STATISTICS_AND_PROBABILITY_for_Senior_Hi.pptx

STATISTICS AND
PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Master Teacher II
Esperanza National High School
Esperanza, Sultan Kudarat, Region XII, Philippines
Email Address: samsudinabdullah42@yahoo.com

REVIEW LESSONS
Measures of Central Tendency (Ungrouped
and Grouped Data)
1. Mean
2. Median
3. Mode
Measures of Variability (Ungrouped and
Grouped data
1. Range
2. Standard Deviation
3. Variance
4. Coefficient of Variation
STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

THE MEAN
Mean(x) is also known as arithmetic
average. It is the sum of the item values
divided by the number of items.
Mean of Grouped Data
If the number of items is too big, it
is best to compute for the measures of
central tendency (Mean, Median and
Mode) using a frequency distribution.

To determine the mean
of a grouped data, use the
formula:
x =
Ʃ𝒇𝒙
𝒏
where:
𝒇 – frequency of the class interval
x – midpoint of the class interval
n – total number of items

Example 1. Calculate the arithmetic mean of the
given distribution on final scores of 100 Grade
11 students in Trigonometry.
x fx
97 291
92 552
87 1,653
82 1,968
77 1,386
72 864
67 536
62 310
57 171
52 104
Scores f
95 – 99 3
90 – 94 6
85 – 89 19
80 – 84 24
75 – 79 18
70 – 74 12
65 – 69 8
60 – 64 5
55 – 59 3
50 – 54 2
n = 100 Ʃfx = 7,835

Solution:
x =
Ʃ𝒇𝒙
𝒏
=
7835
100
= 78.35

Example 2. What is mean of the given distribution
of scores of 75 students in Statistics.
n = 75 Ʃfx = 4,404
Scores f
84 – 90 5
77 – 83 12
70 – 76 8
63 – 69 10
56 – 62 8
49 – 55 2
42 – 48 18
35 – 64 5
28 – 34 3
21 – 27 4
x fx
87 435
80 960
73 584
66 660
59 472
52 104
45 810
38 190
31 93
24 96

Solution:
x =
Ʃ𝒇𝒙
𝒏
=
4404
75
= 58.72

Problem. Compute for the mean of the ages of
GSAT teachers.
Age Group Frequency
60 – 64 2
55 – 59 4
50 – 54 6
45 – 49 12
40 – 44 15
35 – 39 16
30 – 34 12
25 – 29 7
20 – 24 4

Problem. Compute for the mean of the ages of
ENHS teachers. Use the idea of ungrouped and
grouped data. Then compare the results.
60, 62, 54, 40, 33, 35, 22, 23, 55, 57, 25, 26,
34, 44, 41, 44, 44, 44, 44, 45, 59, 58, 52, 50, 36, 33,
34, 37, 39, 22, 29, 30, 31, 32, 33, 34, 40, 45, 56, 63,
45, 25, 27, 28, 39, 34, 45, 37, 61, 60, 33, 32, 31, 22,
23, 24, 25, 26, 27, 28, 30, 50, 48, 52, 55, 62, 60, 33,
34, 44, 44, 44, 45, 46, 42, 37, 39, 40, 42, 44, 23, 24,
61, 50, 53, 55, 56, 57, 58, 59, 33, 30, 29, 33, 50, 28,
27, 45, 45, 44, 44, 56, 56, 57, 40, 44, 45, 24, 25, 26,
30, 31, 27, 27, 30, 24, 25, 41, 43, 42, 50, 53, 55, 54

Solution Using Ungrouped Data
x =
Ʃ𝒙
𝒏
=
𝟔𝟎 + 𝟔𝟐 + 𝟓𝟒 + 𝟒𝟎 + ... + 𝟓𝟒
𝟏𝟐𝟒
=
𝟓𝟎𝟏𝟕
𝟏𝟐𝟒
= 𝟒𝟎. 𝟒𝟔

Class Interval f x fx
62 – 65 3 63.5 190.5
58 – 61 9 59.5 535.5
54 – 57 13 55.5 721.5
50 – 53 9 51.5 463.5
46 – 49 2 47.5 95
42 – 45 24 43.5 1,044
38 – 41 9 39.5 355.5
34 – 37 10 35.5 355
30 – 33 17 31.5 535.5
26 – 29 13 27.5 357.5
22 – 25 15 23.5 352.5
n = 124 ∑fx = 5,006
Solution:
x =
Ʃ𝒇𝒙
𝒏
=
𝟓,𝟎𝟎𝟔
𝟏𝟐𝟒
= 𝟒𝟎. 𝟑𝟕

Solution:
x =
Ʃ𝒇𝒙
𝒏
=
𝟓,𝟎𝟎𝟗
𝟏𝟐𝟒
= 𝟒𝟎. 𝟒𝟎
61 – 63 5 62 310
58 – 60 7 59 413
55 – 57 11 56 616
52 – 54 6 53 318
49 – 51 5 50 250
46 – 48 2 47 94
43 – 45 21 44 924
40 – 42 9 41 369
37 – 39 6 38 228
34 – 36 7 35 245
31 – 33 12 32 384
28 – 30 10 29 290
25 – 27 13 26 338
22 – 24 10 23 230
n = 124 ∑fx = 5,009

THE MEDIAN
Median (Md) is the value of the middle
term when data are arranged in either
ascending or descending order.
Median of Grouped Data
For large quantities of data, the
median is computed using a frequency
distribution with a cumulative frequency
column.

To determine the median of a grouped
data, use the formula:
Md = L +
𝒏
𝟐
−𝑭
𝒇
𝒊 where:
L – the exact lower limit of the median class
n – total number of items
F – “less than” or “equal to” cumulative
frequency preceding the class interval
containing the median
f – frequency of the median class
i – size of the class interval

Example 1. Find the median score of
students of Mr. Dela Cruz Math class.
Scores f
95 – 99 5
90 – 94 11
85 – 89 17
80 – 84 25
75 – 79 20
70 – 74 12
65 – 69 7
60 – 64 3
F
100
95
84
67
42
22
10
3
n = 100

Solution:
Md = L +
𝒏
𝟐
−𝑭
𝒇
𝒊
L = 79.5
n = 100
F = 42
f = 25
i = 99 – 95 + 1 = 5
Md =79.5 +
𝟏𝟎𝟎
𝟐
− 𝟒𝟐
𝟐𝟓
(𝟓)
= 79.5 +
𝟓𝟎 − 𝟒𝟐
𝟐𝟓
(𝟓)
= 79.5 +
𝟖
𝟐𝟓
(𝟓)
= 79.5 +
𝟒𝟎
𝟐𝟓
= 79.5 + 1.6
= 81.1

Example 2. The ages of 115 ENHS teachers
are given below. Find the median age.
Ages f
63 – 69 3
56 – 62 11
49 – 55 18
42 – 48 26
35 – 41 21
28 – 34 15
21 – 27 12
14 – 20 7
7 – 13 2
F
115
112
101
83
57
36
21
9
2
n = 115

Solution:
Md = L +
𝒏
𝟐
−𝑭
𝒇
𝒊
L = 41.5
n = 115
F = 57
f = 26
i = 69 – 63 + 1 = 7
Md = 41.5 +
𝟏𝟏𝟓
𝟐
−𝟓𝟕
𝟐𝟔
(𝟕)
= 41.5 +
𝟓𝟕.𝟓 −𝟓𝟕
𝟐𝟔
(𝟕)
= 41.5 +
𝟎.𝟓
𝟐𝟔
(𝟕)
= 41.5 +
𝟑.𝟓
𝟐𝟔
= 41.5 + 0.135
= 41.635

Problem. Compute for the median of the ages
of ENHS teachers. Use the idea of ungrouped
and grouped data. Then compare the results.
60, 62, 54, 40, 33, 35, 22, 23, 55, 57, 25, 26,
34, 44, 41, 44, 44, 44, 44, 45, 59, 58, 52, 50, 36, 33,
34, 37, 39, 22, 29, 30, 31, 32, 33, 34, 40, 45, 56, 63,
45, 25, 27, 28, 39, 34, 45, 37, 61, 60, 33, 32, 31, 22,
23, 24, 25, 26, 27, 28, 30, 50, 48, 52, 55, 62, 60, 33,
34, 44, 44, 44, 45, 46, 42, 37, 39, 40, 42, 44, 23, 24,
61, 50, 53, 55, 56, 57, 58, 59, 33, 30, 29, 33, 50, 28,
27, 45, 45, 44, 44, 56, 56, 57, 40, 44, 45, 24, 25, 26,
30, 31, 27, 27, 30, 24, 25, 41, 43, 42, 50, 53, 55, 54

Class Interval f F
62 – 65 3 124
58 – 61 9 121
54 – 57 13 112
50 – 53 9 99
46 – 49 2 90
42 – 45 24 88
38 – 41 9 64
34 – 37 10 55
30 – 33 17 45
26 – 29 13 28
22 – 25 15 15
n = 124

Solution:
Md = L +
𝒏
𝟐
−𝑭
𝒇
𝒊
L = 38.5
n = 124
F = 55
f = 9
i = 41 – 38 + 1 = 4
Md = 37.5 +
𝟏𝟐𝟒
𝟐
− 𝟓𝟓
𝟗
(𝟒)
= 37.5 +
𝟔𝟐 − 𝟓𝟓
𝟗
(𝟒)
= 37.5 +
𝟕
𝟗
(𝟒)
= 37.5 +
𝟐𝟖
𝟗
= 37.5 + 3.11
= 40.61

Example 3. Complete the table and compute for the
median score of the Grade 11 students who took the
Precalculus subject.
Scores f CF CP
135 – 139 2
130 – 134 2
125 – 129 4
120 – 124 5
115 – 119 9
110 – 114 8
105 – 109 7
100 – 104 5
95 – 99 3
90 – 94 1
85 – 89 2
80 – 84 1
75 – 79 1
50 100
48 96
46 92
42 84
37 74
28 56
20 40
13 26
8 16
5 10
4 8
2 4
1 2
Note: CF – Cumulative Frequency & CP – Cumulative Percent

Solution:
Md = L +
𝒏
𝟐
−𝑭
𝒇
𝒊
L = 109.5
n = 50
F = 20
f = 8
i = 139 – 135 + 1 = 5
Md = 109.5 +
𝟓𝟎
𝟐
− 𝟐𝟎
𝟖
(𝟓)
= 109.5 +
𝟐𝟓 −𝟐𝟎
𝟖
(𝟓)
= 109.5 +
𝟓
𝟖
(𝟓)
= 109.5 +
𝟐𝟓
𝟖
= 109.5 + 3.125
= 112.625

THE MODE
MODE (Mₒ) is referred to as the
most frequently occurring value in a
given set.
Mode of Grouped Data
In a grouped distribution, the class
interval where the value with the
highest frequency is the modal class.

