STA 2023 EXAM-1
Few Solutions for the
Practice Problems
Mudunuru, Venkateswara Rao
STA 2023
Spring 2016
From Chapters 1, 2, and 3
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 1
Practice Problems Sheet
Important vocabulary from Chapter-1 for Exam-1:
1) For the following data given below in A, B, C. Do the following:
a) Find 10%, 15% trimmed mean.
b) Write the five number summary using the entire data. Calculate IQR.
c) Draw a box and whisker plot.
A) 12, 41, 25, 7, 10, 22, 38, 10, 12, 19, 19, 12, 8, 16, 12, 5
B) 19, 11, 7, 24, 13, 15, 10, 3, 10, 20
C) 79, 50, 82, 91, 87, 118, 80, 93
Statistics Survey Bias Stratified
Inferential Parameter Cluster Sampling Error
Systematic Subjects Convenience Blind/Single Blind
Quantitative Census Individuals Systematic
Qualitative Ordinal Simple Random Continuous
Interval Statistic Discrete Descriptive
Ratio Double Blind Nominal Multi-Stage
Sampling
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 2
2) For the interval data given in tables D, E, F, G. Calculate the following. Write your values
of (b),(c),(d),(e) in the Table format given below:
Note: Remember that the table with the following column headings with details of first two
columns and the required number of rows based on the problem will be given on the exam.
a) Class width.
b) Cumulative frequency
c) Class midpoint (or class
mark)
d) Draw a frequency histogram
e) Class boundaries
f) Draw a ogive graph
g) Relative frequency
h) Draw a relative frequency
histogram
Interval Frequency Class
Midpoint
Class
Boundaries
Relative
Frequency
Cumulative
Frequency
D)
Weight(lb) Client
frequency
30-49 8
50-69 10
70-89 12
90-109 7
E)
Time in
seconds
Freq
1 to 10 15
11 to 20 20
21 to 30 29
31 to 40 25
41 to 50 22
51 to 60 14
Cost
in $
Sales
frequency
10-26 2
27-43 15
44-60 48
61-77 23
F)
G)
Length in
Meters Frequency
1.1 to 1.5 2
1.6 to 2.0 4
2.1 to 2.5 7
2.6 to 3.0 10
3.1 to 3.5 11
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 3
3) Estimate the sample mean, sample standard deviation, Sample variance for the
following interval data in tables D, E, F, and G. Use the Table format given
below.
Note: Remember that the table with the following column headings with details of first
two columns and the required number of rows based on the problem will be given on the
exam.
Interval 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚, 𝒇 𝑴𝒊𝒅𝒑𝒐𝒊𝒏𝒕, 𝒙 𝒐𝒓 𝒎 𝒙𝒇 𝒙𝟐𝒇
4) Draw Stem and Leaf graph for the following and identify the shape. (Symmetric/Right Skewed/Left Skewed)
a) Student GPAs: 1.8, 2.2, 2.6, 3.4, 3.5, 1.9, 3.2, 3.3, 2.4, 4.0, 3.5, 2.9, 3.6, 3.6, 2.7, 3.5, 4.0,
3.3, 3.0, 2.2, and 1.9.
b) Final scores of students:
70, 48, 100, 65, 69, 73, 30, 93, 76, 82, 100, 89, 72, 74, 95, 91, 86, 75, 55, and 76.
