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.
2.
3.
4.
Range
Standard Deviation
Variance
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Example 1. Calculate the arithmetic mean of the
given distribution on final scores of 100 Grade
11 students in Trigonometry.
Scores
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
55 – 59
50 – 54
f
3
6
19
24
18
12
8
5
3
2
n = 100
STATISTICS AND PROBABILITY
x
97
92
87
82
77
72
67
62
57
52
fx
291
552
1,653
1,968
1,386
864
536
310
171
104
Ʃfx = 7,835
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
x =
=
Ʃ𝒇𝒙
𝒏
7835
100
= 78.35
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Example 2. What is mean of the given distribution
of scores of 75 students in Statistics.
Scores
84 – 90
77 – 83
70 – 76
63 – 69
56 – 62
49 – 55
42 – 48
35 – 64
28 – 34
21 – 27
f
5
12
8
10
8
2
18
5
3
4
n = 75
STATISTICS AND PROBABILITY
x
87
80
73
66
59
52
45
38
31
24
fx
435
960
584
660
472
104
810
190
93
96
Ʃfx = 4,404
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
x =
=
Ʃ𝒇𝒙
𝒏
4404
75
= 58.72
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution Using Ungrouped Data
x=
Ʃ𝒙
𝒏
=
𝟔𝟎 + 𝟔𝟐 + 𝟓𝟒 + 𝟒𝟎 + ... + 𝟓𝟒
𝟏𝟐𝟒
STATISTICS AND PROBABILITY
=
𝟓𝟎𝟏𝟕
𝟏𝟐𝟒
= 𝟒𝟎. 𝟒𝟔
SAMSUDIN N. ABDULLAH, Ph.D.
Class Interval
62 – 65
58 – 61
54 – 57
50 – 53
46 – 49
42 – 45
38 – 41
34 – 37
30 – 33
26 – 29
22 – 25
f
3
9
13
9
2
24
9
10
17
13
15
n = 124
x
63.5
59.5
55.5
51.5
47.5
43.5
39.5
35.5
31.5
27.5
23.5
fx
190.5
535.5
721.5
463.5
95
1,044
355.5
355
535.5
357.5
352.5
∑fx = 5,006
Solution:
Ʃ𝒇𝒙
𝟓,𝟎𝟎𝟔
=
=
𝒏
𝟏𝟐𝟒
STATISTICS AND PROBABILITY
x =
𝟒𝟎. 𝟑𝟕
SAMSUDIN N. ABDULLAH, Ph.D.
61 – 63
58 – 60
55 – 57
52 – 54
49 – 51
46 – 48
43 – 45
40 – 42
37 – 39
34 – 36
31 – 33
28 – 30
25 – 27
22 – 24
5
7
11
6
5
2
21
9
6
7
12
10
13
10
n = 124
62
59
56
53
50
47
44
41
38
35
32
29
26
23
310
413
616
318
250
94
924
369
228
245
384
290
338
230
∑fx = 5,009
Solution:
x =
Ʃ𝒇𝒙
𝟓,𝟎𝟎𝟗
= 𝟏𝟐𝟒
𝒏
= 𝟒𝟎. 𝟒𝟎
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
To determine the median of a grouped
data, use the formula:
Md = L +
𝒏
−𝑭
𝟐
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Example 1. Find the median score of
students of Mr. Dela Cruz Math class.
Scores
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
70 – 74
65 – 69
60 – 64
f
5
11
17
25
20
12
7
3
F
100
95
84
67
42
22
10
3
n = 100
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
Md = L +
𝒏
−𝑭
𝟐
𝒇
Md =79.5 +
𝟏𝟎𝟎
− 𝟒𝟐
𝟐
𝟐𝟓
𝒊
L = 79.5
n = 100
F = 42
f = 25
i = 99 – 95 + 1 = 5
= 79.5 +
= 79.5 +
= 79.5 +
𝟓𝟎 − 𝟒𝟐
𝟐𝟓
𝟖
𝟐𝟓
(𝟓)
(𝟓)
(𝟓)
𝟒𝟎
𝟐𝟓
= 79.5 + 1.6
= 81.1
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Example 2. The ages of 115 ENHS teachers
are given below. Find the median age.