To determine the mode of a grouped
data, use the formula:
Mo = Lmo + (
𝒅₁
𝒅₁ + 𝒅₂
)i where:
Lmo – the exact lower limit of the modal class
𝑑₁ – the difference between the frequency of
the modal class and that of the frequency
below the modal class
𝑑₂ – the difference between the frequency of
the modal class and that of the frequency
above the modal class
i – the size of the class interval

Example 1. Determine the modal class and the
modal value for the frequency distribution of
ages of teachers in Esperanza NHS.
Age Group Frequency
60 – 64 2
55 – 59 4
50 – 54 6
45 – 49 12
40 – 44 15
35 – 39 16
30 – 34 12
25 – 29 7
20 – 24 4
Solution:
Lmo = 34.5
d1 = 16 – 12 = 4
d2 = 16 – 15 = 1
i = 39 – 35 + 1 = 5

Mo = 34.5 + (
𝟒
𝟒 + 𝟏
)(5)
= 34.5 +
𝟐𝟎
𝟓
= 34.5 + 4
= 38.5

Example 2. Compute for the modal wage of the
workers in a certain private school

Solution:
Lmo = 1,319.5
d1 = 31 – 24 = 7
d2 = 31 – 12 = 19
i = 1,339 – 1,320 + 1 = 20

Mo = 1,319.50 + (
𝟕
𝟕 + 𝟏𝟗
)(20)
= 1,319.50 +
𝟏𝟒𝟎
𝟐𝟔
= 1,319.50 + 5.385
= 1,324.885

Problem. Find the modal age of ENHS
teachers. Use the idea of ungrouped and
grouped data. Then compare the results.
60, 62, 54, 40, 33, 35, 22, 23, 55, 57, 25, 26,
34, 44, 41, 44, 44, 44, 44, 45, 59, 58, 52, 50, 36, 33,
34, 37, 39, 22, 29, 30, 31, 32, 33, 34, 40, 45, 56, 63,
45, 25, 27, 28, 39, 34, 45, 37, 61, 60, 33, 32, 31, 22,
23, 24, 25, 26, 27, 28, 30, 50, 48, 52, 55, 62, 60, 33,
34, 44, 44, 44, 45, 46, 42, 37, 39, 40, 42, 44, 23, 24,
61, 50, 53, 55, 56, 57, 58, 59, 33, 30, 29, 33, 50, 28,
27, 45, 45, 44, 44, 56, 56, 57, 40, 44, 45, 24, 25, 26,
30, 31, 27, 27, 30, 24, 25, 41, 43, 42, 50, 53, 55, 54

Class Interval f
61 – 63 5
58 – 60 7
55 – 57 11
52 – 54 6
49 – 51 5
46 – 48 2
43 – 45 21
40 – 42 9
37 – 39 6
34 – 36 7
31 – 33 12
28 – 30 10
25 – 27 13
22 – 24 10
n = 124
Solution:
Mo = 42.5 + (
𝟏𝟐
𝟏𝟐 + 𝟏𝟗
)(4)
= 42.5 +
𝟒𝟖
𝟑𝟏
= 42.5 + 1.55
= 44.05

MEASURES OF VARIABILITY describe the spread of
the values about the mean.
1. Range
3. Variance
THE RANGE
The difference between the highest and the
lowest values in a given set of data is the RANGE.
Range = highest value – lowest value
Example 1. Find the range for each set of data given
below.
a) 3, 8, 16, 12, 4, 5, 7, 15
b) 25, 32, 9 18, 12, 30, 28, 22
Answer: 13
Answer: 23

Example 2. Determine the range of data presented in
a frequency distribution below.
a) Class Intervals f
20– 25 13
14– 23 5
8– 13 8
2– 9 10
b)
Class Intervals f
90 – 99 3
80 – 89 7
70 – 79 8
60 – 69 5
50 – 59 2
Range = 25.5 – 1.5 = 24
Range = 99.5 – 49.5 = 50

Population STANDARD DEVIATION is the
measure of the variation of a set of data in terms of
the amounts by which the individual values differ
from their mean. It is the most stable measure of
spread.
Population Standard Deviation of Ungrouped Data
s =
Ʃ𝒅²
𝒏
where:
s – standard deviation
d – deviation from the mean
Ʃ𝒅² – sum of squared deviations
n – number of items

Example 1. Calculate the standard deviation of
the given scores in a quiz: 18, 20, 22, 15, 16, 12,
17, 21, 10, 19.
Solution:
x =
𝟏𝟖 + 𝟐𝟎 + 𝟐𝟐 + 𝟏𝟓 + 𝟏𝟔 + 𝟏𝟐 + 𝟏𝟕 + 𝟐𝟏 + 𝟏𝟎 + 𝟏𝟗
𝟏𝟎
=
𝟏𝟕𝟎
𝟏𝟎
= 17
Scores d d2
18
20
22
15
16
12
17
21
10
19
1
3
5
-2
-1
-5
0
4
-7
2
1
9
25
4
1
25
0
16
49
4
Ʃd2
= 134
s =
Ʃ𝒅²
𝒏
=
𝟏𝟑𝟒
𝟏𝟎
= 𝟏𝟑. 𝟒
s = 3.661

Standard Deviation of Grouped Data
s =
Ʃ𝒇𝒅²
𝒏
where:
s – population standard deviation
d – deviation from the mean
Ʃ𝒇𝒅² – sum of product of
frequency and squared deviations
n – number of items

Example 1. Calculate the standard deviation of
the data presented below.
Class
Intervals
f
252 – 260 3
243 – 251 5
234 – 242 9
225 – 233 12
216 – 224 5
207 – 215 4
198 – 206 2
189 – 197 10
180 – 188 8
171 – 179 2
162 – 170 5

Class Intervals f x fx d d2 fd2
252 – 260 3 256 768 43 1849 5547
243 – 251 5 247 1235 34 1156 5780
234 – 242 9 238 2142 25 625 5625
225 – 233 12 229 2748 16 256 3072
216 – 224 5 220 1100 7 49 245
207 – 215 4 211 844 -2 4 16
198 – 206 2 202 404 -11 121 242
189 – 197 10 193 1930 -20 400 4000
180 – 188 8 184 1472 -29 841 6728
171 – 179 2 175 350 -38 1444 2888
162 – 170 5 166 830 -47 2209 11045
n = 65 Ʃfx = 13823 Ʃfd2 = 45188
Solution:

Class Intervals f
355 – 365 13
344 – 354 5
333 – 343 11
322 – 332 12
311 – 321 15
300 – 310 4
289 – 299 20
278 – 288 8
267 – 277 9
256 – 266 11
245 – 255 5
234 – 244 3
223 – 233 2
212 – 222 9
201 – 211 3
Assignment. Find the mean, median, mode and standard deviation
of the given data.

Answer the following as required. Give your answer in nearest thousandths when
needed.
1. What is the size of the class interval? ____________
2. The range of the data is ____________.
3. The frequency of the median class is ____________.
4. ____________ is the frequency of the modal class.
5. What is the class interval of the median class? ____________
6. Give the class interval of the modal class. ____________
7. Compute for the mean of the data. ____________
8. Solve for the median of the data. ____________
9. What is the mode of the data? ____________
10. The standard deviation of the data is ____________.
11. What is the exact lower limit of the median class? ____________
12. ____________ is the exact lower limit of the modal class.
13. The lower the standard deviation, the ____________ the dispersion of items.
14. Compute for the coefficient of variation of the data. ____________
15. Are the data homogeneous or heterogeneous? ____________

Class
Interval
Freque
ncy
(f)
Class
Mark
(x)
fx Cumula
tive
Frequen
cy
(F)
deviati
on
from
the
mean
(d)
d2 fd2
99 – 105 3 102 306 68 23 529 1587
92 – 98 10 95 950 65 16 256 2560
85 – 91 16 88 1408 55 9 81 1296
78 – 84 8 81 648 39 2 4 32
71 – 77 11 74 814 31 -5 25 275
64 – 70 8 67 536 20 -12 144 1152
57 – 63 9 60 540 12 -19 361 3249
50 – 56 3 53 159 3 -26 676 2028
n = 68 Ʃfx = 5361 Ʃfd2 =12179

1. 7
2. 56
3. 8
4. 16
5. 78 – 84
6. 85 – 91
7. 78.838
8. 80.125
9. 88.5
10.13.383
11.77.5
12.84.5
13.BETTER OR CLOSER
14.16.975%
15.HOMOGENEOUS

Class
Interval
Freque
ncy
(f)
Class
Mark
(x)
fx Cumulat
ive
Frequen
cy
(F)
deviation
from the
mean
(d)
d2 fd2
60 – 64 3 62 186
92 – 98 10
85 – 91 16
78 – 84 8
71 – 77 11
64 – 70 8
57 – 63 9
25 – 29 3 27 81
2

Course Outline in Grade 11 Statistics
and Probability
CHAPTER I. Random Variables and
Probability Distributions
- Random Variables
- Probability of an Event
- Probability Distribution
- Mean of a Discrete Probability
- Variance of a Discrete Probability

CHAPTER II. Normal Distribution
- Normal Curve Distribution
- The z-scores
- Regions of Areas Under the Normal
Curve
- Determining Probabilities
- Percentiles Under Normal Curve
- Applying the Normal Curve Concepts
in Problem Solving

CHAPTER III. Sampling and Sampling
Distribution
- Sampling Techniques Commonly
Used in Research
- Sampling Distribution of Sample
Means
- Mean and Variance of the Sampling
Distribution of Means
- Solving Problems Involving Sampling
- Distribution of the Sample Means

CHAPTER IV. Estimation of Parameters
- Point Estimation of a Population
- Confidence Interval Estimates for the
Population Mean
- Confidence Intervals for the Population
Mean when σ is Unknown
- Point Estimate for the Population
Proportion
- Interval Estimates of Population Proportions
- Interpreting Interval Estimates of
Population Proportions
- Confidence Level and Sample Size

CHAPTER V. Conducting Hypothesis Testing
- Hypothesis Testing
- Elements of Hypothesis Testing
- Hypothesis Testing Using the Traditional
Method
- Small-Sample Tests About a Population
Mean μ
- Significance Tests Using the Probability
Value Approach
- Testing Hypothesis Involving Proportions

CHAPTER VI. Commonly Utilized
Inferential Statistical Tools (Application of
Hypothesis Testing)
- z-test
- t-test
- One Way Analysis of Variance (ANOVA)
- Pearson r (Correlation Analysis)

CHAPTER I
RANDOM VARIABLES
AND PROBABILITY
DISTRIBUTIONS

What is a random variable?
Random Variable is a function that associates a
real number to each element in the sample space. It is
a variable whose values are determined by chance.
A random variable is discrete random variable if
its set of possible outcomes is countable. Mostly,
discrete random variables represent count data, such
as the number of defective chairs produced in a
factory.
A random variable is a continuous random
variable if it takes values on a continuous scale.
Often, continuous random variables represent
measured data, such as heights, weights, and
temperatures.

A. Classify the following random variables as discrete or
continuous.
1. The number of defective computers produced by a
manufacturer
2. The weight of newborns each year in a hospital
3. The number of siblings in a family
4. The amount of paint utilized in a building project
5. The number of dropouts in a school
6. The speed of a car
7. The number of female athletes
8. The time needed to finish the test
9. The amount of sugar in a cup of coffee
10. The number of people who are playing lotto each day
11. The number of accidents per year in an accident prone area

12. The amount of salt and ice to preserve ice cream
13. The number of all public school students in the world
14. The magnitude of several earthquakes
15. The number of private school teachers in the
Philippines
16. The body temperature of a patient
17. The size of a Flat TV screen
18. The number of households in a subdivision
19. The heights of students
20. The vital statistics a female candidate
21. The number of used clothes for the refugees
22. The number of eggs in one tray

23. The length of the top of a table
24. The amount of sugar needed to bake
25. The number of students in the TVL track
26. The width of a blackboard
27. The sticks of chalk in a box
28. The number of coins in my pocket
29. The number of Korean teachers here at ENHS
30. The kilogram of fruits in a table
31. The storm signals of typhoons
32. The distance between school and market
33. The angle of elevation
34. The height of flagpole
35. The thickness of a book

A. Classify the following random variables as discrete or
continuous.
1. The number of defective computers produced by a
manufacturer
2. The weight of newborns each year in a hospital
3. The number of siblings in a family
4. The amount of paint utilized in a building project
5. The number of dropouts in a school
6. The speed of a car
7. The number of female athletes
8. The time needed to finish the test
9. The amount of sugar in a cup of coffee
10. The number of people who are playing lotto each day
11. The number of accidents per year in an accident prone area
Discrete
Continuous
Discrete
Continuous
Discrete
Continuous
Discrete
Continuous
Continuous
Discrete
Discrete

12. The amount of salt and ice to preserve ice cream
13. The number of all public school students in the world
14. The intensity of several earthquakes striking
Mindanao
15. The number of private school teachers in the
Philippines
16. The body temperature of a patient
17. The size of a Flat TV screen
18. The heights of students
19. The number of households in a subdivision
20. The vital statistics a female candidate
21. The number of used clothes for the refugees
22. The number of eggs in one tray
Continuous
Discrete
Continuous
Discrete
Continuous
Continuous
Continuous
Discrete
Continuous
Discrete
Discrete

23. The length of the top of a table
24. The amount of sugar needed to bake
25. The number of students in the TVL track
26. The width of a blackboard
27. The sticks of chalk in a box
28. The coins in my pocket
29. The Korean teachers here at ENHS
30. The kilogram of fruits in a table
31. The storm signals of typhoons
32. The distance between school and market
33. The angle of elevation
34. The height of flagpole
35. The thickness of a book
Continuous
Continuous
Discrete
Continuous
Discrete
Discrete
Discrete
Continuous
Continuous
Continuous
Continuous
Continuous
Continuous

B.1. Suppose three cell phones are tested at random. Let D represent
the defective cell phones and N represent the non-defective cell
phones. Assume X be the random variable representing the number of
defective cell phones. Complete the table below to show the values of
the random variable.
Possible Outcomes Value of the Random Variable X
(number of defective cell phones)
NNN 0
NND 1
NDN 1
DND 2
DDN 2
DNN 1
NDD 2
DDD 3
Possible
Outcomes
Value of the Random Variable X
(number of defective cell phone)
The values of a random variable X are 0, 1, 2 and 3.