5) Draw the Pie Chart for the following data in tables H, I, J
Note: Remember that the table with the following column headings with details of first
two columns and the required number of rows based on the problem will be given on the
exam
6) Extract the data from the stem and leaf plots given below in K, L:
Category 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚, 𝒇 𝑷𝒆𝒓𝒆𝒏𝒕𝒂𝒈𝒆 (%) 𝑫𝒆𝒈𝒓𝒆𝒆𝒔
H) I) J)
ICU
Type
Frequency
Medical 12
Surgery 6
Cardiac 5
Other 2
Expenses Cost in $
Housing & Food 130
Savings 25
Miscellaneous 20
Insurance 35
Major Students
Maths 45
Biology 21
Nursing 55
Business 45
Stem Leaf
2 3,4,5,6,8,9
3 5,7
4 1,4,4,6,6,8,9
5 5,8,9
K) Stem Leaf
7 0, 3, 3, 5, 6, 7
8 2, 4, 5, 5, 4
9 0, 0, 5, 3, 7, 1
L)
Key: 2|3 is 0.23
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 4
REMEMBER:
Graph Type X-Axis Y-Axis Histogram or
Frequency Histogram
Class
Boundaries/Midpoints Frequency
Relative Frequency
histogram
Class
Boundaries/Midpoints Relative Frequency
Ogive Graph Class Upper
Boundaries Cumulative frequency
𝑪𝒉𝒂𝒑𝒕𝒆𝒓 − 𝟑 𝑸𝒖𝒊𝒄𝒌 𝒇𝒐𝒓𝒎𝒖𝒍𝒂𝒔 𝑳𝒊𝒔𝒕𝒆𝒅 𝑫𝒂𝒕𝒂
𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊
𝒏𝒊=𝟏
𝒏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
𝒔 = √∑(𝒙𝒊 − �̅�)𝟐
𝒏 − 𝟏 𝑶𝑹
𝒔 =√∑ 𝒙𝒊
𝟐 −(∑ 𝒙𝒊)𝟐
𝒏𝒏 − 𝟏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆
𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔
𝑮𝒓𝒐𝒖𝒑𝒆𝒅 𝒐𝒓 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂. 𝑵𝒐𝒕𝒆: 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂 𝒊𝒔 𝒔𝒂𝒎𝒆 𝒂𝒔 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂 𝒘𝒊𝒕𝒉 𝒂𝒏 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒂𝒍
𝒔𝒕𝒆𝒑 𝒐𝒇 𝒇𝒊𝒏𝒅𝒊𝒏𝒈 𝒕𝒉𝒆 𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒆𝒂𝒄𝒉 𝒄𝒍𝒂𝒔𝒔
𝒂𝒏𝒅 𝒕𝒓𝒆𝒂𝒕𝒊𝒏𝒈 𝒊𝒕 𝒂𝒔 𝒙𝒊 𝒐𝒇 𝒕𝒉𝒆 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂
𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊𝒇𝒊
𝒌𝒊=𝟏
∑ 𝒇𝒊𝒌𝒊=𝟏
=∑ 𝒎𝒊𝒇𝒊
𝒌𝒊=𝟏
∑ 𝒇𝒊𝒌𝒊=𝟏
=∑ 𝒎𝒊𝒇𝒊
𝒌𝒊=𝟏
𝒏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
𝒔 = √∑(𝒙𝒊−�̅�)𝟐𝒇𝒊
𝒏−𝟏 𝑶𝑹 𝒔 = √∑ 𝒙𝒊
𝟐𝒇𝒊−(∑ 𝒙𝒊𝒇𝒊)
𝟐
𝒏
𝒏−𝟏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔 © Ven M
udun
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 5
FEW SOLUTIONS FOR YOUR HELP. PLEASE NOTE THAT YOU MUST SHOW THE
FORMULAS USED FOR EVERY PROBLEM.
1) [FOR TABLE: A] Ordering the data in ascending form, we get
5 7 8 10 10 12 12 12
12 16 19 19 22 25 38 41
Here we have 16 data values.
a) TRIMMED MEAN
10% Trimmed Mean: We have 10% of 16 = 1.6 ~ 2 { (10
100) × 16 = 1.6 }
Hence Trim 2 data points from both the ends of the ordered data. So we have 12
remaining points.
5, 7, 8, 10, 10, 12, 12, 12, 12, 16, 19, 19, 22, 25, 38, 41
The Mean of 12 points is the 10% Trimmed mean and is obtained by summing
the remaining 12 data points and dividing by 12.