Ages
63 – 69
56 – 62
49 – 55
42 – 48
35 – 41
28 – 34
21 – 27
14 – 20
7 – 13
STATISTICS AND PROBABILITY
f
3
11
18
26
21
15
12
7
2
n = 115
F
115
112
101
83
57
36
21
9
2
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
Md = L +
𝒏
−𝑭
𝟐
𝒇
Md = 41.5 +
𝟏𝟏𝟓
− 𝟓𝟕
𝟐
𝟐𝟔
𝒊
L = 41.5
n = 115
F = 57
f = 26
i = 69 – 63 + 1 = 7
= 41.5 +
= 41.5 +
= 41.5 +
𝟓𝟕.𝟓 − 𝟓𝟕
𝟐𝟔
𝟎.𝟓
𝟐𝟔
(𝟕)
(𝟕)
(𝟕)
𝟑.𝟓
𝟐𝟔
= 41.5 + 0.135
= 41.635
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Class Interval
62 – 65
58 – 61
54 – 57
50 – 53
46 – 49
42 – 45
38 – 41
34 – 37
30 – 33
26 – 29
22 – 25
STATISTICS AND PROBABILITY
f
3
9
13
9
2
24
9
10
17
13
15
n = 124
F
124
121
112
99
90
88
64
55
45
28
15
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
Md = L +
𝒏
−𝑭
𝟐
𝒇
Md = 37.5 +
𝒊
L = 38.5
n = 124
F = 55
f=9
i = 41 – 38 + 1 = 4
STATISTICS AND PROBABILITY
= 37.5 +
= 37.5 +
= 37.5 +
𝟏𝟐𝟒
− 𝟓𝟓
𝟐
𝟗
𝟔𝟐 − 𝟓𝟓
𝟗
𝟕
𝟗
(𝟒)
(𝟒)
(𝟒)
𝟐𝟖
𝟗
= 37.5 + 3.11
= 40.61
SAMSUDIN N. ABDULLAH, Ph.D.
Example 3. Complete the table and compute for the
median score of the Grade 11 students who took the
Precalculus subject.
Scores
135 – 139
130 – 134
125 – 129
120 – 124
115 – 119
110 – 114
105 – 109
100 – 104
95 – 99
90 – 94
85 – 89
80 – 84
75 – 79
f
2
2
4
5
9
8
7
5
3
1
2
1
1
CF
50
48
46
42
37
28
20
13
8
5
4
2
1
CP
100
96
92
84
74
56
40
26
16
10
8
4
2
Note: CF – Cumulative Frequency & CP – Cumulative Percent
Solution:
Md = L +
𝒏
−𝑭
𝟐
𝒇
𝒊
L = 109.5
n = 50
F = 20
f=8
i = 139 – 135 + 1 = 5
Md = 109.5 +
= 109.5 +
= 109.5 +
= 109.5 +
𝟓𝟎
− 𝟐𝟎
𝟐
𝟖
𝟐𝟓 − 𝟐𝟎
𝟖
𝟓
𝟖
(𝟓)
(𝟓)
(𝟓)
𝟐𝟓
𝟖
= 109.5 + 3.125
= 112.625
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
(
)i
𝒅₁ + 𝒅₂
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
STATISTICS AND PROBABILITY
Solution:
Lmo = 34.5
d1 = 16 – 12 = 4
d2 = 16 – 15 = 1
i = 39 – 35 + 1 = 5
SAMSUDIN N. ABDULLAH, Ph.D.