2. Suppose three coins are tossed. Let Y be the random variable
representing the number of tails that occur. Find the values of
the random variable Y. Complete the table below.
Possible Outcomes Value of the Random Variable Y
(number of tails)
HHH 0
THH 1
HTH 1
HHT 1
HTT 2
THT 2
TTH 2
TTT 3
The values of the random variable Y are 0, 1, 2 and 3.
Possible
Outcomes
Value of the Random Variable Y
(number of tails)

Suppose four coins are
tossed. Let X be the random
variable representing the
number of HEADS that occur.
Find the values of the random
variable X. Complete the table.
Quiz # 2

Possible Outcomes
Value of the Random
Variable X
The values of the random variable X are ____________________________.

Possible Outcomes
Value of the Random
Variable X
(Number of Heads that
occur)
TTTT 0
HTTT 1
THTT 1
TTHT 1
TTTH 1
HHTT 2
TTHH 2
THHT 2
HTTH 2
THTH 2
HTHT 2
HHHT 3
THHH 3
HTHH 3
HHTH 3
HHHH 4
The values of the random variable X are 0, 1, 2, 3 & 4.

3. Two balls are drawn in succession without replacement from
an urn containing 5 red balls and 6 blue balls. Let Z be the
random variable representing the number of blue balls. Find
the values of the random variables Z. Complete the table.
Possible Outcomes Value of the Random Variable Z
(number of blue balls)
RR 0
RB 1
BR 1
BB 2
Note: Using the idea of a combination (₁₁C₂ = 55), there are 55 outcomes of the sample
space. In that combinations, Blue doesn’t occur if you pick up all RED. Sometimes,
BLUE occurs only once or twice.
Thus, the values of the random variable Y are 0, 1 and 2.

4. A random experiment consists of selecting two balls in
succession from an urn containing two black balls and one white
ball. Specify the sample space for this experiment. Let K be the
random variable that represents the number of black balls. What
are the values of K?
Solution:
n(S) = nCr =
𝑛!
𝑟! 𝑛−𝑟 !
= ₃C₂ =
3!
2! 3−2 !
=
3(2!)
2!1!
= 3
S = {(Black, Black), (Black, White), (White, Black)}
No Black 0
1 Black 2
2 Black 1
The random variable K has values of 0, 1 and 2.

5. A random experiment consists of selecting two balls in
succession from an urn containing four black balls and two white
balls. Specify the sample space for this experiment. Let M be the
random variable that represents the number of black balls. What
are the values of M?
Solution:
n(S) = nCr =
𝑛!
𝑟! 𝑛−𝑟 !
= ₆C₂ =
6!
2!4!
=
6(5)(4!)
2(1)(4!)
= 15
S = {W₁W₂, W₁B₁, W₁B₂, W₁B₃, W₁B₄, W₂B₁, W₂B₂, W₂B₃, W₂B₄,
B₁B₂, B₁B₃, B₁B₄, B₂B₃, B₂B₄, B₃B₄}
0 Back 1
1 Black 8
2 Black 6
The random variable M has values of 0, 1 and 2.

Lesson 1 Sample Space and Events
A sample space denoted by S is the se of all possible outcomes of an experiment. Each
possible outcome or element of the set is called a point or a sample point. In other words, an
element of the set is called a point or a sample point in the sample space.
An event is any subset of a sample space.
Examples:
1. Experiment of Tossing a Coin
S = {h, t}
2. Experiment of Tossing Two Coins
S = {(h, h), (h, t), (t, h), (t, t)}
3. Experiment of Rolling a Die
S = {1, 2, 3, 4, 5, 6}
PROBABILITY OF AN EVENT

4. ExperimentofRollingTwoDice(Oneisred,theotherisgreen.)
Thesamplespaceofthisexperimentisillustratedbelow.
R/G 1 2 3 4 5 6
1 {(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
4 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)}

6. Five coins are tossed. Let X be the random variable that represents the number of TAILS.
Enumerate the outcomes of the sample space and determine the possible values of the
random variable X.
0 TAIL HHHHH
1 TAIL THHHH HTHHH HHTHH HHHTH HHHHT
2 TAILS TTHHH THTHH THHTH THHHT HTTHH HHTTH HHHTT HTHTH
HHTHT HTHHT
3 TAILS HHTTT HTHTT HTTHT HTTTH THHTT TTHHT TTTHH THTHT
TTHTH THTTH
4 TAILS TTTTH THTTT TTHTT TTTHT TTTTH
5 TAILS TTTTT
The values of a random variable X are 0, 1, 2, 3, 4 and 5.
Note: There are 32 outcomes of the sample space since
tossing five coins will give you an equation 2⁵ = 32.

Exercise:
A. List the outcomes of the sample space of the following
experiments. Then find the cardinality of the sample space.
1. Tossing three coins
S = {TTT, TTH, THH, THT, HHT, HTH, HTT, HHH}
n(S) = 8
2. Rolling a die and tossing a coin simultaneously.
S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}
n(S) = 12
3. Tossing a coin and spinning the spinner with 8 numbers.
S = {H1, H2, H3, H4, H5, H6, H7, H8, T1, T2, T3, T4, T5, T6, T7, T8}
n(S) = 16
4. Getting a defective item when two items are randomly selected from a box
of two defective and three non-defective items.
S = {D₁D₂, D₁N₁, D₁N₂, D₁N₃, D₂N₁, D₂N₂, D₂N₃, N₂N₃, N₁N₂, N₁N₃}
n(S) = 10
5. Drawing a spade from a standard deck of cards
n(S) = 52
6. Drawing a card greater than 7 from a deck of cards
n(S) = 52
2³ = 8
6¹(2¹) = 6(2) = 12
₅C₂ = 10
2(8) = 16

B. Find the cardinality of the sample of each experiment.
1. Tossing a Coin
n(S) = 2¹ = 2
2. Tossing Two Coins
n(S) = 2² = 4
3. Tossing Three Coins
n(S) = 2³ = 8
4. Rolling a Die
n(S) = 6¹ = 6
5. Rolling Two Dice
n(S) = 6² = 36
6. Rolling Three Dice
n(S) = 6³ = 216
7. Rolling a Die and Tossing a Coin Simultaneously
n(S) = 6¹(2¹) = 6(2) = 12

8. Rolling Two Dice and Two Coins Simultaneously
n(S) = 6²(2²) = 36(4) = 144
9. Rolling a Die and Tossing Three Coins Simultaneously
n(S) = 6¹(2³) = (6)(8) = 48
10. Rolling Three Dice and Tossing Three Coins
n(S) = (6³)(2³) = (216)(8) = 1,728
11. Drawing a Standard Deck of Cards
n(S) = 52
12. Drawing Three Balls from a Box Containing Ten Balls
n(S) = ₁₀C₃ =
10!
3!7!
=
10(9)(8)(7!)
(3)(2)(1)(7!)
=
720
6
= 120
13. Drawing Four Marbles from an Urn Containing 15 Marbles
n(S) = ₁₅C₄ =
15!
4!11!
=
15(14)(13)(12)(11!)
(4)(3)(2)(1)(11!)
=
32,760
24
= 1,365
14. Drawing Two Apples from a Basket Containing 8 Apples
n(S) = ₈C₂ =
𝟖!
𝟐!𝟔!
=
(𝟖)(𝟕)(𝟔!)
(𝟐)(𝟏)(𝟔!)
=
𝟓𝟔
𝟐
= 28

Examples:
1. Whatistheprobabilityofgettinganevennumberintheexperimentofrollingadie?
P(A)=
𝒏(𝑨)
𝒏(𝑺)
=
3
6
=
1
2
Solution:
S={1,2,3,4,5,6} n(S)=6
A={2,4,6} n(A)=3
Review Problems on Probability of an Event

2.Whatistheprobabilitythatthesumofthefacesofthetwodiceis8?
Solution:
F ={(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
n(F) =5
n(S) =36
P(F)=
5
36
=
5
36

Find the probability of the following events.
Event (E) P(E)
1 Getting an even number in a single roll of a die 1
2
2 Getting a sum of 6 when two dice are rolled 5
36
3 Getting an ace when a card is drawn from a deck 1
13
4 The probability that all children are boys if a couple has three children 1
4
5 Getting an odd number and a tail when a die is rolled and a coin is tossed
simultaneously
1
4
18
7 Getting a black card and 10 when a card is drawn from a deck 1
26
8 Getting a red queen when a card is drawn from a deck 1
26
9 Getting doubles when two dice are rolled 1
6
10 Getting a red ball from a box containing 3 red and 6 black balls 1
3

Find the probability of the following events.
Event (E) P(E)
1 Getting an even number in a single roll of a die 1
2
36
3 Getting an ace when a card is drawn from a deck 1
13
4 The probability that all children are boys if a couple has three children 1
4
5 Getting an odd number and a tail when a die is rolled and a coin is tossed
simultaneously
1
4
18
7 Getting a black card and 10 when a card is drawn from a deck 1
26
8 Getting a red queen when a card is drawn from a deck 1
26
9 Getting doubles when two dice are rolled 1
6
10 Getting a red ball from a box containing 3 red and 6 black balls 1
3
E = {2, 4, 6}
E = {(1, 5), (2, 4), (5, 1), (4, 2), (3, 3)}
E = {A of Spade, A of Club, A of Heart, A of Diamond}}
S = {GGG, GBG, BBG, BBB}
E = {(5, 6), (6, 5)}
E = {1T, 3T, 5T}
E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
E = {Q of Diamond, Q of Heart}
E = {10 of Spade, 10 of Club}

Review Problems on Probability
A. From a standard deck of 52 cards, what is the
probability of
1. picking a black card?
2. picking a face card?
3. not picking a face card?
4. picking a black and face card?
5. not picking a black and face card?
6. picking a red and nonface cards?
7. picking an ace card?
8. not picking an ace card?
1/2
3/13
10/13
3/26
23/26
5/13
1/13
12/13

Quiz (1/4 sheet of paper)
A. Find the cardinality of each sample space.
1. Tossing six coins
2. Tossing a pair of coins and spinning a
spinner with 10 numbers simultaneously
3. Rolling a pair of dice and drawing a card
from standard deck simultaneously
4. Tossing three coins and rolling two dice
simultaneously
5. Drawing five balls in a box containing 12
balls
64
40
1,872
288
792

