frequency distribution that includes cumulative frequencies as one of the columns. 2. Specify the horizontal and vertical scales.  The horizontal scale consists of the upper class boundaries.  The vertical scale measures cumulative frequencies. 3. Plot points that represent the upper class boundaries and their corresponding cumulative frequencies. 28
Constructing an Ogive 4. Connect the points in order from left to right. 5. The graph should start at the lower boundary of the first class (cumulative frequency is zero) and should end at the upper boundary of the last class (cumulative frequency is equal to the sample size).
29
Example: Ogive Construct an ogive for the Internet usage frequency distribution. Frequency,
Class
Class boundaries
f
Cumulative frequency
7 – 18
6.5 – 18.5
6
6
19 – 30
18.5 – 30.5
10
16
31 – 42
30.5 – 42.5
13
29
43 – 54
42.5 – 54.5
8
37
55 – 66
54.5 – 66.5
5
42
67 – 78
66.5 – 78.5
6
48
79 – 90
78.5 – 90.5
2
50
30
Solution: Ogive Internet Usage Cumulative frequency
60 50 40 30 20 10 0
6.5
18.5
30.5
42.5
54.5
66.5
78.5
90.5
Time online (in minutes)
From the ogive, you can see that about 40 subscribers spent 60 minutes or less online during their last session. The greatest increase in usage occurs between 30.5 minutes and 42.5 minutes. 31
Graphing Quantitative Data Sets Stem-and-leaf plot • Each number is separated into a stem and a leaf. • Similar to a histogram. • Still contains original data values. 26
Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45
2 3
1 5 5 6 7 8 0 6 6
4
5 32
Example: Constructing a Stem-and-Leaf Plot The following are the numbers of text messages sent last month by the cellular phone users on one floor of a college dormitory. Display the data in a stem-and-leaf plot. 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
33
Solution: Constructing a Stem-and-Leaf Plot 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
• The data entries go from a low of 78 to a high of 159. • Use the rightmost digit as the leaf. For instance, 78 = 7 | 8 and 159 = 15 | 9 • List the stems, 7 to 15, to the left of a vertical line. • For each data entry, list a leaf to the right of its stem. 34
Solution: Constructing a Stem-and-Leaf Plot Include a key to identify the values of the data.
From the display, you can conclude that more than 50% of the cellular phone users sent between 110 and 130 text messages. 35
Graphing Quantitative Data Sets Dot plot • Each data entry is plotted, using a point, above a horizontal axis
Data: 21, 25, 25, 26, 27, 28, 30, 36, 36, 45 26 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
36
Example: Constructing a Dot Plot Use a dot plot organize the text messaging data. 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
• So that each data entry is included in the dot plot, the horizontal axis should include numbers between 70 and 160. • To represent a data entry, plot a point above the entry's position on the axis. • If an entry is repeated, plot another point above the previous point. 37
Solution: Constructing a Dot Plot 155 159 144 129 105 145 126 116 130 114 122 112 112 142 126 118 118 108 122 121 109 140 126 119 113 117 118 109 109 119 139 139 122 78 133 126 123 145 121 134 124 119 132 133 124 129 112 126 148 147
From the dot plot, you can see that most values cluster between 105 and 148 and the value that occurs the most is 126. You can also see that 78 is an unusual data value. 38
Graphing Qualitative Data Sets Pie Chart • A circle is divided into sectors that represent categories. • The area of each sector is proportional to the frequency of each category.