𝟒
Mo = 34.5 + (
)(5)
𝟒+𝟏
𝟐𝟎
= 34.5 +
𝟓
= 34.5 + 4
= 38.5
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Example 2. Compute for the modal wage of the
workers in a certain private school
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
Lmo = 1,319.5
d1 = 31 – 24 = 7
d2 = 31 – 12 = 19
i = 1,339 – 1,320 + 1 = 20
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
𝟕
Mo = 1,319.50 + (
)(20)
𝟕 + 𝟏𝟗
𝟏𝟒𝟎
= 1,319.50 +
𝟐𝟔
= 1,319.50 + 5.385
= 1,324.885
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Class Interval
61 – 63
58 – 60
55 – 57
52 – 54
49 – 51
46 – 48
43 – 45
40 – 42
37 – 39
34 – 36
31 – 33
28 – 30
25 – 27
22 – 24
STATISTICS AND PROBABILITY
f
5
7
11
6
5
2
21
9
6
7
12
10
13
10
n = 124
Solution:
Mo = 42.5 + (
𝟏𝟐
)(4)
𝟏𝟐 + 𝟏𝟗
= 42.5 +
𝟒𝟖
𝟑𝟏
= 42.5 + 1.55
= 44.05
SAMSUDIN N. ABDULLAH, Ph.D.
MEASURES OF VARIABILITY describe the spread of
the values about the mean.
1. Range
2. Standard Deviation
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
Answer: 13
b) 25, 32, 9 18, 12, 30, 28, 22 Answer: 23
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Example 2. Determine the range of data presented in
a frequency distribution below.
a)
b)
Class Intervals
20 – 25
14 – 23
8 – 13
2–9
f
13
5
8
10
Range = 25.5 – 1.5 = 24
Class Intervals
90 – 99
80 – 89
70 – 79
60 – 69
50 – 59
f
3
7
8
5
2
Range = 99.5 – 49.5 = 50
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
Scores
18
20
22
15
16
12
17
21
10
19
𝟏𝟎
d
1
3
5
-2
-1
-5
0
4
-7
2
d2
1
9
25
4
1
25
0
16
49
4
2
Ʃd = 134
STATISTICS AND PROBABILITY
s=
=
=
𝟏𝟕𝟎
𝟏𝟎
= 17
Ʃ𝒅²
𝒏
𝟏𝟑𝟒
𝟏𝟎
= 𝟏𝟑. 𝟒
s = 3.661
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Example 1. Calculate the standard deviation of
the data presented below.
Class
Intervals
f
252 – 260
243 – 251
234 – 242
225 – 233
216 – 224
207 – 215
198 – 206
189 – 197
180 – 188
171 – 179
162 – 170
3
5
9
12
5
4
2
10
8
2
5
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
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
Assignment. Find the mean, median, mode and standard deviation
of the given data.
Class Intervals
f
355 – 365
344 – 354
333 – 343
322 – 332
311 – 321
300 – 310
289 – 299
278 – 288
267 – 277
256 – 266
245 – 255
234 – 244
223 – 233
212 – 222
201 – 211
13
5
11
12
15
4
20
8
9
11
5
3
2
9
3
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? ____________
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Class
Interval
Freque Class
ncy
Mark
(f)
(x)
fx
Cumula deviati
tive
on
Frequen from
cy
the
(F)
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
STATISTICS AND PROBABILITY
Ʃfd2 =12179
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Class
Interval
Freque
ncy
(f)
Class
Mark
(x)
fx
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
Cumulat deviation
ive
from the
mean
Frequen
cy
(d)
(F)
d2
fd2
2
Course Outline in Grade 11 Statistics
and Probability
CHAPTER I. Random Variables
Probability Distributions
-
and
Random Variables
Probability of an Event
Probability Distribution
Mean of a Discrete Probability
Variance of a Discrete Probability
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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)
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
CHAPTER I
RANDOM VARIABLES
AND PROBABILITY
DISTRIBUTIONS
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
A. Classify the following random variables as discrete or
continuous.
