Download  Chapter 3 Numerically Summarizing Data

Chapter 3 Numerically Summarizing Data
Chapter 3.1 Measures of Central Tendency
Objective A: Mean, Median and Mode
Three measures of central of tendency: the mean, the median, and the mode.
A1. Mean
The mean of a variable is the sum of all data values divided by the number of observations.
Population mean:  
x
i
N
where
xi
is each data value and N is the population size (the
number of
observations in the population).
Sample mean: x 
x
i
n
where
xi
is each data value and n in the sample size (the number of
observations in the sample).
Example 1: Population: 12 16 23 17 32 27 14 16
Compute the population mean and sample mean from a simple random sample of size 4.
Does the sample mean equal to the population mean? Does the population mean or sample
mean stay the same? Explain.
(a) Population mean: (Round the mean to one more decimal place than that in the raw
data) 𝜇 =
=
=
∑𝑥𝑖
N=8
𝑁
12+16+23+17+32+27+14+16
8
157
8
= 19.625
≈ 19.6
(b) Sample mean:
From a lottery method, 23 16 14 17 were selected.
x=
=
∑𝑥𝑖
n=4
𝑛
23+16+14+17
4
=
70
5
= 17.5
(c) Does the sample mean equal to the population mean?
No.
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(d) Does the population mean or sample mean stay the same?
Explain.
𝜇 stays the same.
x varies from sample to sample.
A2. Median
The median, M, is the value that lies in the middle of the data when arranged in ascending
order.
Example 1: Find the median of the data given below.
4 12 32 24 9 18 28 10 36
Reorder:
4
9
10
12
18
24
28
32
36
n = 9 = odd
The location of the median is at =
n 1
2
=
9 1
= 5th position.
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The median is 18.
Example 2: Find the median of the data given below.
$35.34 $42.09 $38.72 $43.28 $39.45 $49.36 $30.15 $40.88
Reorder: $30.15 $35.34 $38.72 $39.45 $40.88 $42.09
$43.28 $49.36
N = 8 = even
The location of median is between
n
n
and  1 which is between
2
2
8
8
and=  1 4th and 5th position.
2
2
The median =
39.45+40.88
2
=
80.33
2
= 40.165
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A3. Mode
Mode is the most frequent observation in the data set.
Example 1: Find the mode of the data given below.
76 60 81 72 60 80 68 73 80 67
Reorder:
60 60 67 68 72 73 76 80 80 81
Mode = 60 and 80
Example 2: Find the mode of the data given below.
A C D C B C A B B F B W F D B W D
A D C D
Reorder:
AAABBBBBCCCCDDDDDFFWW
Mode = B and D
Example 3: The following data represent the G.P.A. of 12 students.
2.56 3.21 3.88 2.44 1.96 2.85 2.32 3.38 1.86 3.04 2.75 2.23
Find the mean, median, and mode G.P.A.
Reorder: 1.86 1.96 2.23 2.32 2.44 2.56 2.75 2.85 3.04
3.21 3.38 3.88
(a) mean
x=
=
=
∑𝑥𝑖
n = 12
𝑛
18.6+1.96+2.23+2.32+2.44+2.56+2.75+2.85+3.04+3.21+3.38+3.88
32.48
12
12
≈ 2.7067
≈ 2.707
(b) median
Reorder: 1.86 1.96 2.23 2.32 2.44 2.56 2.75 2.85 3.04
3.21 3.38 3.88
median = (2.56+2.75)/2 = 2.655
(c) mode
Reorder: 1.86 1.96 2.23 2.32 2.44 2.56 2.75 2.85 3.04
3.21 3.38 3.88
mode = None
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Objective B: Relation Between the Mean, Median and Distribution Shape
- The mean is sensitive to extreme data. For continuous data, if the distribution shape
is a bell-shaped curve, the mean is a better measure of central tendency because it
includes all data values in a data set.
- The median is resistant to extreme data. For continuous data, if the distribution
shape is skewed to the right or left, the median is a better measure of central
tendency.
- The mode is used to represent the measure of central tendency for qualitative data.
Definition: A numerical summary of data is said to be resistant if extreme values (very large or
small) relative to the data do not affect its value substantially.
Mean or Median versus Skewness
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Chapter 3.2 Measures of Dispersion
Measurement of dispersion is a numerical measure that can quantify the spread of data.
In this section, the three numerical measures of dispersion that we will discuss are the range,
variance, and standard deviation. In the later section, we will discuss another measure of
dispersion called interquartile range (IQR).
Objective A:Range, Variance and Standard Deviation
A1.Range
Range = R = largest data value - smallest data value
The range is not resistant because it is affected by extreme values in the data set.
A2. Variance and Standard Deviation
Variance is based on the deviation about the mean. Since the sum of deviation about the mean
is zero, we cannot use the average deviation about the mean as a measure of spread.
We use the average squared deviation instead.
The population variance,  2 , of a variable is the sum of the squared deviations about the
population mean,  , divided by the number of observations in the population, N .