39
Example: Constructing a Pie Chart The numbers of motor vehicle occupants killed in crashes in 2005 are shown in the table. Use a pie chart to organize the data. (Source: U.S. Department of Transportation, National Highway Traffic Safety Administration) Vehicle type Cars Trucks Motorcycles Other
Killed 18,440 13,778 4,553 823 40
Solution: Constructing a Pie Chart • Find the relative frequency (percent) of each category. Vehicle type Frequency, f Cars
18,440
Trucks
13,778
Motorcycles Other
4,553 823
Relative frequency 18440 0.49 37594 13778 0.37 37594 4553 0.12 37594 823 0.02 37594
37,594 41
Solution: Constructing a Pie Chart • Construct the pie chart using the central angle that corresponds to each category. To find the central angle, multiply 360º by the category's relative frequency. For example, the central angle for cars is 360(0.49) ≈ 176º
42
Solution: Constructing a Pie Chart
Vehicle type
Frequency,
Relative frequency
f
360º(0.49)≈176º
Cars
18,440
0.49
Trucks
13,778
0.37
4,553
0.12
823
0.02
Motorcycles Other
Central angle
360º(0.37)≈133º 360º(0.12)≈43º
360º(0.02)≈7º
43
Solution: Constructing a Pie Chart
Relative Vehicle type frequency
Central angle
Cars
0.49
176ยบ
Trucks
0.37
133ยบ
Motorcycles
0.12
43ยบ
Other
0.02
7ยบ
From the pie chart, you can see that most fatalities in motor vehicle crashes were those involving the occupants of cars. 44
Graphing Qualitative Data Sets
Frequency
Pareto Chart • A vertical bar graph in which the height of each bar represents frequency or relative frequency. • The bars are positioned in order of decreasing height, with the tallest bar positioned at the left.
Categories 45
Example: Constructing a Pareto Chart In a recent year, the retail industry lost $41.0 million in inventory shrinkage. Inventory shrinkage is the loss of inventory through breakage, pilferage, shoplifting, and so on. The causes of the inventory shrinkage are administrative error ($7.8 million), employee theft ($15.6 million), shoplifting ($14.7 million), and vendor fraud ($2.9 million). Use a Pareto chart to organize this data. (Source: National Retail Federation and Center for Retailing Education, University of Florida)
46
Solution: Constructing a Pareto Chart Cause
$ (million)
Admin. error
7.8
Employee theft
15.6
Shoplifting
14.7
Vendor fraud
2.9
From the graph, it is easy to see that the causes of inventory shrinkage that should be addressed first are employee theft and shoplifting. 47
Section 2.3 Measures of Central Tendency
48
Measures of Central Tendency Measure of central tendency • A value that represents a typical, or central, entry of a data set. • Most common measures of central tendency: Mean Median Mode
49
Measure of Central Tendency: Mean Mean (average) • The sum of all the data entries divided by the number of entries. • Sigma notation: Σx = add all of the data entries (x) in the data set. x • Population mean: N
• Sample mean:
x x n 50
Example: Finding a Sample Mean The prices (in dollars) for a sample of roundtrip flights from Chicago, Illinois to Cancun, Mexico are listed. What is the mean price of the flights? 872 432 397 427 388 782 397
51
Solution: Finding a Sample Mean 872 432 397 427 388 782 397 • The sum of the flight prices is Σx = 872 + 432 + 397 + 427 + 388 + 782 + 397 = 3695 • To find the mean price, divide the sum of the prices by the number of prices in the sample x 3695 x 527.9 n 7
The mean price of the flights is about $527.90. 52
Measure of Central Tendency: Median Median • The value that lies in the middle of the data when the data set is ordered. • Measures the center of an ordered data set by dividing it into two equal parts. • If the data set has an odd number of entries: median is the middle data entry. even number of entries: median is the mean of the two middle data entries. 53
Example: Finding the Median The prices (in dollars) for a sample of roundtrip flights from Chicago, Illinois to Cancun, Mexico are listed. Find the median of the flight prices. 872 432 397 427 388 782 397
54
Solution: Finding the Median 872 432 397 427 388 782 397 • First order the data. 388 397 397 427 432 782 872 • There are seven entries (an odd number), the median is the middle, or fourth, data entry. The median price of the flights is $427. 55
Example: Finding the Median The flight priced at $432 is no longer available. What is the median price of the remaining flights? 872 397 427 388 782 397
56
Solution: Finding the Median 872 397 427 388 782 397 • First order the data. 388 397 397 427 782 872 • There are six entries (an even number), the median is the mean of the two middle entries. 397 427 Median 412 2
The median price of the flights is $412. 57
Measure of Central Tendency: Mode Mode • The data entry that occurs with the greatest frequency. • If no entry is repeated the data set has no mode. • If two entries occur with the same greatest frequency, each entry is a mode (bimodal).