B. On rolling a die, what is the probability of
having
1. a 3?
2. an even number?
3. zero?
4. a number greater than 4?
5. a number lying between 0 and 7?
6. a number less than 4?
7. an odd number?
8. a prime number?
9. a composite number?
10. a multiple of 3?
1/6
1/2
0
1/3
1
1/2
1/2
1/2
1/3
1/3

C. From standard deck of cards, what is the
probability of:
1. picking a red card?
2. picking a face card?
3. picking a nonface card?
4. picking a black and 9 card?
5. not picking a black and 9 card?
6. picking a club card?
7. not picking a club card?
8. picking a red face card?
9. not picking a red face card?
10. picking any card?
1/2
3/13
10/13
1/26
25/26
1/13
12/13
3/26
23/26
1

A box contains 4 white balls, 3 red balls,
and 3 green balls. If three balls are
drawn at random, what is the probability
that
1. they are all white?
2. two are red and one is green?
3. exactly two are green?
4. none is white?
5. they are of different colors?
6. none is red?
1/30
3/40
7/40
1/6
3/10
7/24

Solutions of C
n(S) = ₁₀C₃ =
10!
3!7!
=
10(9)(8)(7!)
(3)(2)(1)(7!)
=
720
6
= 120
1. n(E) = ₄C₃ =
4!
3!1!
=
(4)(3!)
3!(1)
= 4 P(E) =
4
120
=
1
30
2. n(E) = (₃C₂)( ₃C₁) = (
3!
2!1!
)(
3!
1!2!
) =
(3)(2!)
2!(1)
•
(3)(2!)
1!(2!)
= 3(3) = 9
P(E) =
9
120
=
3
40
3. n(E) = (₃C₂)( ₇C₁) = (
3!
2!1!
)(
7!
1!6!
) =
(3)(2!)
2!(1)
•
(7)(6!)
1!(6!)
= 3(7) = 21
P(E) =
21
120
=
7
40

Solutions of C
n(S) = ₁₀C₃ =
10!
3!7!
=
10(9)(8)(7!)
(3)(2)(1)(7!)
=
720
6
= 120
4. n(E) = ₆C₃ =
6!
3!3!
=
(6)(5)(4)(3!)
(3!)(3)(2)(1)
=
120
6
= 20 P(E) =
20
120
=
1
6
5. n(E) = (₄C₁)( ₃C₁)(₃C₁ ) = (
4!
1!3!
)(
3!
1!2!
)(
3!
1!2!
)
=
(4)(3!)
(1)(3!)
•
(3)(2!)
1!(2!)
•
(3)(2!)
1!(2!)
= 4(3)(3) =36
P(E) =
36
120
=
3
10
6. n(E) = ( ₇C₃) =
7!
3!4!
=
(7)(6)(5)(4!)
3(2)(1)(4!)
=
210
6
= 35
P(E) =
35
120
=
7
24

Exercises:
A. Determine whether the given values can serve
as the values of a probability distribution of the random
variable X that can take on only the values 1, 2, 3, and 4.
Explain your answer.
1. P(1) =
1
19
, P(2) =
10
19
, P(3) =
5
19
, P(4) =
5
19
1
19
+
10
19
+
5
19
+
5
19
=
21
19
>1
2. P(1) = 0.25, P(2) = 0.75, P(3) = 0.25, P(4) = 0.25
0.25 + 0.75 + 0.25 + 0.25 = 1.5
3. P(1) = 0.15, P(2) = 0.27, P(3) = 0.29, P(4) = 0.29
0.15 + 0.27 + 0.29 + 0.29 = 1
It cannot
It cannot
It can

Exercises:
A. Determine whether the given values can
serve as the values of a probability distribution of
the random variable X that can take on only the
values 1, 2, 3, and 4. Explain your answer.
1. P(1) =
1
19
, P(2) =
10
19
, P(3) =
5
19
, P(4) =
5
19
1
19
+
10
19
+
5
19
+
5
19
=
21
19
>1 They cannot
2. P(1) = 0.25, P(2) = 0.75, P(3) = 0.25,
P(4) = 0.25
0.25 + 0.75 + 0.25 + 0.25 = 1.5 They cannot

3. P(1) = 0.15, P(2) = 0.27, P(3) = 0.29,
P(4) = 0.29
0.15 + 0.27 + 0.29 + 0.29 = 1
They can
4. P(1) =
1
5
, P(2) =
2
5
, P(3) =
1
5
, P(4) =
1
5
1
5
+
2
5
+
1
5
+
1
5
= 1 They can
5. P(1) = 0.35, P(2) = 0.15, P(3) = 0.05,
P(4) = 0.45
0.35 + 0.15 + 0.05 + 0.45 = 1 They can

6. P(1) = 0.25, P(2) = 0.21, P(3) = 0.19,
P(4) = 0.18
0.25 + 0.21 + 0.19 + 0.18 = 0.83
7. P(1) =
1
8
, P(2) =
3
8
, P(3) =
3
8
, P(4) =
1
8
1
8
+
3
8
+
3
8
+
1
8
= 1
8. P(1) =
5
17
, P(2) =
21
34
, P(3) =
5
34
,
P(4) =
1
17
5
17
+
21
34
+
5
34
+
1
17
= 38/34
They cannot
They can
They cannot

9. P(1) = 0.22, P(2) = 0.11, P(3) = 0.17,
P(4) = 0.50
10. P(1) = 0.05, P(2) = 0.11,
P(3) = 0.18, P(4) = 0.18

B. For each of the following, determine whether it
can serve as the probability distribution of a random
variable X. Explain your answer.
1. P(X) =
1
8
for x = 1, 2, 3, …, 8 It can
1
8
+
1
8
+
1
8
+
1
8
+
1
8
+
1
8
+
1
8
+
1
8
=
8
8
= 1
2. P(X) =
1
6
for x = 1, 2, 3, …, 9
3. P(X) =
3+𝑥
3 −𝑥
for x = 1, 2, 3, 4
4. P(X) =
12
25𝑥
for x = 1, 2, 3, 4
5. P(X) =
𝑥 −2
5
for x = 1, 2, 3, 4, 5

Seatwork (1 whole) (Show your solution).
A box contains 5 yellow ball, 4
brown balls, 4 orange balls and 3 black
balls. If four balls are drawn at random,
what is the probability that
1. they are all yellow?
2. three are brown and one is black?
3. exactly two are orange?
4. none is black?
5. they are of different colors?

Decision-making is an important aspect in
business, education, insurance, and other real-
life situations. Many decisions are made by
assigning probabilities to all possible outcomes
pertaining to the situation and then evaluating
the results. This situation requires the use of
random variable and probability distribution.
Discrete Probability Distribution or
Probability Function consists of the values a
random variable can assume and the
corresponding probabilities of the values.

Properties of a Probability
Distribution
1. The probability of each value of the
random variable must be between or equal to
0 and 1. In symbol, we write it as 0 ≤ P(E) ≤ 1.
2. The sum of the probabilities of all
values of the random variables must be equal
to 1. In symbol, we write it as Ʃ P(E) = 1.

CONSTRUCTING PROBABILITY
DISTRIBUTION and ITS
CORRESPONDING HISTOGRAM
Example 1.
Four coins are tossed. Let Z be the random
variable representing the number of heads that
occur. Construct probability distribution of
Discrete Random Variable Z.

Random
Variable Z
Possible Outcomes of Each
Event
P(Z)
0 HEAD TTTT 1
16
1 HEAD HTTT THTT TTHT TTTH 1
4
2 HEADS HHTT HTHT TTHH
THHT THTHT HTTH
3
8
3 HEADS HHHT HHTH HTHH
THHH
1
4
4 HEADS HHH 1
16
Solution:
n(S) = 2⁴ = 16

Number of Heads
P(Z)
0 1 2 3 4
1
16
1
4
3
8
1
4
1
16
Probability Distribution

0 1 2 3 4
0.2
0.1
0.4
0.3
Number of Tails (Z)
Probability
P(Z)

Example 2. Three coins are tossed. Let Y be the
random variable representing the number of
tails that occur. Construct probability
distribution of a discrete random variable.
Number of Tails Y 0 1 2 3
Probability P(Y) 𝟏
𝟖
𝟑
𝟖
𝟑
𝟖
𝟏
𝟖

0 1 2 3
0.2
0.1
0.4
0.3
Number of Tails (Y)
Probability
P(Y)

Solution:
n(S) = 5C3 = 10
S = {N1N2N3, D1N1N2, D1N1N3, D1N2N3, D2N1N2,
D2N1N3, D2N2N3, D1D2N1, D1D2N2, D1D2N3}
Number of Defective
Computer (X)
0 1 2
Probability P(x)
𝟏
𝟏𝟎
𝟑
𝟓
𝟑
𝟏𝟎

0 1 2
0.4
0.2
0.8
0.6
Number of Tails (X)
Probability
P(X)

X 1 5 8 7 9
P(X) 𝟏
𝟑
𝟏
𝟑
𝟏
𝟑
𝟏
𝟑
𝟏
𝟑
X 0 2 4 6 8
P(X) 𝟏
𝟔
𝟏
𝟔
𝟏
𝟑
𝟏
𝟔
𝟏
𝟔
X 1 2 3 5
P(X) 𝟏
𝟒
𝟏
𝟖
𝟏
𝟒
𝟏
𝟖
X 4 8 12 15 17
P(X) 𝟏
𝟓
𝟏
𝟖
𝟏
𝟖
𝟏
𝟓
𝟏
𝟖
X 1 3 5 7
P(X) 0.35 0.25 0.22 0.12
1)
2)
3)
4)
5)
Determine whether the table presents a probability
distribution. Explain your answer.

Solve the following problems.
1. The daily demand for copies of a movie magazine
at a variety store has the probability distribution as
follows.
Number of Copies X Probability P(X)
0 0.06
1 0.14
2 0.16
3 0.14
4 0.12
5 0.10
6 0.08
7 0.07
8 0.06
9 0.04
10 0.03

Questions:
1. What is the probability that three or more
copies will be demanded in a particular day? 0.64
2. What is the probability that the demand
will be at least two but not more than six? 0.60
is between four and eight? 0.25
is less than nine? 0.93
5. What is the probability that the number
of demand is even number? 0.45
is more than five? 0.28

Mean of a Discrete Probability Distribution
Preparatory Lessons:
A. Given the values of the variables x and y,
evaluate the following summations:
x₁ = 4, x₂ = 2, x₃ = 5, x₄ = 1
y₁ = 2, y₂ = 1, y₃ = 0, y₄ = 2
1. Ʃx = 4 + 2 + 5 + 1 = 12
2. Ʃy = 2 + 1 + 0 + 2 = 5
3. Ʃxy = 4(2) + 2(1) + 5(0) + 1(2) = 12
4. Ʃ(x + y) = (4 + 2) + (2 + 1) + (5 + 0) + (1 + 2) = 17
5. Ʃ4xy = 4(4)(2) + 4(2)(1) + 4(5)(0) + 4(1)(2) = 48

B. The following are the scores of 40
students in a test. Compute the mean score.
Score Number of Students
42 8
50 12
53 9
38 7
46 4
Solution:
x =
𝟒𝟐(𝟖) + 𝟓𝟎(𝟏𝟐) + 𝟓𝟑(𝟗) + 𝟑𝟖(𝟕) + 𝟒𝟔(𝟒)
40
=
𝟏𝟖𝟔𝟑
𝟒𝟎
= 46.575

C. Consider rolling a die. What is the average
number of spots that would appear?
Number of Spots X Probability P(x) x·P(X)
1 1
6
1
6
2 1
6
2
6
3 1
6
3
6
4 1
6
4
6
5 1
6
5
6
6 1
6
6
6
Mean =
𝟐𝟏
𝟔
= 3.5