1. The number of defective computers produced by a
manufacturer Discrete
2. The weight of newborns each year in a hospital Continuous
3. The number of siblings in a family Discrete
4. The amount of paint utilized in a building project Continuous
5. The number of dropouts in a school Discrete
6. The speed of a car Continuous
7. The number of female athletes Discrete
8. The time needed to finish the test Continuous
9. The amount of sugar in a cup of coffee Continuous
10. The number of people who are playing lotto each day Discrete
11. The number of accidents per year in an accident prone area
Discrete
12. The amount of salt and ice to preserve ice cream
Continuous
13. The
number of all public school students in the world
Discrete
14. The intensity of several earthquakes striking
Mindanao Continuous
15. The number of private school teachers in the
Philippines Discrete
16. The body temperature of a patient Continuous
17. The size of a Flat TV screen Continuous
18. The heights of students Continuous
19. The number of households in a subdivision Discrete
20. The vital statistics a female candidate Continuous
21. The number of used clothes for the refugees Discrete
22. The number of eggs in one tray Discrete
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
23. The length of the top of a table Continuous
24. The amount of sugar needed to bake Continuous
25. The number of students in the TVL track Discrete
26. The width of a blackboard Continuous
27. The sticks of chalk in a box Discrete
28. The coins in my pocket Discrete
29. The Korean teachers here at ENHS Discrete
30. The kilogram of fruits in a table Continuous
31. The storm signals of typhoons Continuous
32. The distance between school and market Continuous
33. The angle of elevation Continuous
34. The height of flagpole Continuous
35. The thickness of a book Continuous
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
Possible
Outcomes
of the Random Variable X
Value Value
of the
Random Variable X
(number of defective cell phones)
(number of defective cell phone)
NNN
0
NND
1
NDN
1
DND
2
DDN
2
DNN
1
NDD
2
DDD
3
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
Possible
Outcomes
Outcomes
Value Value
of the
Random
Variable
of the
Random Variable
Y Y
(numberof
oftails)
tails)
(number
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.
Quiz # 2
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Possible Outcomes
Value of the Random
Variable X
The values of the random variable X are ____________________________.
Value of the Random
Possible Outcomes
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 !
3!
= ₃C₂ = 2! 3−2 !=
3(2!)
2!1!
=3
S = {(Black, Black), (Black, White), (White, Black)}
No Black
1 Black
2 Black
0
2
1
The random variable K has values of 0, 1 and 2.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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) =
𝑛!
C
=
=
n r
𝑟! 𝑛−𝑟 !
6!
6(5)(4!)
₆C₂ =
= 2(1)(4!) = 15
2!4!
S = {W₁W₂, W₁B₁, W₁B₂, W₁B₃, W₁B₄, W₂B₁, W₂B₂, W₂B₃, W₂B₄,
B₁B₂, B₁B₃, B₁B₄, B₂B₃, B₂B₄, B₃B₄}
0 Back
1 Black
2 Black
1
8
6
The random variable M has values of 0, 1 and 2.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
PROBABILITY
AN
EVENT
Lesson
1
Sample OF
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}
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
4. Experiment of Rolling Two Dice (One is red, the other is green.)
The sample space of this experiment is illustrated below.
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)}
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
HHTHT
THTHH
HTHHT
THHTH
THHHT
HTTHH
HHTTH
HHHTT
HTHTH
3 TAILS
HHTTT
TTHTH
HTHTT
THTTH
HTTHT
HTTTH
THHTT
TTHHT
TTTHH
THTHT
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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}
2³ = 8
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
6¹(2¹) = 6(2) = 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
2(8) = 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
₅C₂ = 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
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
10!
n(S) = ₁₀C₃ =
3!7!
=
10(9)(8)(7!)
(3)(2)(1)(7!)
=
720
6
= 120
13. Drawing Four Marbles from an Urn Containing 15 Marbles
15!
15(14)(13)(12)(11!) 32,760
n(S) = ₁₅C₄ = 4!11! = (4)(3)(2)(1)(11!) = 24 = 1,365
14. Drawing Two Apples from a Basket Containing 8 Apples
𝟖!
(𝟖)(𝟕)(𝟔!) 𝟓𝟔
n(S) = ₈C₂ =𝟐!𝟔! = (𝟐)(𝟏)(𝟔!) = 𝟐 = 28
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Review Problems on Probability of an Event
Examples:
1. What is the probability of getting an even number in the experiment of rolling a die?
Solution:
S = {1, 2, 3, 4, 5, 6}
A = {2, 4, 6}
P(A) =
n(S) = 6
n(A) = 3
𝒏(𝑨)
𝒏(𝑺)
=
3
6
1
=2
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
2. What is the probabil ty that the sum of the faces of the two dice is 8?
Solution:
F = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
n(F) = 5
n(S) = 36
P(F) =
5
36
5
= 36
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Find the probability of the following events.