58
Example: Finding the Mode The prices (in dollars) for a sample of roundtrip flights from Chicago, Illinois to Cancun, Mexico are listed. Find the mode of the flight prices. 872 432 397 427 388 782 397
59
Solution: Finding the Mode 872 432 397 427 388 782 397 • Ordering the data helps to find the mode. 388 397 397 427 432 782 872 • The entry of 397 occurs twice, whereas the other data entries occur only once. The mode of the flight prices is $397. 60
Example: Finding the Mode At a political debate a sample of audience members was asked to name the political party to which they belong. Their responses are shown in the table. What is the mode of the responses? Political Party Democrat Republican Other Did not respond
Frequency, f 34 56 21 9 61
Solution: Finding the Mode Political Party Democrat Republican Other Did not respond
Frequency, f 34 56 21 9
The mode is Republican (the response occurring with the greatest frequency). In this sample there were more Republicans than people of any other single affiliation. 62
Comparing the Mean, Median, and Mode • All three measures describe a typical entry of a data set. • Advantage of using the mean: The mean is a reliable measure because it takes into account every entry of a data set. • Disadvantage of using the mean: Greatly affected by outliers (a data entry that is far removed from the other entries in the data set).
63
Example: Comparing the Mean, Median, and Mode Find the mean, median, and mode of the sample ages of a class shown. Which measure of central tendency best describes a typical entry of this data set? Are there any outliers? Ages in a class 20
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
64
Solution: Comparing the Mean, Median, and Mode Ages in a class 20
20
20
20
20
20
21
21
21
21
22
22
22
23
23
23
23
24
24
65
Mean:
x 20 20 ... 24 65 x 23.8 years n 20
Median:
21 22 21.5 years 2
Mode:
20 years (the entry occurring with the greatest frequency) 65
Solution: Comparing the Mean, Median, and Mode Mean ≈ 23.8 years
Median = 21.5 years
Mode = 20 years
• The mean takes every entry into account, but is influenced by the outlier of 65. • The median also takes every entry into account, and it is not affected by the outlier. • In this case the mode exists, but it doesn't appear to represent a typical entry.
66
Solution: Comparing the Mean, Median, and Mode Sometimes a graphical comparison can help you decide which measure of central tendency best represents a data set.
In this case, it appears that the median best describes the data set. 67
Weighted Mean Weighted Mean • The mean of a data set whose entries have varying weights.
( x w) • x where w is the weight of each entry x w
68
Example: Finding a Weighted Mean You are taking a class in which your grade is determined from five sources: 50% from your test mean, 15% from your midterm, 20% from your final exam, 10% from your computer lab work, and 5% from your homework. Your scores are 86 (test mean), 96 (midterm), 82 (final exam), 98 (computer lab), and 100 (homework). What is the weighted mean of your scores? If the minimum average for an A is 90, did you get an A?
69
Solution: Finding a Weighted Mean x∙w
Score, x
Weight, w
Test Mean
86
0.50
86(0.50)= 43.0
Midterm
96
0.15
96(0.15) = 14.4
Final Exam
82
0.20
82(0.20) = 16.4
Computer Lab
98
0.10
98(0.10) = 9.8
Homework
100
0.05
100(0.05) = 5.0
Source
Σw = 1
Σ(x∙w) = 88.6
( x w) 88.6 x 88.6 w 1
Your weighted mean for the course is 88.6. You did not get an A. 70
Mean of Grouped Data Mean of a Frequency Distribution • Approximated by
( x f ) x n
n f
where x and f are the midpoints and frequencies of a class, respectively
71
Finding the Mean of a Frequency Distribution In Words 1. Find the midpoint of each class.