10% Trimmed mean, �̅� =∑ 𝑥𝑖
𝑛=
177
12= 14.75
15% Trimmed Mean: We have 15% of 16 = 2.4 ~ 2 { (15
100) × 16 = 2.4 }
Hence Trim 2 data points from both the ends of the ordered data. So we have 12
remaining points. Which in turn gives us the same trimmed mean as we have for 10%
above.
b) Five Number Summary: 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 5;
𝑄1 =10 + 10
2= 10; 𝑄2 =
12 + 12
2= 12; 𝑄3 =
19 + 22
2= 20.5
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 = 41; 𝐼𝑄𝑅 = 𝑄3 – 𝑄1 = 20.5 – 10 = 10.5
c) Box Plot:
5 10 12 20.5 41 © Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 6
Outlier Identification:
Any point which does not fall in the interval (𝑄1 − 1.5 ∗ 𝐼𝑄𝑅, 𝑄3 + 1.5 ∗ 𝐼𝑄𝑅) is considered as
an outlier.
i.e., ( 𝟏𝟎 – (𝟏. 𝟓 ∗ 𝟏𝟎. 𝟓), 𝟐𝟎. 𝟓 + (𝟏. 𝟓 ∗ 𝟏𝟎. 𝟓)) = ( −𝟓. 𝟕𝟓, 𝟑𝟔. 𝟐𝟓 ). Hence the potential outliers in
the data are 38 and 41.
[FOR TABLE: B] Ordering the data in ascending form, we get
3, 7, 10, 10, 11, 13, 15, 19, 20, 24
a) TRIMMED MEAN
10% Trimmed Mean: We have 10% of 10 = 1 { (10
100) × 10 = 1}
Hence Trim 1 data point from both the ends of the ordered data. So we have 8
remaining points.
3, 7, 10, 10, 11, 13, 15, 19, 20, 24
The Mean of 8 points is the 10% Trimmed mean and is obtained by summing the
remaining 8 data points and dividing by 8.
10% Trimmed mean, �̅� =∑ 𝑥𝑖
𝑛=
105
8= 13.125
15% Trimmed Mean: We have 15% of 10 = 1.5 ~ 2 { (15
100) × 10 = 1.5 }
Hence Trim 2 data points from both the ends of the ordered data. So we have 6
remaining points.
3, 7, 10, 10, 11, 13, 15, 19, 20, 24
The Mean of 6 points is the 15% Trimmed mean and is obtained by summing
these remaining 6 data points and dividing by 6.
15% Trimmed mean, �̅� =∑ 𝑥𝑖
𝑛=
78
6= 13
b) Five Number Summary: 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 3;
𝑄1 = 10; 𝑄2 =11 + 13
2= 12; 𝑄3 = 19
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 = 24; 𝐼𝑄𝑅 = 𝑄3 – 𝑄1 = 19 – 10 = 9
c) Box Plot:
3 10 12 19 24 © Ve
n Mud
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 7
Fre
qu
ency
Midpoints
Outlier Identification:
Any point which does not fall in the interval (𝑄1 − 1.5 ∗ 𝐼𝑄𝑅, 𝑄3 + 1.5 ∗ 𝐼𝑄𝑅) is
considered as an outlier.
i.e., ( 𝟏𝟎 – (𝟏. 𝟓 ∗ 𝟗), 𝟏𝟗 + (𝟏. 𝟓 ∗ 𝟗)) = ( −𝟑. 𝟓, 𝟑𝟐. 𝟓 )
Hence there are NO potential outliers in the data.
1) [FOR TABLE: D]: 𝐶𝐿𝐴𝑆𝑆 𝑊𝐼𝐷𝑇𝐻: (49 − 30) + 1 = 20
NOTE: Using the data in columns draw the requested graphs. Check slides for more detail.