I. Find the mean, median and mode of each set
of data. Show your solution if any. Round off
your answers in 4 decimal palaces.
1) 10, 8, 7, 15, 20, 8, 8
Solution:
x =
𝟏𝟎 + 𝟖 +𝟕 + 𝟏𝟓 + 𝟐𝟎 + 𝟖 + 𝟖
𝟕
=
𝟕𝟔
𝟕
𝑴𝒅 = 8
x = 10.8571 𝑴𝒐 = 8

I. Find the mean, median and mode of each set
of data. Show your solution if any. Round off
your answers in 4 decimal palaces.
2) 150, 80, 95, 115, 250, 300, 125, 130,
150, 150
Solution:
x =
𝟏𝟓𝟎+𝟖𝟎+𝟗𝟓+𝟏𝟏𝟓+𝟐𝟓𝟎+𝟑𝟎𝟎+𝟏𝟐𝟓+𝟏𝟑𝟎+𝟏𝟓𝟎+𝟏𝟓𝟎
𝟏𝟎
=
𝟏𝟓𝟒𝟓
𝟏𝟎
𝑴𝒅 =
𝟏𝟑𝟎+𝟏𝟓𝟎
𝟏𝟎
=
𝟐𝟖𝟎
𝟐
= 140
= 154.5000 𝑴𝒐 = 150

II. Solve for x , 𝑴𝒅 and 𝑴𝒐 of the following set
of scores.
Solution:
x =
𝟐𝟓(𝟏𝟎)+𝟐𝟑(𝟓)+𝟐𝟎(𝟒)+𝟏𝟓(𝟏𝟏)
𝟑𝟎
=
𝟔𝟏𝟎
𝟑𝟎
𝑴𝒅 =
𝟐𝟑+𝟐𝟎
𝟐
=
𝟒𝟑
𝟐
= 21.5000
= 20.3333 𝑴𝒐 = 15
Scores Frequency
25 10
23 5
20 4
15 11

Formula for the Mean of the Probability
Distribution
µ = Ʃx · P(x)
Examples:
1. The probabilities that a customer will buy 1, 2, 3,
4, or 5 items in a grocery store are
𝟑
𝟏𝟎
,
𝟏
𝟏𝟎
,
𝟏
𝟏𝟎
,
𝟐
𝟏𝟎
,
𝟑
𝟏𝟎
. What
is the average number of items that customer will buy?
Solution:
µ = 1(
𝟑
𝟏𝟎
) + 2(
𝟏
𝟏𝟎
) + 3(
𝟏
𝟏𝟎
) + 4(
𝟐
𝟏𝟎
) + 5(
𝟑
𝟏𝟎
)
=
𝟑
𝟏𝟎
+
𝟐
𝟏𝟎
+
𝟑
𝟏𝟎
+
𝟖
𝟏𝟎
+
𝟏𝟓
𝟏𝟎
µ = 3.1

2. The probabilities that a surgeon operates
on 3, 4, 6, 7 or 8 patients in any day are 0.15, 0.10,
0.20, 0.25, and 0.30, respectively. Find the average
number of patients that a surgeon operates on a
day.
3. Suppose the casino realizes that it is losing
money in the long term and decides to adjust the
payout levels by subtracting $1.00 from each price.
The new probability distribution for each outcome
is provided by the following table.:
Outcome -$2.00 -$1.00 $2.00 $3.00
Probability 0.30 0.40 0.20 0.10

Variance of a Discrete Probability Distribution
σ² = Ʃ(x - µ)² · P(x) or
σ² = Ʃx² · P(x) - µ²
Standard Deviation of a Discrete Probability
Distribution
σ = Ʃ(x − µ)² · P(x) or
σ = Ʃx² · P(x) − µ²

Example:
Find the variance and standard deviation of a given Discrete Probability
Distribution below.
x P(x) x.P(x) x - µ (x - µ)² (x - µ)².P(x)
1 0.20 0.20 -4.48 20.0704 4.014080
3 0.15 0.45 -2.48 6.1504 0.922560
5 0.13 0.65 -0.48 0.2304 0.029952
7 0.25 1.75 1.52 2.3104 0.577607
9 0.27 2.43 3.52 12.3904 3.345408
Ʃx.P(x) = 5.48 Ʃ(x - µ)².P(x) =
8.8896
ơ²= 8.8896 (Variance)
ơ = 8.8896 = 2.9815 (Standard Deviation)

Assignment (1 /2 CW)
Complete the table. Then, find the mean, variance and standard deviation of a
given Discrete Probability Distribution below.
x P(x) x.P(x) x - µ (x - µ)² (x - µ)².P(x)
1 0.10
2 0.18
5 0.22
6 0.19
7 0.15
11 0.16
Ʃx.P(x) = Ʃ(x - µ)².P(x) =

Assignment (1 /2 CW)
Complete the table. Then, find the variance and standard deviation of a given
Discrete Probability Distribution below.
x P(x) x.P(x) x - µ (x - µ)² (x - µ)².P(x)
1 0.10 0.10 -4.51 20.3401 2.03401
2 0.18 0.36 -3.51 12.3201 2.21762
5 0.22 1.10 -0.51 0.2601 0.05722
6 0.19 1.14 0.49 0.2401 0.04562
7 0.15 1.05 1.49 2.2201 0.33302
11 0.16 1.76 5.49 30.1401 4.82242
Ʃx.P(x) = 5.51 Ʃ(x - µ)².P(x) =
9.5099
ơ²= 9.5099 (Variance)
ơ = 9.5099 = 3.0838 (Standard Deviation)

CHAPTER II
NORMAL
DISTRIBUTION

NORMAL CURVE is a bell-shaped curve which shows the
probability distribution of a continuous random variable. It
represents a normal distribution. It has a mean µ = 0 and standard
deviation ơ = 1. Its skewness is 0 and its kurtosis is 3.

Properties of the Normal Probability Distribution
1. The distribution curve is bell-shaped.
2. The curve is symmetrical about its center.
3. The mean, the median, and the mode coincide at the center.
4. The width of the curve is determined by the standard deviation of the
distribution.
5. The tails of the curve flatten out indefinitely along the horizontal axis, always
approaching the axis but never touching it. That is, the curve is asymptotic to the
base line.
6. The area under the curve is 1. Thus, it represents the probability or proportion
or the percentage associated with specific sets of measurement values.

Skewness talks about the degree of
symmetry of a curve. It is asymmetry in a
statistical distribution, in which the curve
appears distorted or skewed either to the
left or to the right. It can be quantified to
define the extent to which a distribution
differs from a normal distribution.
Kurtosis, on the other hand, talks about
the degree of peakedness of a curve. It
refers to the pointedness or flatness of a
peak in the distribution curve.

Skewed to
the Left
Skewed to
the Right
Skewness is less
than zero (negative).
Skewness is greater
than zero (positive).

Types of Kurtosis

If the kurtosis of a curve is greater than
zero (positive), the distribution is said to be
Leptokurtic. This means that the distribution is
taller and thinner than the normal curve.
If the kurtosis of a curve is less than zero
(negative), the distribution is said to be
Platykurtic. This indicates that the distribution
is flatter and wider than the normal curve.
A normal distribution (normal curve) is
said to be Mesokurtic.

The skewness of
a normal curve is 0
and its kurtosis is
3.

A. Determine the area BELOW the following.
1. z = 2
2. z = 2.9
3. z = -1.5
4. z = 2.14
5. z = -2.8
6. z = -2.15
7. z = -0.12
8. z = 1.67
9. z = -0.76
10. z = 0.1

B. Determine the area ABOVE the following.
1. z = 2.5
2. z = -2.5
3. z = 1.25
4. z = -0.15
5. z = 2.13
6. z = -2.15
7. z = -0.03
8. z = -1.64
9. z = 1.96
10. z = 2.33

C. Determine the area of the region indicated
by the following. Draw a normal curve for
each.
1. -1 < z < 1
2. -2 < z < 2
3. -1.5 < z < 2.5
4. 0.18 < z < 3.2
5. -3 < z < 1.65
6. -0.1 < z < 1.47
7. -2.33 < z < 1.64
8. -2.88 < z < 3
9. -1.96 < z < 1.96
10. -2.96 < z < -0.01

A. Determine the area of the region indicated by the
following.
1. -1 < z < 1
2. -2 < z < 2
3. -1.5 < z < 2.5
4. 0.18 < z < 3
5. -3 < z < 1.65
B. Determine the area of the region indicated by the
following.
1. Below z = -2.76
2. Above z = -1.27
3. Below z = 1.09
4. Above z = 1.55
5. Below z = 2.13

Find the area of the shaded region of the normal curve.
1.
A = 0.3413 or 34.13%

2.
A = 2(0.4938)
= 0.9876 or 98.76%

3.
2.
A = 0.5 – 0.3944
= 0.1056 or 10.56%
-1.25

A = 0.4938 + 0.2734
= 0.7672 or 76.72%
4.

5.
A = (0.50 – 0.3944) + (0.4772 – 0.1915)
= 0.1056 + 0.2857
= 0.3913 or 39.13%

A = 0.5 – 0.3944
= 0.1056 or 10.56%

A = 0.5 – 0.3944 + 0.4772 – 0.3159
= 0.1056 + 0.1613
= 0.2669 or 26.69%

A = 0.5 – 0.3944 + 0.3413 + 0.5 – 0.3159
= 0.1056 + 0.3413 + 0.1841
= 0.6310 or 63.10%

A = 0.5 – 0.4970
= 0.003 or 0.30%
-2.75

A = 0.5 – 0.4970 + 0.3944
= 0.003 + 0.3944
= 0.3974 or 39.74%
-2.75

A = 0.5 – 0.4970 + 0.3944 + 0.5 – 0.4394
= 0.003 + 0.3944 + 0.0606
= 0.458 or 45.80%
-2.75

A = 1 – 2(0.4750)
= 1 – 0.95
= 0.05 or 5%

Applications of
Normal Curve

The following formula is used when
sample size is not given:

A. The scores of students in the first quarter
examination for Mathematics has a mean (µ) 32
and standard deviation (σ) of 5. Find the z-
scores corresponding to each of the following.
1. 37
2. 22
3. 33
4. 28
5. 40
6. 27
7. 34
8. 30
9. 32
10. 25
17 22 27 32 37 42 47

Solutions:
1. z =
𝒙 − 𝝁
σ
=
𝟑𝟕 −𝟑𝟐
𝟓
=
𝟓
𝟓
= 1
2. z =
𝒙 − 𝝁
σ
=
𝟐𝟐 −𝟑𝟐
𝟓
=
−𝟏𝟎
𝟓
= -2
3. z =
𝒙 − 𝝁
σ
=
𝟑𝟑 −𝟑𝟐
𝟓
=
𝟏
𝟓
= 0.2
4. z =
𝒙 − 𝝁
σ
=
𝟐𝟖 −𝟑𝟐
𝟓
=
−𝟒
𝟓
= -0.8
5. z =
𝒙 − 𝝁
σ
=
𝟒𝟎 −𝟑𝟐
𝟓
=
𝟖
𝟓
= 1.6

6. z =
𝒙 − 𝝁
σ
=
𝟐𝟕 −𝟑𝟐
𝟓
=
−𝟓
𝟓
= -1
7. z =
𝒙 − 𝝁
σ
=
𝟑𝟒 −𝟑𝟐
𝟓
=
𝟐
𝟓
= 0.4
8. z =
𝒙 − 𝝁
σ
=
𝟑𝟎 −𝟑𝟐
𝟓
=
−𝟐
𝟓
= -0.4
9. z =
𝒙 − 𝝁
σ
=
𝟑𝟐 −𝟑𝟐
𝟓
=
𝟎
𝟓
= 0
10. z =
𝒙 − 𝝁
σ
=
𝟐𝟓 −𝟑𝟐
𝟓
=
−𝟕
𝟓
= -1.4

B. The scores of a group of students in a
standardized test are normally distributed with a
mean of 60 and standard deviation of 8. Answer
the following.
1. How many percent of the students got below
72?
2. What part of the group scored between 58
and 76?
3. If there were 250 students who took the test,
about how many students scored higher than
64?
4. How many percent of the students got above
65?