2
 (x  )

2
i
Definition Formula
N
 
2
x
2
i
 x 

2
i
N
Computational Formula
N
The sample variance, s 2 , of a variable is the sum of the squared deviations about the sample
mean, x , divided by the number of observations in the sample minus 1, n  1 .
s2 
s 
2
 (x  x )
2
i
Definition Formula
n 1
x
i
2
 x 

2
i
n 1
n
Computational Formula
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In order to use the sample variance to obtain an unbiased estimate of the population variance,
we divide the sum of the squared deviations about the sample mean by n  1 .We call n  1 the
degree of freedom because the first n  1 observations have freedom to be whatever value they
wish, but the nth value has no freedom in order to force
The population standard deviation,
 ( x  x ) to be zero.
i
 , is the square root of the population variance or
σ  Population Variance
The sample standard deviation, s , is the square root of the sample variance or
s  Sample Variance
To avoid round-off error, never use the rounded value of the variance to compute the standard
deviation. Keep a few more decimal places for an intermediate step calculation.
Example 1: Use the definition formula to find the population variance and standard deviation.
Population: 4, 10, 12, 13, 21
Definition formula 𝛿 2 =
𝜇=
∑(𝑥𝑖 −𝜇)
where 𝜇 =
𝑁
4+10+12+13+21
5
=
5
𝑁
= 12
xi  
4 - 12 = -8
10 – 12 = -2
12 – 12 = 0
13 – 12 = 1
21 – 12 = 9
xi
4
10
12
13
21
Population variance: 𝜎 2 =
60
∑𝑥𝑖
150
5
( xi   ) 2
(-8)² = 64
(-2)² = 4
0² = 0
1² = 1
9² = 81
 ( xi   ) 2  150
= 30
Population standard deviation: √𝜎 2 = 𝜎 = √30≈ 5.5
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Example 2: Use the definition formula to find the sample variance and standard deviation.
Sample: 83, 65, 91, 84
Definition formula 𝑠 2 =
Sample mean: x =
∑(𝑥𝑖 − x )2
𝑛−1
83+65+91+84
4
=
where x =
323
4
∑𝑥𝑖
𝑛
= 80.75 ≈ 80.8
xi  
xi
( xi   ) 2
83
83 – 80.75 = 2.25
(2.25)² = 5.0625
65
65 – 80.75 = -15.75
(-15.75)² = 248.0625
91
91 – 80.75 = 10.25
(10.25) 2  105.0625
84
84 – 80.75 = 3.25
(3.25) 2  10.5625
 ( xi  x) 2  368.75
Sample variance: 𝑠 2 =
368.75
4−1
≈ 122.9166 ≈ 122.9
Sample standard deviation: s = √122.9166 ≈ 11.08677 ≈ 11.1
Example 3: Use StatCrunch to find the sample variance and standard deviation.
Sample: 83, 65, 91, 84 (same data set as Example 2)
Step 1:
Click StatCrunch navigation button under the Course Home page 
Click StatCrunch website  Click Open StatCrunch
Input the raw data in var 1 column  Click Stat Click Summary Stats Columns
Step 2:
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Click var1 under Select column(s): Under Statistics:, choose Variance and Std. dev. (click
them while holding Ctrl key on the keyboard)  Click Compute!
Variance and standard deviation are computed.
s 2  122.9
s  11.1
For more detailed instructions, please download
“Q3.2.20 “by clicking the StatCrunch Handout navigation button of the course homepage.
Note : For a small data set, students are expected to calculate the standard deviation by
hand.
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Objective C : Empirical Rule
The figure below illustrates the Empirical Rule
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Example 1:
SAT Math scores have a bell-shaped distribution with a mean of 515 and a
standard deviation of 114. (Source: College Board, 2007)
(a) What percentage of SAT scores is between 401 and 629?

  1
  1
 515  114
 629
515
 515  114
 401
According to the Empirical Rule, approximately 68% of the data will lie within 1
standard deviation of the mean.
68% of SAT scores is between 401 and 629.
(b) What percentage of SAT scores is between 287 and 743?
 2
2
 1
401
1
515
  2
515  2(114)
 287
629
  2
515  2(114)
 743
According to the Empirical Rule, approximately 95% of the data will lie within 2
standard deviations of the mean.
95% of SAT scores is between 287 and 743.
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(c) What percentage of SAT scores is less than 401 or greater than 629?
–
100%
68%
401
629
 1 1
=
401 515 629
32%
(d) What percentage of SAT scores is between 515 and 743?
95%
287
743
95÷ 2 = 47.5
2
1
=
515 629 743
47.5%
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(e) About 99.7% of SAT scores will be between what scores?
According to the Empirical Rule, approximately 99.7% of the data will lie
within 3 standard deviations of the mean.
(𝜇 − 3𝜎, 𝜇 + 3𝜎)
= (515-3(114), 515 + 3(114))
= (173, 857)
Chapter 3.4
Measures of Position and Outliers
Measures of position determine the relative position of a certain data value within the entire
set of data.
Objective A : z -scores
The z-score represents the distance that a data value is from the mean in terms of the
number of standard deviations.
Population
z -score:
z
x
Sample
z -score:
z
xx
s