Frequency Histogram:
Weight(lb) Client
frequency
Class
Mid-Point
Class
Boundaries
Relative
Frequency
(%)
Cumulative
Frequency
30-49 8 39.5 29.5 49.5 21.62 8
50-69 10 59.5 49.5 69.5 27.03 18
70-89 12 79.5 69.5 89.5 32.43 30
90-109 7 99.5 89.5 109.5 18.92 37
37
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 8
Rel
ativ
e F
req
uen
cy
Midpoints
Cu
mu
lati
ve
Fre
qu
ency
Class Upper Boundaries
Relative Frequency Histogram
Ogive Graph:
[FOR TABLE: G] CLASS WIDTH: (1.5-1.1) + 0.1 = 0.5
Length in
Meters Frequency
Class
Midpoint
Class
Boundaries
Relative
Frequency
(%)
Cumulative
Frequency
1.1 to 1.5 2 1.3 1.05 1.55 5.88 2
1.6 to 2.0 4 1.8 1.55 2.05 11.76 6
2.1 to 2.5 7 2.3 2.05 2.55 20.59 13
2.6 to 3.0 10 2.8 2.55 3.05 29.41 23
3.1 to 3.5 11 3.3 3.05 3.55 32.35 34
34 © Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 9
Fre
qu
ency
Midpoints Midpoints
Rel
ativ
e F
req
uen
cy
Class Upper Boundaries
Cu
mu
lati
ve
Fre
qu
ency
3) [TABLE: D] For Sample Mean, Sample Standard Deviation and Sample
Variance. Follow these steps:
Weight(lb) Client
frequency Midpoint, m 𝒇. 𝒎 𝒇. 𝒎𝟐
30-49 8 39.5 316 12482
50-69 10 59.5 595 35402.5
70-89 12 79.5 954 75843
90-109 7 99.5 696.5 69301.75
37 2561.5 193029.3
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 10
You MUST mention the formulas and steps [given in green color below] in your exam)
Sample mean:
𝑁𝑂𝑇𝐸: �̅� =∑ 𝑓𝑖𝑚𝑖
𝑘𝑖=1
∑ 𝑓𝑖𝑘𝑖=1
=∑ 𝑓𝑖𝑚𝑖
𝑘𝑖=1
𝑛 𝑜𝑟 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑠𝑎𝑚𝑒 𝑎𝑠
∑ 𝑥𝑖𝑓𝑖𝑘𝑖=1
∑ 𝑓𝑖𝑘𝑖=1
�̅� =∑ 𝑓𝑖𝑚𝑖
𝑘𝑖=1
𝑛=
2561.5
37= 69.23
Sample Standard Deviation,
𝑠 =√∑ 𝑥𝑖
2𝑓𝑖 −(∑ 𝑥𝑖𝑓𝑖)2
𝑛𝑛 − 1
=√193029.3 −
(2561.5)2
3737 − 1
= 20.88
Sample Variance: 𝑠2 = (15.531)2 = 436.04
[TABLE: G] For Sample Mean, Sample Standard Deviation and Sample Variance.
Follow these steps:
Length in
Meters Frequency Midpoint,
m 𝒇. 𝒎 𝒇. 𝒎𝟐
1.1 to 1.5 2 1.3 2.6 3.38
1.6 to 2.0 4 1.8 7.2 12.96
2.1 to 2.5 7 2.3 16.1 37.03
2.6 to 3.0 10 2.8 28 78.4
3.1 to 3.5 11 3.3 36.3 119.79
34 90.2 251.56
Sample mean:
𝑁𝑂𝑇𝐸: �̅� =∑ 𝑓𝑖𝑚𝑖
𝑘𝑖=1
∑ 𝑓𝑖𝑘𝑖=1
=∑ 𝑓𝑖𝑚𝑖
𝑘𝑖=1
𝑛 𝑜𝑟 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑠𝑎𝑚𝑒 𝑎𝑠
∑ 𝑥𝑖𝑓𝑖𝑘𝑖=1
∑ 𝑓𝑖𝑘𝑖=1
�̅� =∑ 𝑓𝑖𝑚𝑖
𝑘𝑖=1
𝑛=
90.2
34= 2.65 © Ve
n Mud
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 11
Sample Standard Deviation,
𝑠 =√∑ 𝑥𝑖
2𝑓𝑖 −(∑ 𝑥𝑖𝑓𝑖)2
𝑛𝑛 − 1
=√251.56 −
(90.2)2
3434 − 1
= 0.61
Sample Variance: 𝑠2 = (15.531)2 = 0.372
5) For Table –H: You must compute the Last Two Column values.