Solution:
1. z =
𝒙 − 𝝁
σ
=
𝟕𝟐 −𝟔𝟎
𝟖
=
𝟏𝟐
𝟖
= 1.5
Referring to the z-table, the area
below z = 1.5 is 0.9332. Therefore, about
93.32% of the group got below 72.

2. z =
𝒙 − 𝝁
σ
=
𝟓𝟖 −𝟔𝟎
𝟖
=
−𝟐
𝟖
= -0.25
z =
𝒙 − 𝝁
σ
=
𝟕𝟔 −𝟔𝟎
𝟖
=
𝟏𝟔
𝟖
= 2
A = 0.0987 + 0.4772
= 0.5759 or 57.59%
Thus, there were 57.59% of the students who
scored between 58 and 76.

3. z =
𝒙 − 𝝁
σ
= z =
𝟔𝟒 −𝟔𝟎
𝟖
=
𝟒
𝟖
= 0.5
A = 0.5 – 0.1915
= 0.3085
250(0.3085) = 77.125 or 77
Thus, there were 77 students who got
higher than 64.

4. z =
𝒙 − 𝝁
σ
= z =
𝟔𝟓 −𝟔𝟎
𝟖
=
𝟓
𝟖
= 0.63
A = 0.5 – 0.2357
= 0.2643 or 26.43%
Thus, there were 26.43% of the students
who got above 65.

C. A highly selective university only admits the top 5%
of the total examinees in their entrance examination.
The results of this year’s entrance examination follow
a normal distribution with a mean of 285 and a
standard deviation of 12. What is the least score of an
examinee who can be admitted to the university?
Solution:
z =
𝒙 −𝟐𝟖𝟓
𝟏𝟐
A = (1 – 0.05 ) – 0.5
= 0.95 – 0.5
= 0.45
1.65 =
𝒙 −𝟐𝟖𝟓
𝟏𝟐
x – 285 = 1.65(12)
= 19.8 + 285
X = 304.8 or 305

Learning the Probability Notations Under the
Normal Curve
P(a < z < b) denotes the probability that the z-
score is between a and b.
P(z >a) denotes the probability that the z-score is
greater than a.
P(z < a) denotes the probability that the z-score is
less than a.
P(a ≤ z ≤ b) = P(a < a < b)

The Central Limit Theorem is of fundamental
importance in Statistics because it justifies the use of
normal curve methods for a wide range of problems.
This theorem applies automatically to sampling from
infinite population.
The following formula is used when sample is
given.
z =
𝒙 − 𝝁
𝓸
𝒏
where:
𝑥 = sample mean
μ = population mean
σ = population standard deviation
n = sample size

The following formula is used when sample
is not given given.
z =
𝒙 − 𝝁
σ
where:
𝑥 = sample mean
μ = population mean
σ =population standard deviation

Quiz (1/2 CW)
A. Find the following: Draw a
normal curve for each problem
1. P(z < -2.52) =
2. P(z > 2.17) =
3. P(1.23 < z < 2.21) =
4. P(-0.23 < z < -1.41) =
5. P(-2.03 < z < 1.08) =

Problems
1. The average time it takes a group of college
students to complete a certain examination is 46.2
minutes. The standard deviation is 8 minutes. Assume
that the variable is normally distributed.
a. What is the probability that a randomly
selected college student will complete the examination
in less than 43 minutes?
b. If 50 randomly selected college students take
the examination, what is the probability that the mean
time it takes the group to complete the test will be
more than 43 minutes?

a) Given:
x = 43 minutes
μ = 46.2 minutes
σ = 8
Solution:
P(x < 43) = ?
z =
𝒙 − 𝝁
σ
=
43 − 46.2
8
=
−3.2
8
= -0.40
P(x < 43) = P(z < -0.40)
= 0.500 – 0.1554
= 0.3446 or 34.46%
Thus, the probability that a
randomly selected college student
will complete the test in less than
43 minutes is 34.46%.

b) Given:
x = 43 minutes
μ = 46.2 minutes
σ = 8
n = 50
Solution:
P(x > 43) = ?
z =
𝒙 − 𝝁
𝓸
𝒏
=
43 − 46
𝟖
𝟓𝟎
=
−3.2
𝟖
𝟕.𝟎𝟕
=
−𝟑.𝟐
𝟏.𝟏𝟑
= -2.83
P(x > 43) = P(z > -0.2.83)
= 0.4977 + 0.500
= 0.9977 or 99.77%
Thus, the probability that 50
randomly selected college students
will complete the test in more than
43 minutes is 99.77%.

2. The entrance examination scores of incoming
freshmen in a state college are normally distributed
with a mean of 78 and a standard deviation of 10.
What is the probability that a randomly selected
student has a score
a. below 78?
b. below 76?
c. between 75 to 80?
d. above 95?
e. What is the probability that the 45
randomly selected freshmen can have a mean of
greater than 76?

a) Given:
x = 78
μ = 78
σ = 10
Solution:
P(x < 78) = ?
z =
𝒙 − 𝝁
σ
=
78 − 78
10
=
0
10
= 0
P(x < 78) = P(z < 0)
= 0.50 or 50%
Thus, the probability of a randomly
selected student to have a score of
less than 78 is 50%.

b) Given:
x = 76
μ = 78
σ = 10
Solution:
P(x < 76) = ?
z =
𝒙 − 𝝁
σ
=
76 − 78
10
=
−2
10
= -0.2
P(x < 78) = P(z < -0.2)
= 0.0793 or 7.93%
selected student to have a score
less than 76 is 7.93%.

c) Given:
x₁ = 75
x₂ = 80
μ = 78
σ = 10
Solution:
P(75 < x < 80) = ?
z =
𝒙 − 𝝁
σ
=
75 − 78
10
=
−3
10
= -0.3
z =
80 − 78
10
=
2
10
= 0.2
P(75 < x < 80) = 0.1179 + 0.0793
= 0.1972 or 19.72%
selected student to have a score
between 75 and 80 is 19.72%.

d) Given:
x = 95
μ = 78
σ = 10
Solution:
P(x > 95) = ?
z =
𝒙 − 𝝁
σ
=
95 − 78
10
=
17
10
= 1.7
P(x > 95) = P(x > 1.7)
= 0.500 – 0.4554
= 0.0446 or 4.46%
Thus, the probability of a randomly selected
student to have a score above 95 4.46%.

e) Given:
x = 76
μ = 78
σ = 10
n = 45
Solution:
P(x > 95) = ?
z =
𝒙 − 𝝁
ℴ
𝒏
=
76 − 78
𝟏𝟎
𝟒𝟓
=
−2
𝟏𝟎
𝟔.𝟕𝟏
=
−𝟐
𝟏.𝟒𝟗
= -1.34
P(x > 76) = P(x > 1.34)
= 0.4099 + 0.5000
= 0.9099 or 90.99%
Thus, the probability that the 45 randomly
selected freshmen can have a mean of greater
than 76 is 90.99%.

3. Suppose from the 1,000 incoming freshmen who
took the entrance examination, it was found out that
their mean score was 80 and the standard deviation
was 12.
a. How many students passed the test if the
passing score is set at 75?
b. What scores comprise the middle 95% of all
scores?
c. What scores comprise above 95% of all
scores?
d. What scores comprise below 89% of all
scores?

a) Given:
x = 75
μ = 80
ℴ = 12
Solution:
P(x > 75)
z =
𝒙 − 𝝁
σ
=
𝟕𝟓 − 𝟖𝟎
𝟏𝟐
=
−𝟓
𝟏𝟐
= -0.42
P(x > 75) = P(z > -0.42)
= 0.1628 + 0.5000
= 0.6628
0.6628 (1000) = 662.8 or 663
Thus, there were 663 freshmen who passed the
entrance examination..

CHAPTER III
SAMPLING AND
SAMPLING
DISTRIBUTION

POPULATION
SAMPLE
Sampling is a process of
getting the sample.

Statistic versus Parameter
Statistics – a branch of
Mathematics. It is a subject offered in a
school.
Statistic – a datum in a collection
of statistics. It is a characteristic of a
sample. It is used to estimate the value
of a population. The average grade of
students would be an example of a
statistic.

Statistic versus Parameter
Sample Statistic – any quantity computed
from a sample taken from a population with the
intention of using this quantity to estimate same
but unknown quantities of the population. The
examples would be sample mean and sample
variance.
Parameter – a useful component of
statistical analysis. It refers to the characteristics
that are used to define a given population.
Statistic describes a sample while
parameter describes a population. In other
words, statistic is used to estimate a parameter.

Examples of a Parameter
Population mean (µ)
Population standard deviation (σ)
Population variance (σ²)
Examples of a Statistic
Population mean (µ)
Population standard deviation (σ)
Population variance (σ²)

Say something about the following figures.
10
8
80
55
24
38
40
17
29
26
32.7
Sample Mean
34 34
32
32
31
30
33
35
33
33
32.7
Figure 1
Figure 2

Figure 1 Figure 2
Mean 32.7 Mean 32.7
Standard Error 6.92989 Standard Error 0.4726
Median 27.5 Median 33
Mode None Mode 33
Standard Deviation 21.9142 Standard Deviation 1.4944
Sample Variance 480.233 Sample Variance 2.2333
Kurtosis 1.3037 Kurtosis -0.1518
Skewness 1.13241 Skewness -0.3595
Range 72 Range 5
Minimum 8 Minimum 30
Maximum 80 Maximum 35
Sum 327 Sum 327
Count 10 Count 10
Descriptive Statistics of the two given sets of sample data

Random Sampling refers
to the sampling technique in
which each member of the
population is given equal
chance from a population is
called sample and the process
of taking samples is called
sampling.

Since survey research has a larger scope of
respondents, sampling technique is very
necessary. For instance, the population of the
research is 6,033 students, teachers, parents and
school administrators. It doesn’t mean that all of
these 6,033 target respondents will be given a
survey questionnaire. Sampling technique should
be done systematically so that expenses and time
will be minimized but the generality and reliability
of the information will be maintained.

Probability Sampling Methods
1. Simple Random Sampling
- Fishbowl method
- Lottery Method
2. Systematic Sampling
3. Stratified Sampling
4. Cluster Sampling
5. Multistage Sampling

Simple Random Sampling (SRS) is
a basic sampling technique where a
researcher selects a group of a sample
for study from a larger group
(population). Each individual is chosen
entirely by chance and each member of
the population has an equal chance of
being included in the sample.

Systematic Sampling is a statistical
method involving the selection of
elements from an ordered sampling
frame. The most common form of
systematic sampling is an
equiprobability method. In this
approach, progression through the list
is treated circularly, with a return to the
top once the end of the list is passed.

Stratified Sampling is a
method of sampling in which
the researcher divides the
population into separate
groups, called strata. Then, a
probability sampling is drawn
from each group.

Cluster Sampling is a sampling
technique used when mutually
homogeneous yet internally
heterogeneous groupings are evident
in a statistical population. It is often
used in marketing research. In this
sampling technique, the total
population is divided into groups called
clusters a simple random sample of the
group is selected.

Multistage Sampling is the taking
of samples in stages using smaller and
smaller sampling units at each stage. It
can be a complex form of cluster
sampling since it is a type of sampling
which involves dividing the populations
into groups. A combination of
stratified, cluster and simple random
sampling is used in multistage sampling
technique.