In Symbols (lower limit)+(upper limit) x 2
2. Find the sum of the products of the midpoints and the frequencies.
( x f )
3. Find the sum of the frequencies.
n f
4. Find the mean of the frequency distribution.
( x f ) x n 72
Example: Find the Mean of a Frequency Distribution Use the frequency distribution to approximate the mean number of minutes that a sample of Internet subscribers spent online during their most recent session. Class
Midpoint
Frequency, f
7 – 18
12.5
6
19 – 30
24.5
10
31 – 42
36.5
13
43 – 54
48.5
8
55 – 66
60.5
5
67 – 78
72.5
6
79 – 90
84.5
2 73
Solution: Find the Mean of a Frequency Distribution Class
Midpoint, x Frequency, f
(x∙f)
7 – 18
12.5
6
12.5∙6 = 75.0
19 – 30
24.5
10
24.5∙10 = 245.0
31 – 42
36.5
13
36.5∙13 = 474.5
43 – 54
48.5
8
48.5∙8 = 388.0
55 – 66
60.5
5
60.5∙5 = 302.5
67 – 78
72.5
6
72.5∙6 = 435.0
79 – 90
84.5
2
84.5∙2 = 169.0
n = 50
Σ(x∙f) = 2089.0
( x f ) 2089 x 41.8 minutes n 50 74
The Shape of Distributions Symmetric Distribution • A vertical line can be drawn through the middle of a graph of the distribution and the resulting halves are approximately mirror images.
75
The Shape of Distributions Uniform Distribution (rectangular) • All entries or classes in the distribution have equal or approximately equal frequencies. • Symmetric.
76
The Shape of Distributions Skewed Left Distribution (negatively skewed) • The “tail” of the graph elongates more to the left. • The mean is to the left of the median.
77
The Shape of Distributions Skewed Right Distribution (positively skewed) • The “tail” of the graph elongates more to the right. • The mean is to the right of the median.
78
Section 2.3 Summary • Determined the mean, median, and mode of a population and of a sample • Determined the weighted mean of a data set and the mean of a frequency distribution • Described the shape of a distribution as symmetric, uniform, or skewed and compared the mean and median for each
79
Section 2.4 Measures of Variation
80
Range Range • The difference between the maximum and minimum data entries in the set. • The data must be quantitative. • Range = (Max. data entry) – (Min. data entry)
81
Example: Finding the Range A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the range of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42
82
Solution: Finding the Range • Ordering the data helps to find the least and greatest salaries. 37 38 39 41 41 41 42 44 45 47 minimum
maximum
• Range = (Max. salary) – (Min. salary) = 47 – 37 = 10
The range of starting salaries is 10 or $10,000.
83
Deviation, Variance, and Standard Deviation Deviation • The difference between the data entry, x, and the mean of the data set. • Population data set: Deviation of x = x – μ • Sample data set: Deviation of x = x – x
84
Example: Finding the Deviation A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the deviation of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42
Solution: • First determine the mean starting salary. x 415 41.5 N 10 85
Solution: Finding the Deviation • Determine the deviation for each data entry.
Salary ($1000s), x Deviation: x – μ 41
41 – 41.5 = –0.5
38
38 – 41.5 = –3.5
39
39 – 41.5 = –2.5
45
45 – 41.5 = 3.5
47
47 – 41.5 = 5.5
41
41 – 41.5 = –0.5
44
44 – 41.5 = 2.5
41
41 – 41.5 = –0.5
37
37 – 41.5 = –4.5
42
42 – 41.5 = 0.5
Σx = 415
Σ(x – μ) = 0 86
Deviation, Variance, and Standard Deviation Population Variance
( x ) • N 2
2
Sum of squares, SSx
Population Standard Deviation 2 ( x ) 2 • N
87
Finding the Population Variance & Standard Deviation In Words 1. Find the mean of the population data set.
In Symbols x N
2. Find deviation of each entry.
x–μ
3. Square each deviation.
(x – μ)2
4. Add to get the sum of squares.
SSx = Σ(x – μ)2
88
Finding the Population Variance & Standard Deviation In Words
In Symbols 2 ( x ) 2
5. Divide by N to get the population variance.
6. Find the square root to get the population standard deviation.
( x ) 2 N
N
89
Example: Finding the Population Standard Deviation A corporation hired 10 graduates. The starting salaries for each graduate are shown. Find the population variance and standard deviation of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42 Recall Îź = 41.5.