NOTE: These steps need not be shown in exam
Percentage for Surgery: 6
25× 100 = 24%
Degrees for Surgery: 24
100× 360 = 86.4
For Table –J:
ICU Type Frequency
Medical 12
Surgery 6
Cardiac 5
Other 2
ICU Type Frequency Percentage Degree
Medical 12 48 172.8
Surgery 6 24 86.4
Cardiac 5 20 72
Other 2 8 28.8
25 100 360
Major Students Percentage Degree
Maths 45 27.11 97.60
Biology 21 12.65 45.54
Nursing 55 33.13 119.27
Business 45 27.11 97.60
166
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 12
4) Draw Stem and Leaf graph for the following and identify the shape.
(Symmetric/Right Skewed/Left Skewed)
a) Student GPAs: 1.8, 2.2, 2.6, 3.4, 3.5, 1.9, 3.2, 3.3, 2.4, 4.0, 3.5, 2.9, 3.6, 3.6, 2.7,
3.5, 4.0, 3.3, 3.0, 2.2, and 1.9
Stem and Leaf Plot:
Clearly, Data is almost symmetrical (or bell shaped)
6) [FOR TABLE-K]:
The data points from the stem-leaf plot are:
0.23, 0.24, 0.25, 0.26, 0.28, 0.29, 0.35, 0.37, 0.41, 0.44, 0.44, 0.46, 0.46, 0.48, 0.49,
0.55, 0.58, 0.59
[FOR TABLE-L]:
The data points from the stem-leaf plot are:
70, 73, 73, 75, 76, 77, 82, 84, 85, 85, 84, 90, 90, 95, 93, 97, 91
Stem Leaf
1 8,9,9
2 2,6,4,9,7,2
3 4,5,2,3,5,6,5,3,0
4 0,0
Key: 1|8 is 1.8
© Ven M
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© Venkateswara Rao Mudunuru Spring 2016 Exam-1 Few Solutions For Practice Sheet 13
Note: Formulas will not be provided nor any cheat sheets are allowed on the exams.
𝑪𝒉𝒂𝒑𝒕𝒆𝒓 − 𝟑 𝑸𝒖𝒊𝒄𝒌 𝒇𝒐𝒓𝒎𝒖𝒍𝒂𝒔 𝑳𝒊𝒔𝒕𝒆𝒅 𝑫𝒂𝒕𝒂
𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊
𝒏𝒊=𝟏
𝒏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
𝒔 = √∑(𝒙𝒊 − �̅�)𝟐
𝒏 − 𝟏 𝑶𝑹
𝒔 =√∑ 𝒙𝒊
𝟐 −(∑ 𝒙𝒊)𝟐
𝒏𝒏 − 𝟏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆
𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔
𝑮𝒓𝒐𝒖𝒑𝒆𝒅 𝒐𝒓 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂.
𝑵𝒐𝒕𝒆: 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂 𝒊𝒔 𝒔𝒂𝒎𝒆 𝒂𝒔 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂 𝒘𝒊𝒕𝒉 𝒂𝒏 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒂𝒍 𝒔𝒕𝒆𝒑 𝒐𝒇 𝒇𝒊𝒏𝒅𝒊𝒏𝒈 𝒕𝒉𝒆 𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒆𝒂𝒄𝒉 𝒄𝒍𝒂𝒔𝒔
𝒂𝒏𝒅 𝒕𝒓𝒆𝒂𝒕𝒊𝒏𝒈 𝒊𝒕 𝒂𝒔 𝒙𝒊 𝒐𝒇 𝒕𝒉𝒆 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂
𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊𝒇𝒊
𝒌𝒊=𝟏
∑ 𝒇𝒊𝒌𝒊=𝟏
=∑ 𝒎𝒊𝒇𝒊
𝒌𝒊=𝟏
∑ 𝒇𝒊𝒌𝒊=𝟏
=∑ 𝒎𝒊𝒇𝒊
𝒌𝒊=𝟏
𝒏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
𝒔 = √∑(𝒙𝒊 − �̅�)𝟐𝒇𝒊
𝒏 − 𝟏 𝑶𝑹
𝒔 =√∑ 𝒙𝒊
𝟐𝒇𝒊 −(∑ 𝒙𝒊𝒇𝒊)𝟐
𝒏𝒏 − 𝟏
𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔 © Ven M
udun
uru