Non-probability Sampling
Methods
1. Quota Sampling
2. Convenience Sampling
3. Purposive Sampling
4. Self-Selection Sampling
5. Snowball Sampling
6. Judgemental Sampling

Problem:
A researcher is conducting a study about the effect of
student absenteeism on academic performance of students.
The main respondents of the study are the students from all
grade levels. The number of sub-population per grade level
is as follows:
Grade 7 – 1209
Grade 8 – 1083
Grade 9 – 985
Grade 10 – 889
Grade 11 – 1087
Grade 12 – 780
What appropriate sampling technique can be applied? How
many samples do we have? How many samples from each
grade level?

Stratified Random Sampling using Slovin’s
Equation
Slovin’s Equation
n =
𝑵
𝟏 + 𝑵𝒆𝟐
where:
n = desired sample
N = population
e = margin of error = 5% = 0.05

Solution:
Grade 7 – 1209
Grade 8 – 1083
Grade 9 – 985
Grade 10 – 889
Grade 11 – 1087
Grade 12 – 780
6033
n =
𝑵
𝟏 + 𝑵𝒆𝟐
=
𝟔𝟎𝟑𝟑
𝟏 + 𝟔𝟎𝟑𝟑(𝟎.𝟎𝟓)𝟐
=
𝟔𝟎𝟑𝟑
𝟏 + 𝟔𝟎𝟑𝟑(𝟎.𝟎𝟎𝟐𝟓)
=
𝟔𝟎𝟑𝟑
𝟏 + 𝟏𝟓.𝟎𝟖𝟐𝟓
=
𝟔𝟎𝟑𝟑
𝟏𝟔.𝟎𝟖𝟐𝟓
n = 375
Proportional Percentage:
𝟑𝟕𝟓
𝟔𝟎𝟎𝟑
= 0.0622

Grade 7 – 1209 x 0.0622 = 75
Grade 8 – 1083 x 0.0622 = 67
Grade 9 – 985 x 0.0622 = 61
Grade 10 – 889 x 0.0622 = 55
Grade 11 – 1087 x 0.0622 = 68
Grade 12 – 780 x 0.0622 = 49
375
Then, apply the simple random
sampling technique in choosing the
individual respondent per group.

Quiz (1 whole):
A researcher is conducting a study about the full
implementation of Senior High School (SHS) curriculum in
Sultan Kudarat. The following are the sub-population of the
study:
Students – 3050
Teachers – 550
Parents – 320
Principals – 150
Compute for the total number of sample as well
as the sample per group.

Solution:
Students – 3050
Teachers – 550
Parents – 320
Principals – 150
4070
n =
𝑵
𝟏 + 𝑵𝒆𝟐
=
𝟒𝟎𝟕𝟎
𝟏 + 𝟒𝟎𝟕𝟎(𝟎.𝟎𝟓)𝟐
=
𝟒𝟎𝟕𝟎
𝟏 + 𝟒𝟎𝟕𝟎(𝟎.𝟎𝟎𝟐𝟓)
=
𝟒𝟎𝟕𝟎
𝟏 + 𝟏𝟎.𝟏𝟕𝟓
=
𝟒𝟎𝟕𝟎
𝟏𝟏.𝟏𝟕𝟓
n = 364
𝟑𝟔𝟒
𝟒𝟎𝟕𝟎
= 0.0894

Students – 3050 x 0.0894 = 273
Teachers – 550 x 0.0894 = 49
Grade 9 – 320 x 0.0894 = 29
Grade 10 – 150 x 0.0894 = 13
364
Then, apply the simple random
sampling technique in choosing the
individual respondent per group.

Population
USM - Kabacan – 1580
MSU - Maguindanao – 1398
CCSPC – 1409
SKSU – 1216

Solution:
USM - Kabacan – 1580
MSU - Maguindanao – 1398
CCSPC – 1409
SKSU – 1216
5603
n =
𝑵
𝟏 + 𝑵𝒆𝟐
=
𝟓𝟔𝟎𝟑
𝟏 + 𝟓𝟔𝟎𝟑(𝟎.𝟎𝟓)𝟐
=
𝟓𝟔𝟎𝟑
𝟏 + 𝟓𝟔𝟎𝟑(𝟎.𝟎𝟎𝟐𝟓)
=
𝟓𝟔𝟎𝟑
𝟏 + 𝟏𝟒.𝟎𝟎𝟕𝟓
=
𝟓𝟔𝟎𝟑
𝟏𝟓.𝟎𝟎𝟕𝟓
n = 373
𝟑𝟕𝟑
𝟓𝟔𝟎𝟑
= 0.0666

USM - Kabacan – 1580x0.0666 = 105
MSU - Maguindanao – 1398x0.0666 = 93
CCSPC – 1409x0.0666 = 94
SKSU – 1216x0.0666 = 81
373
Then, apply the simple random sampling
technique in choosing the individual respondent per
group.

II. A researcher is conducting a study about the
implementation of Solid Waste Management in the City
Divisions of Region XII. The following are the sub-population
of the study:
General Santos City – 4050
Koronadal City – 2890
Cotabato City – 3060
Tacurong City – 2079
Kidapawan City – 1980
Compute for the total number of sample as well
as the sample per group.

Assignment (1 whole)
Direction: Use the idea of a Normal Curve and the Central
Limit Theorem to solve the following problems. Illustrate the
shaded region of a normal curve representing your answer.
1. The IQ scores of children in a special education class are
normally distributed with a mean of 95 and a standard
deviation of 10.
a. What is the probability that one of the children has an
IQ score below 100?
b. What is the probability that a child has an IQ score
above 120?
c. What are the chances that a child has an IQ score of
140?
d. How many children have IQ scores above 100 if there
are 30 of them in class?

Select your answers from the following:
1. Mean
2. Median
3. Mode
4. Range
6. Variance
7. Coefficient of Variation
8. Kurtosis
9. Skewness
10. Scatteredness
11. Frequency
12. Percentage
13. t-distribution
curve
14.Normal Curve
15. Statistics
16. Zero
17. Bell-Shaped
18. Research
19. Statistics and
Probability
20. Simple Random
Sampling

Two Types of Statistics
1. Descriptive Statistics is concerned with the gathering,
classification and presentation of data and the collection
of summarizing values to describe group characteristics
of data. The most commonly used summarizing values to
describe group characteristics of data are percentage,
measures of central tendency (mean, mode, median);
measures of variability (range, standard deviation,
variance, coefficient of variation); of skewness and
kurtosis. Examples of descriptive statistics are the class
average of examination, range of student scores, average
salary, means of managerial satisfaction and average
return of investment.

2. Inferential Statistics pertains to the
methods dealing with making inference,
estimates or prediction about a large set of
data using the information gathered.
Commonly used inferential statistical tools or
techniques are testing hypothesis using the z-
test, t-test, analysis of variance (ANOVA),
simple linear correlation (Pearson r),
Spearman’s Rho, chi-square (x²) and
regression.

Two Forms of Hypothesis
1. Null Hypothesis (Ho) is the hypothesis to
be tested and it represents what the
investigation doubts to be true.
2. Alternative Hypothesis (Ha) is the
operational statement of the theory that
the experimenter or researcher believes
to be true and wishes to be true.

Two Types of Hypothesis Testing
1. One-tailed (directional) test occurs when the
researcher has the prior expectation about
the sample value he expects to observe.
2. Two-tailed (non-directional) test occurs when
the alternative hypothesis does not specify a
directional difference for the parameter of
interest. This test is applied when the researcher
doesn’t have the prior expectation regarding the
value he expects to see in the sample.

Two Types of Hypothesis Testing
1. One-tailed (directional) test occurs when the
researcher has the prior expectation about
the sample value he expects to observe.

2. Two-tailed (non-directional) test occurs when the
alternative hypothesis does not specify a directional
difference for the parameter of interest. This test is
applied when the researcher doesn’t have the prior
expectation regarding the value he expects to see in
the sample.

What is a Hypothesis?
A hypothesis is basically a statement
about the target population. This is
formulated as a result of years of observation
and researches. New researches may result
from one’s desire to determine whether or
not a researcher’s hypothesis is supported
when a sample data are subjected to rigorous
scientific statistical methods.
A statistical hypothesis is an assertion or
conjecture concerning one or more
populations

Steps in Hypothesis Testing
Step 1. Formulate the null and alternative
hypotheses.
Step 2. Set the level of significance (α).
Step 3. Select the appropriate test statistic
(statistical tool).
Step 4. Establish the critical (rejection) region.
Step 5. Compute the value of the test statistic
from the sample data.
Step 6. State your conclusion.

Testing a Hypothesis About a Single Mean
Using Large Samples (z-test)
z =
𝑥 − 𝜇
ℴ
𝑛

Examples:
1. In a recent survey of nurses in Region
XII, it was found out that the average monthly
net income of nurses is ₱ 8,048.25. Suppose a
researcher wants to test this figure by a
random sample of 158 nurses in Region XII to
determine whether the monthly net income
has changed. Suppose further the average net
monthly income of the 158 nurses is ₱ 9,568.40
and the population standard deviation was
found to be ₱ 1,563.42.

Solution:
I. Ho: x = ₱8,048.25
Ha: x > ₱8,048.25
II. α = 0.05
III. z-test (right-tailed)
IV. The z-critical value = 1.65
V. Computation:
z =
𝑥 − 𝜇
ℴ
𝑛
=
9568.40 − 8048.25
1563.42
158
=
1520.15
1563.42
12.57
=
1520.15
124.38
z = 12.22
VI. Decision Making/Conclusion
Since that z-computed value of 12.22 is
greater than the z-critical value of 1.65, we have
to reject the null hypothesis. Thus, the current
average salary of nurses in Region XII which is
₱9,568.40 is significantly higher than ₱8,048.40.

2. The owner of a factory that sells a
particular bottled fruit juice claims that the
average capacity of their product is 250 mL. To
test the claim, a consumer group gets a sample
of 100 such bottles, calculates the capacity of
each bottle, and then finds the mean capacity
to be 248 mL. The standard deviation is 5 mL. Is
the claim true at 1% significant level?

Solution:
I. Ho: x = 250 mL
Ha: x < 250 mL
II. α = 0.01
III. z-test (left-tailed)
IV. The z-critical value = -2.33
V. Computation:
z =
𝑥 − 𝜇
ℴ
𝑛
=
248 − 250
5
100
=
−2
5
10
=
−2
0.5
z = -4
Since that z-computed value of -4 is less than
the z-critical value of -2.33, we have to reject the null
hypothesis. Thus, the 248 mL is significantly lower than
250 mL. The claim is not true.

3. A researcher claims that there is a
significant difference on the Mathematics
performance of male and female students.
A population of male students in Grade 10
has a mean of 38.25 and a standard
deviation of 10.5. To prove his claim, a
sample of 81 female students in the same
grade level is found to have a mean of
36.80. Is the claim of a researcher true?
Use the 5% level of significance.

Solution:
I. Ho: x = 38.25
Ha: x ≠ 38.25
II. α = 0.05
III. z-test (two-tailed)
V. Computation:
z =
𝑥 − 𝜇
ℴ
𝑛
=
36.80− 38.25
10.5
81
=
−1.45
10.5
9
=
−1.45
1.17
z = -1.24
Since that z-computed value of -1.24 is
greater than the z-critical value of -1.65, we have
to accept the null hypothesis. The claim of a
researcher is not true. Thus, there is no significant
difference on the Mathematics performance of
male and female students.

Confidence Coefficients of z-Distribution
(z-test)
TypesofTest/SignificantLevel 0.01 0.05 0.10
One-Tailed/One-SidedTest 2.33 1.65 1.29
Two-Tailed/Two-SidedTest 2.58 1.96 1.65

Direction: Fill-in the boxes with the correct answers regarding hypothesis
testing. Second row serves as your example.
zcomp value Inequality
Symbol
Zcritical value Decision Interpretation
-5.256 < -2.33 Reject Ho. There is a significant
difference between the
sample mean and
population mean.
1 15.783 1.65
2 -1.678 -1.96
3 -2.05 -2.33
4 0.247 1.65
5 -4.097 -2.33
6 7.89 1.96
7 -5.079 -1.65
8 -2.32 -2.33
9 1.98 1.96
10 40.235 1.96

Direction: Fill-in the boxes with the correct answer regarding hypothesis testing.
Second row serves as your example.
zcomp value Inequality
Symbol
Zcritical value Decision Interpretation
-5.256 < -2.33 Reject Ho. There is a significant
sample mean and
population mean.
1 15.783 > 1.65 Reject Ho. There is a significant
sample mean and
population mean.
2 -1.678 > -1.96 Accept Ho. There is no significant
sample mean and
population mean.