Event (E)
P(E)
1
Getting an even number in a single roll of a die
2
Getting a sum of 6 when two dice are rolled
3
Getting an ace when a card is drawn from a deck
4
The probability that all children are boys if a couple has three children
5
Getting an odd number and a tail when a die is rolled and a coin is tossed
simultaneously
6
Getting a sum of 11 when two dice are rolled
7
Getting a black card and 10 when a card is drawn from a deck
8
Getting a red queen when a card is drawn from a deck
9
Getting doubles when two dice are rolled
10
Getting a red ball from a box containing 3 red and 6 black balls
STATISTICS AND PROBABILITY
1
2
5
36
1
13
1
4
1
4
1
18
1
26
1
26
1
6
1
3
SAMSUDIN N. ABDULLAH, Ph.D.
Find the probability of the following events.
Event (E)
1
Getting an even number in a single roll of a die
2
Getting a sum of 6 when two dice are rolled
P(E)
E = {2, 4, 6}
E = {(1, 5), (2, 4), (5, 1), (4, 2), (3, 3)}
3
Getting an ace when a card is drawn from a deck
E = {A of Spade, A of Club, A of Heart, A of Diamond}}
4
The probability that all children are boys if a couple has three children
S = {GGG, GBG, BBG, BBB}
5
Getting an odd number and a tail when a die is rolled and a coin is tossed
simultaneously
E = {1T, 3T, 5T}
6
Getting a sum of 11 when two dice are rolled
7
Getting a black card and 10 when a card is drawn from a deck
E = {(5, 6), (6, 5)}
E = {10 of Spade, 10 of Club}
8
Getting a red queen when a card is drawn from a deck
E = {Q of Diamond, Q of Heart}
9
Getting doubles when two dice are rolled
E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
10
Getting a red ball from a box containing 3 red and 6 black balls
STATISTICS AND PROBABILITY
1
2
5
36
1
13
1
4
1
4
1
18
1
26
1
26
1
6
1
3
SAMSUDIN N. ABDULLAH, Ph.D.
Review Problems on Probability
A. From a standard deck of 52 cards, what is the
probability of
1. picking a black card? 1/2
2. picking a face card? 3/13
3. not picking a face card? 10/13
4. picking a black and face card? 3/26
5. not picking a black and face card? 23/26
6. picking a red and nonface cards? 5/13
7. picking an ace card? 1/13
8. not picking an ace card? 12/13
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Quiz (1/4 sheet of paper)
A. Find the cardinality of each sample space.
1. Tossing six coins 64
2. Tossing a pair of coins and spinning a
spinner with 10 numbers simultaneously 40
3. Rolling a pair of dice and drawing a card
from standard deck simultaneously 1,872
4. Tossing three coins and rolling two dice
simultaneously
288
5. Drawing five balls in a box containing 12
balls 792
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
B. On rolling a die, what is the probability of
having
1. a 3? 1/6
2. an even number? 1/2
3. zero? 0
4. a number greater than 4? 1/3
5. a number lying between 0 and 7? 1
6. a number less than 4? 1/2
7. an odd number? 1/2
8. a prime number? 1/2
9. a composite number? 1/3
10. a multiple of 3? 1/3
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
C. From standard deck of cards, what is the
probability of:
1. picking a red card? 1/2
2. picking a face card? 3/13
3. picking a nonface card? 10/13
4. picking a black and 9 card? 1/26
5. not picking a black and 9 card? 25/26
6. picking a club card? 1/13
7. not picking a club card? 12/13
8. picking a red face card? 3/26
9. not picking a red face card? 23/26
10. picking any card? 1
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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? 1/30
2. two are red and one is green? 3/40
3. exactly two are green? 7/40
4. none is white? 1/6
5. they are of different colors? 3/10
6. none is red?7/24
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solutions of C
n(S) = ₁₀C₃
1. n(E) = ₄C₃ =
10!