Example 1: The average 20- to 29-year-old man is 69.6 inches tall, with a standard deviation of
3.0 inches, while the average 20- to 29-year-old woman is 64.1 inches tall, with a standard
deviation of 3.8 inches. Who is relatively taller, a 67-inch man or 62-inch woman?
Man: µ = 69.6 inches 𝜎 = 3.0 𝑖𝑛𝑐ℎ𝑒𝑠 x = 67 inches
Z=
𝑥− 𝜇
𝜎
=
67−69.6
3.0
= -0.87
0.87 standard deviation below the mean.
Woman: µ = 64.1 inches
Z=
𝑥− 𝜇
𝜎
=
62−64.1
3.8
𝜎 = 3.8 inches
x = 67 inches
= -0.55
0.55 standard deviation below the mean
Therefore, the 62- inch woman is relatively taller than the 67- inch man.
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Objective B:Percentiles and Quartiles
B1. Percentiles
The k th percentile,
Pk , of a set of data is a value such that k percent of the observations are
less than or equal to the value.
Example 1: Explain the meaning of the 5th percentile of the weight of males 36 months
of age is 12.0 kg.
5% of 36-month-old males weigh 12.0 kg or less.
95% of 36-month-old males weigh more than 12.0 kg.
The most common percentiles are quartiles.
The first quartile, Q1 , is equivalent to P .
25
The second quartile, Q2 , is equivalent to
The third quartile, Q3 , is equivalent to
P50 .
P75 .
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Example 2: Determine the quartiles of the following data.
46 45 58 71 42 66 72 42 61 49 80
Ascending order:
42 42 45 46 49 58 61 66 71 72 80
M = 58
𝑄2 = 58
Lower half of the data:
42 42 45 46 49
𝑄1 = 45
Upper half of the data:
61 66 71 72 80
𝑸𝟑 = 71
B2. Interquartile
The interquartile range, IQR, is the measure of dispersion that is based on quartiles. The range
and standard deviation are affected by extreme values. The IQR is resistant to extreme values.
Example 1: One variable that is measured by online homework systems is the amount of time
a student spends on homework for each section of the text. The following is a
summary of the number of minutes a student spends for each section of the text
for the fall 2007 semester in a College Algebra class at Joliet Junior College.
Q1  42 Q2  51.5
Q3  72.5
(a) Provide an interpretation of Q1 .
Q1  42 :
25% of the students spend 42 minutes or less on homework for each section, and 75% of
Q2  51.5:
Q3  72.5:
the students spend more than 42 minutes.
50% of the students spend 51.5 minutes or less on homework for each section, and 50% of
the students spend more than 51.5 minutes.
75% of the students spend 72.5 minutes or less on homework for each section, and 25% of
the students spend more than 72.5 minutes.
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(b) Determine and interpret the interquartile range.
IQR = 𝑄3 - 𝑄1 = 72.5 – 42 = 30.5 minutes
The middle of 50% of all students has a range of 30.5 minutes of time
spent on homework.
(c) Do you believe that the distribution of time spent doing homework is
skewed or symmetric? Why?
Skewed right. The difference between Q2 and Q1 is less than the
difference between Q3 and Q2 .
Objective C:Outliers
Extreme observations are called outliers; they may occur by error in the measurement or during
data entry or from errors in sampling.
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Chapter 3.5 The Five-Number Summary and Boxplots
Objective A :The Five-Number Summary
Example 1: The number of chocolate chips in a randomly selected 21 name-brand cookies
were recorded. The results are shown below.
28 23 28 31 27 29 2419 26 23 21 25 22 23
21 23 33 28 33 21 30
Find the five-number Summary.
M  25
Ascending order:
19 21 21 21 22 23 23 23 23 24 25 26 27 28 28 28 29 30 31 33 33
Lower half of the data: 19 21 21 22 23 23 23 23 24
𝑄1 =
22+23
2
= 22.5
Upper half of the data: 26 27 28 28 28 29 30 31 33 33
𝑄3 =
28+29
2
= 28.5
Five-number summary:
Minimum = 19, Q1 = 22.5, Q2 = 25, Q3 = 28.5, Maximum = 33
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Objective B: Boxplot The five-number summary can be used to construct a graph called the
boxplot.
Objective C:Using a Boxplot to describe the shape of a distribution
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Example 1: Use the side-by-side boxplots shown to answer the questions that follow.
(a) To the nearest integer, what is the median of variable x ?
≈ 15
(b) To the nearest integer, what is the first quartile of variable y ?
≈ 22
(c) Which variable has more dispersion? Why?
The y variable has more dispersion because the IQR on y is wider than the IQR on the x variable.
(d) Does the variable x have any outliers? If so, what is the value of the outlier?
Yes, there is an asterisk on the right side of the boxplot. Outlier ≈ 30
(e) Describe the shape of the variable y . Support your position.
Since there is a longer whisker on the left and Q2  Q1 is bigger than Q3  Q2, the shape of
the distribution is skewed to the left.
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