3 -2.05 > -2.33 AcceptHo. There isnosignificant
differencebetweenthe
sample meanand
population mean.
4 0.247 < 1.65 AcceptHo. There isnosignificant
sample meanand
population mean.
5 -4.097 < -2.33 RejectHo. There isasignificant
sample meanand
population mean.

6 7.89 > 1.96 RejectHo. There isasignificant
sample meanand
populationmean.
7 -5.079 < -1.65 RejectHo. There isasignificant
sample meanand
populationmean.
8 -2.32 > -2.33 AcceptHo. There isnosignificant
sample meanand
populationmean.

9 1.98 > 1.96 RejectHo. Significant
10 40.235 > 1.96 RejectHo. Thereisasignificant
samplemeanand
populationmean.

Another Problem on Hypothesis Testing
A researcher wants to prove that the average
monthly salary of the private school teachers is
significantly different from the average monthly
salary of the public school teachers. The average
salary of the public school teacher is Pph24,500 and
a population standard deviation of Php4,480.15. A
sample of 150 private school teachers was
considered and found to have an average monthly
salary of Php15,000.
Is the claim of a researcher true? Use
hypothesis testing to justify your answer.

Solution:
I. Ho: x = 24,500.00
Ha: x < 24,500
II. α = 0.05
III. z-test (Left-tailed)
V. Computation:
z =
𝑥 − 𝜇
ℴ
𝑛
=
15,000−24,500
4,480.15
150
=
−9,500
4,480.15
12.2474
=
−9,500
365.8042
z = -25.9702
less than the z-critical value of -1.65, we have to
reject the null hypothesis. The claim of a researcher
is true. Thus, the monthly average salary of private
school teachers is significantly lower than the
monthly salary of private school teachers.

1. Given: µ = 594.41
ℴ = 87.16
samples: 578, 605, 599, 790, 554, 615, 568, 498, 598, 625, 618, 608, 589, 580, 589
Question: Is the sample mean significantly different from the population mean?
Solution:
I. Ho: x = 594.41
Ha: x ≠ 594.41
II. α = 0.05
III. z-test (Two-tailed)
V. Computation:
x =
578+599+605+589+790+554+615+568+498+598+625+618+608+589
15
=
9014
15
= 600.9333
z =
600.9333 − 594.41
87.16
15
=
6.5233
87.16
3.8730
=
6.5233
22.5046
= 0.2899

IV. . Since that the z-comp = 0.2899 is less than z-
critical = 1.96, we must reject the null hypothesis.
Thus, the sample mean is not significantly different
from the population mean.

2. A teacher claims that the learning performance
of male and females students in Mathematics is
comparable. In a recently concluded standardized
test in Mathematics , male students were found to
have a population mean of 48.25 and a standard
deviation of 5.25. To prove his claim, a teacher
randomly chose his samples of female students
and their scores were as follows: 35, 35, 44, 49,
50, 53, 54, 45, 35, 38, 29, 30, 38, 40, 30, 35, 36,
28, 36, 30.
Is the claim of a teacher true using 1% level
of significance?

Solution:
I. Ho: x = 48.25
Ha: x ≠ 48.25
II. α = 0.01
V. Computation:
z =
𝑥 − 𝜇
ℴ
𝑛
=
38.50−48.25
5.25
20
=
−9.75
5.25
4.4721
=
−9,75
1.1739
z = -8.30565
less than the z-critical value of -2.58, we have to
reject the null hypothesis. The claim of a researcher
is not true. Thus, the learning performance of male
students is significantly higher than female students
in Mathematics.

3. In a recently concluded English proficiency
examination, a population of male students
was found to have a mean of 70.08 and a
standard deviation of 12.86. A sample of
female students registered the following raw
scores: 90, 75, 68, 80, 68, 70, 68, 68, 78, 85,
83, 65, 71, 82, 58, 68, 76, 80, 85, 78, 78, 80,
85. Using the 5% level of significance, are
female students more proficient in English
compared with male students?

Solution:
I. Ho: x = 70.08
Ha: x ≠ 70.08
II. α = 0.05
V. Computation:
z =
𝑥 − 𝜇
ℴ
𝑛
=
75.6087−70.08
12.86
23
=
5.5287
12.86
4.7958
=
5.5287
2.6815
z = 2.0618
Since that z-computed value of
2.0618 is greater than the z-critical value
of 1.96, we have to reject the null
hypothesis. Female students are more
proficient in English compared with male
students.

THE DIFFERENCE BETWEEN THE z-Distribution
CURVE (NORMAL CURVE) AND t-Distribution
Curve
The confidence coefficients of the
z-distribution are constant with the
given confidence level regardless of
the number of sample while the
confidence coefficients of the t-
distribution change depending upon to
the degrees of freedom.

Testing a Hypothesis About a Single Mean
Using Small Samples (t-test)
t =
𝑥 − 𝜇
𝑠
𝑛

1. A certain brand of laundry soap is
advertised to have a net weight of 500 grams. If
the net weights of a random sample of 10
boxes are 495, 503, 507, 498, 490, 505, 510,
502, 493, and 506 grams, can it be concluded
that the average net weight of the boxes is less
than the advertised amount? Use 3% level of
significance.

Solution:
I. Ho: x = 500 grams
Ha: x ≠ 500 grams
II. α = 0.01
III. t-test (two-tailed)
IV. tcritical (df = 9) = 3.250
V. Computation:
x =
495+503+507+498+490+505+510+502+493+506
10
=
5009
10
x = 500.9
t =
𝑥 − 𝜇
𝑠
𝑛
=
− 250
5
100
=
−2
5
10
t =
500.9 − 500
6.61
10
=
0.9
2.09
z = 0.4306
Since that t-computed value of 0.4306
is less than the t-critical value of 3.250, we have to
accept the null hypothesis. Thus, the net weights
of a sample of 10 boxes of soap are statistically
equal to the advertised brand of soap.

Testing a Hypothesis About Two
Sample Means (t-test)
; Where:
x₁ = first sample mean
x₂ = second sample mean
s₁ = standard deviation of a first sample
s₂ = standard deviation of a second sample
n₁ = number of the first sample
n₂ = number of the second sample
t =
𝒙₁−𝒙₂
𝒔₁²
𝒏₁
+
𝒔₂²
𝒏₂

Problems:
1. The pre-test results of the two
sections in Mathematics are as follows:
Section A: 25, 20, 24, 25, 26, 28, 20, 18
Section B: 23, 21, 23, 26, 25, 27, 19, 17, 19
Using 5% level of significance, is there
a significant difference in the pre-test
scores of Section A and Section B?

I. Hₒ: x₁ = x₂
Hₐ: x₁ ≠ x₂
II. α = 0.05
III. t-test (two-tailed)
III. df = 8 + 9 – 2 = 15
tcritical = 2.1315
V. Computation:
x₁ =
25+20+24+25+26+28+20+18
8
= 23.25
x₂ =
23+21 23+26+25+27+19+17+19
9
= 22.22
t =
𝒙₁−𝒙₂
𝒔₁²
𝒏₁
+
𝒔₂²
𝒏₂
=
23.25−22.2222
12.2142
8
+
11.9444
9
=
1.0278
1.5268+1.3272
=
1.0278
2.854
=
1.0278
1.6894
tcomp = 0.6084

VI. Decision
Since that the t-computed value = 0.6084 is less
than the t-critical value = 2.1315, we have to accept
the null hypothesis. Therefore, there is no significant
difference on the pre-test scores of Section A and
Section B.

THE NATURE OF STATISTICS
Statistics refers to the methods in collection,
presentation, analysis and interpretation of data.
Data Gathering or Collection may be done through
interview, questionnaires, tests, observation, registration
and experiments.
Presentation of Data refers to the organization of
data into tables, graphs, charts or paragraphs. Hence,
presentation of data may be tabular, graphical or textual.
Analysis of Data pertains to the process of
extracting from the given data relevant and noteworthy
information and this uses statistical tools or techniques.
Interpretation of Data refers to the drawing of
conclusions or inferences from the analyzed data.

IDENTIFYING THE
STATISTICAL TOOL
APPLICABLE FOR THE
GIVEN STATEMENT OF
THE PROBLEM

1. SOP: What is the profile of STEM teachers in terms of teaching
experience and educational attainment?
2. SOP: To what extent is the problem solving skills of grade 7
students?
3. SOP: Is there a significant gender difference on the performance
of students in their Geometry subjects?
4. SOP: What is the impact of the reading interest on students’
literary comprehension?

5. SOP: What is the effect of teachers’ educational qualifications
on the learning performance of students in Mathematics?
6. SOP: Is there a significant difference in the learning
performance of the students exposed in the three different
methods of teaching: Traditional, Game-Based, and Activity-
Oriented?
7. SOP: Is there a significant difference between the responses of
the women and men in the legalization of the divorce in the
Philippines?
8. SOP: Are the public school teachers more competent compared
to the private school teachers?

9. SOP: What is the profile of the NQuESH takers in terms of
administrative experience and educational attainment?
10. SOP: What is the level of the reading comprehension of grade
7 students?
11. SOP: Is there a significant difference between the performance
of the students in the two previous grading periods?
12. SOP: Is there a significant relationship between the reading
interest and literary comprehension of the students?

13. SOP: Is the learning performance of the students in
Mathematics significantly influenced by the educational
qualification of their teachers?
14. SOP: Is there a significant difference in the learning
performance of the students exposed in the three different
methods of teaching: Traditional, CAI, and PWA?
15. SOP: Is there a significant relationship between the responses
of the women and men in the legalization of the divorce in the
Philippines?
16. SOP: Are the public school teachers more satisfied with their
jobs compared to the private school teachers?

1. To what degree is the student
absenteeism in the following causes:
1.1 Physical/school factors,
1.2 Health problems,
1.3 Personal attitudes,
1.4 Family-related issues,
1.5 Teacher-related reasons,
1.6 Subject-related matters,
1.7 Classroom atmosphere,
1.8 Peer relationship,
1.9 Financial constraints, and
1.10 Obsession in the computer or
online games/social networking sites?
17.

2. What is the level of academic performance of low
performing students in the following tool subjects:
2.1 Filipino,
2.2 English,
2.3 Mathematics, and
2.4 Science?
3. Is there a significant difference in the
attitudes of students towards absenteeism
when they are grouped according to:
3.1 Grade 7,
3.2 Grade 8,
3.3 Grade 9, and
3.4 Grade 10?
18.
19.

4. Is the assessment of the respondents towards
absenteeism significantly different according to
the following types of respondents:
4.1 Low performing students,
4.2 Their respective parents or guardians, and
4.3 Their close friends?
5. Does the academic
performance of struggling
students in the tool subjects
significantly differ from each
other?
20.
21.

6. Is there a significant
difference in the attitudes
of male and female
students towards
absenteeism?
7. Is there a significant
relationship between the
causes of absenteeism and
academic performance of the
struggling students?
22.
23.

8. What intervention
programs can be
proposed to
minimize, if not
totally eradicate
absenteeism among
the low performing
students?
24.

9. Is there a
significant difference
between the
academic
performance of TVL
and HUMSS
students?
25.

Thank you so much
From
SAMSUDIN N. ABDULLAH, Ph.D.
Master Teacher II

STATISTICS_AND_PROBABILITY_for_Senior_Hi.pptx

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