90
Solution: Finding the Population Standard Deviation • Determine SSx • N = 10
Deviation: x – μ
Squares: (x – μ)2
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
38
38 – 41.5 = –3.5
(–3.5)2 = 12.25
39
39 – 41.5 = –2.5
(–2.5)2 = 6.25
45
45 – 41.5 = 3.5
(3.5)2 = 12.25
47
47 – 41.5 = 5.5
(5.5)2 = 30.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
44
44 – 41.5 = 2.5
(2.5)2 = 6.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
37
37 – 41.5 = –4.5
(–4.5)2 = 20.25
42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5
Salary, x
91
Solution: Finding the Population Standard Deviation Population Variance
( x ) 88.5 8.9 • N 10 2
2
Population Standard Deviation • 2 8.85 3.0 The population standard deviation is about 3.0, or $3000. 92
Deviation, Variance, and Standard Deviation Sample Variance
( x x ) • s n 1
2
2
Sample Standard Deviation •
2 ( x x ) s s2 n 1
93
Finding the Sample Variance & Standard Deviation In Words
In Symbols x n
1. Find the mean of the sample data set.
x
2. Find deviation of each entry.
xx
3. Square each deviation.
( x x )2
4. Add to get the sum of squares.
SS x ( x x )2
94
Finding the Sample Variance & Standard Deviation In Words 5. Divide by n – 1 to get the sample variance.
6. Find the square root to get the sample standard deviation.
In Symbols 2 ( x x ) s2 n 1
( x x ) 2 s n 1
95
Example: Finding the Sample Standard Deviation The starting salaries are for the Chicago branches of a corporation. The corporation has several other branches, and you plan to use the starting salaries of the Chicago branches to estimate the starting salaries for the larger population. Find the sample standard deviation of the starting salaries. Starting salaries (1000s of dollars) 41 38 39 45 47 41 44 41 37 42
96
Solution: Finding the Sample Standard Deviation • Determine SSx • n = 10
Deviation: x – μ
Squares: (x – μ)2
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
38
38 – 41.5 = –3.5
(–3.5)2 = 12.25
39
39 – 41.5 = –2.5
(–2.5)2 = 6.25
45
45 – 41.5 = 3.5
(3.5)2 = 12.25
47
47 – 41.5 = 5.5
(5.5)2 = 30.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
44
44 – 41.5 = 2.5
(2.5)2 = 6.25
41
41 – 41.5 = –0.5
(–0.5)2 = 0.25
37
37 – 41.5 = –4.5
(–4.5)2 = 20.25
42
42 – 41.5 = 0.5
(0.5)2 = 0.25
Σ(x – μ) = 0
SSx = 88.5
Salary, x
97
Solution: Finding the Sample Standard Deviation Sample Variance
( x x ) 88.5 9.8 • s n 1 10 1 2
2
Sample Standard Deviation
88.5 3.1 • s s 9 2
The sample standard deviation is about 3.1, or $3100. 98
Interpreting Standard Deviation • Standard deviation is a measure of the typical amount an entry deviates from the mean. • The more the entries are spread out, the greater the standard deviation.
99
Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics: • About 68% of the data lie within one standard deviation of the mean. • About 95% of the data lie within two standard deviations of the mean. • About 99.7% of the data lie within three standard deviations of the mean.
100
Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) 99.7% within 3 standard deviations 95% within 2 standard deviations
68% within 1 standard deviation
34%
34%
2.35%
2.35%
13.5%
x 3s
x 2s
13.5%
x s
x
x s
x 2s
x 3s 101
Example: Using the Empirical Rule In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64 inches, with a sample standard deviation of 2.71 inches. Estimate the percent of the women whose heights are between 64 inches and 69.42 inches.
102
Solution: Using the Empirical Rule • Because the distribution is bell-shaped, you can use the Empirical Rule.
34%
13.5% 55.87
x 3s
58.58
x 2s
61.29
x s
64
x
66.71
x s
69.42
x 2s
72.13
x 3s
34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall. 103
Chebychev’s Theorem • The portion of any data set lying within k standard deviations (k > 1) of the mean is at least: 1 1 2 k 1 3 • k = 2: In any data set, at least 1 2 or 75% 2
4
of the data lie within 2 standard deviations of the mean. 1 8 • k = 3: In any data set, at least 1 2 or 88.9% 3 9
of the data lie within 3 standard deviations of the mean. 104
Example: Using Chebychev’s Theorem The age distribution for Florida is shown in the histogram. Apply Chebychev’s Theorem to the data using k = 2. What can you conclude?
105
Solution: Using Chebychev’s Theorem
k = 2: μ – 2σ = 39.2 – 2(24.8) = -10.4 (use 0 since age can’t be negative) μ + 2σ = 39.2 + 2(24.8) = 88.8
At least 75% of the population of Florida is between 0 and 88.8 years old. 106
Standard Deviation for Grouped Data Sample standard deviation for a frequency distribution •
( x x ) 2 f s n 1
where n= Σf (the number of entries in the data set)
• When a frequency distribution has classes, estimate the sample mean and standard deviation by using the midpoint of each class.
107
Example: Finding the Standard Deviation for Grouped Data You collect a random sample of the number of children per household in a region. Find the sample mean and the sample standard deviation of the data set.
Number of Children in 50 Households
1
3
1
1
1
1
2
2
1
0
1
1
0
0
0
1
5
0
3
6
3
0
3
1
1
1
1
6
0
1
3
6
6
1
2
2
3
0
1
1
4
1
1
2
2
0
3
0
2
4
108
Solution: Finding the Standard Deviation for Grouped Data • First construct a frequency distribution. • Find the mean of the frequency distribution. xf 91 x 1.8 n 50
The sample mean is about 1.8 children.
x
f
xf
0
10
0(10) = 0
1
19
1(19) = 19
2
7
2(7) = 14
3
7
3(7) =21
4
2
4(2) = 8
5
1
5(1) = 5
6
4
6(4) = 24
Σf = 50 Σ(xf )= 91
109
Solution: Finding the Standard Deviation for Grouped Data • Determine the sum of squares. x
f
xx
( x x )2
0
10
0 – 1.8 = –1.8
(–1.8)2 = 3.24
3.24(10) = 32.40
1
19
1 – 1.8 = –0.8
(–0.8)2 = 0.64
0.64(19) = 12.16
2
7
2 – 1.8 = 0.2
(0.2)2 = 0.04
0.04(7) = 0.28
3
7
3 – 1.8 = 1.2
(1.2)2 = 1.44
1.44(7) = 10.08
4
2
4 – 1.8 = 2.2
(2.2)2 = 4.84
4.84(2) = 9.68
5
1
5 – 1.8 = 3.2
(3.2)2 = 10.24
10.24(1) = 10.24
6
4
6 – 1.8 = 4.2
(4.2)2 = 17.64
17.64(4) = 70.56
( x x )2 f
( x x )2 f 145.40 110
Solution: Finding the Standard Deviation for Grouped Data • Find the sample standard deviation. x 2 x
( x x )2
( x x ) f 145.40 s 1.7 n 1 50 1
( x x )2 f
The standard deviation is about 1.7 children.
111
Section 2.5 Measures of Position
112
Quartiles • Fractiles are numbers that partition (divide) an ordered data set into equal parts. • Quartiles approximately divide an ordered data set into four equal parts. First quartile, Q1: About one quarter of the data fall on or below Q1. Second quartile, Q2: About one half of the data fall on or below Q2 (median). Third quartile, Q3: About three quarters of the data fall on or below Q3. 113
Example: Finding Quartiles The test scores of 15 employees enrolled in a CPR training course are listed. Find the first, second, and third quartiles of the test scores. 13 9 18 15 14 21 7 10 11 20 5 18 37 16 17
Solution: • Q2 divides the data set into two halves. Lower half
Upper half
5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q2 114
Solution: Finding Quartiles • The first and third quartiles are the medians of the lower and upper halves of the data set. Lower half
Upper half
5 7 9 10 11 13 14 15 16 17 18 18 20 21 37
Q1
Q2
Q3
About one fourth of the employees scored 10 or less, about one half scored 15 or less; and about three fourths scored 18 or less. 115
Interquartile Range Interquartile Range (IQR) • The difference between the third and first quartiles. • IQR = Q3 – Q1
116
Example: Finding the Interquartile Range Find the interquartile range of the test scores. Recall Q1 = 10, Q2 = 15, and Q3 = 18
Solution: • IQR = Q3 – Q1 = 18 – 10 = 8 The test scores in the middle portion of the data set vary by at most 8 points.
117
Box-and-Whisker Plot Box-and-whisker plot • Exploratory data analysis tool. • Highlights important features of a data set. • Requires (five-number summary): Minimum entry First quartile Q1 Median Q2 Third quartile Q3 Maximum entry 118
Drawing a Box-and-Whisker Plot 1. Find the five-number summary of the data set. 2. Construct a horizontal scale that spans the range of the data. 3. Plot the five numbers above the horizontal scale. 4. Draw a box above the horizontal scale from Q1 to Q3 and draw a vertical line in the box at Q2. 5. Draw whiskers from the box to the minimum and maximum entries. Box Whisker Minimum entry
Whisker
Q1
Median, Q2
Q3
Maximum entry
119
Example: Drawing a Box-and-Whisker Plot Draw a box-and-whisker plot that represents the 15 test scores. Recall Min = 5 Q1 = 10 Q2 = 15 Q3 = 18 Max = 37 Solution: 5
10
15
18
37
About half the scores are between 10 and 18. By looking at the length of the right whisker, you can conclude 37 is a possible outlier. 120
Percentiles and Other Fractiles Fractiles Quartiles
Deciles Percentiles
Summary Divides data into 4 equal parts Divides data into 10 equal parts Divides data into 100 equal parts
Symbols Q1, Q2, Q3
D1, D2, D3,‌, D9 P1, P2, P3,‌, P99
121
Example: Interpreting Percentiles The ogive represents the cumulative frequency distribution for SAT test scores of college-bound students in a recent year. What test score represents the 72nd percentile? How should you interpret this? (Source: College Board Online)
122
Solution: Interpreting Percentiles
The 72nd percentile corresponds to a test score of 1700. This means that 72% of the students had an SAT score of 1700 or less.
123
The Standard Score Standard Score (z-score) • Represents the number of standard deviations a given value x falls from the mean μ.
value - mean x • z standard deviation
124
Example: Comparing z-Scores from Different Data Sets In 2007, Forest Whitaker won the Best Actor Oscar at age 45 for his role in the movie The Last King of Scotland. Helen Mirren won the Best Actress Oscar at age 61 for her role in The Queen. The mean age of all best actor winners is 43.7, with a standard deviation of 8.8. The mean age of all best actress winners is 36, with a standard deviation of 11.5. Find the z-score that corresponds to the age for each actor or actress. Then compare your results.
125
Solution: Comparing z-Scores from Different Data Sets • Forest Whitaker
z
x
• Helen Mirren
z
x
45 43.7 0.15 8.8
0.15 standard deviations above the mean
61 36 2.17 11.5
2.17 standard deviations above the mean
126
Solution: Comparing z-Scores from Different Data Sets
z = 0.15
z = 2.17
The z-score corresponding to the age of Helen Mirren is more than two standard deviations from the mean, so it is considered unusual. Compared to other Best Actress winners, she is relatively older, whereas the age of Forest Whitaker is only slightly higher than the average age of other Best Actor winners. 127
هذا الجزء توضيحي للطالب الغير قادرين على االستيعاب الجيد
128
Median
Coefficient of variation (c.v.)
بالتوفيق
138