=
3!7!
4!
3!1!
=
=
(4)(3!)
3!(1)
2. n(E) = (₃C₂)( ₃C₁) =
P(E) =
10(9)(8)(7!)
(3)(2)(1)(7!)
=
720
6
=4
= 120
P(E) =
3!
(3)(2!)
3!
( )(1!2!) =
2!1!
2!(1)
•
4
120
=
(3)(2!)
1!(2!)
1
30
= 3(3) = 9
9
3
=
120 40
3!
7!
3. n(E) = (₃C₂)( ₇C₁) = ( )( )
2!1! 1!6!
21
7
P(E) =
=
120 40
STATISTICS AND PROBABILITY
=
(3)(2!)
2!(1)
•
(7)(6!)
1!(6!)
= 3(7) = 21
SAMSUDIN N. ABDULLAH, Ph.D.
Solutions of C
n(S) = ₁₀C₃
4. n(E) = ₆C₃ =
5.
10!
=
3!7!
6!
3!3!
=
=
10(9)(8)(7!)
(3)(2)(1)(7!)
(6)(5)(4)(3!)
(3!)(3)(2)(1)
=
120
6
=
720
6
= 120
= 20
P(E) =
20
120
=
1
6
4!
3!
3!
n(E) = (₄C₁)( ₃C₁)(₃C₁ ) = (1!3!)(1!2!)(1!2!)
(4)(3!) (3)(2!) (3)(2!)
=
•
•
= 4(3)(3) =36
(1)(3!)
1!(2!)
1!(2!)
36
3
P(E) =
=
120 10
6. n(E) = ( ₇C₃) =
P(E) =
7!
3!4!
35
120
=
=
(7)(6)(5)(4!)
3(2)(1)(4!)
=
210
6
= 35
7
24
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
10
5
5
1. P(1) = , P(2) = , P(3) = , P(4) =
It cannot
19
1
19
+
10
19
19
+
5
19
+
5
19
19
=
19
21
>1
19
2. P(1) = 0.25, P(2) = 0.75, P(3) = 0.25, P(4) = 0.25
It cannot
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
It can
0.15 + 0.27 + 0.29 + 0.29 = 1
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
10
5
5
1. P(1) = , P(2) = , P(3) = , P(4) =
19
1
10
+
19 19
+
5
19
+
19
5
21
=
19 19
19
>1
19
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
1
2
1
1
4. P(1) = , P(2) = , P(3) = , P(4) =
1
5
5
2
1
+
5
5
5
1
+
5
5
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,
They cannot
P(4) = 0.18
0.25 + 0.21 + 0.19 + 0.18 = 0.83
1
3
3
1
7. P(1) = , P(2) = , P(3) = , P(4) =
1
8
3
8
8
3
8
+ + +
8. P(1) =
5
,
17
1
=
8
8
1
P(2) =
21
,
34
1
P(4) =
17
5
21 5
1
+ + + = 38/34
17
34
34
17
8
8
They can
P(3) =
5
,
34
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
1. P(X) =
for x = 1, 2, 3, …, 8
It can
2.
3.
8
1 1 1
+ +
8 8 8
1
P(X) =
6
3+𝑥
P(X) =
3 −𝑥
1
8
1
8
1
8
1
8
1
8
8
8
+ + + + + = =1
for x = 1, 2, 3, …, 9
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Decision-making is an important aspect in
business, education, insurance, and other reallife 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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
n(S) = 2⁴ = 16
Random
Possible Outcomes of Each
Variable Z
Event
0 HEAD TTTT
1 HEAD
HTTT
THTT
2 HEADS
HHTT
THHT
HHHT
THHH
HHH
HTHT TTHH
THTHT
HTTH
HHTH HTHH
3 HEADS
4 HEADS
TTHT
TTTH
P(Z)
1
16
1
4
3
8
1
4
1
16
Probability Distribution
Number of Heads
P(Z)
0
1 2
3 4
1 1 3 1 1
16 4 8 4 16
Probability P(Z)
0.4
0.3
0.2
0.1
0
STATISTICS AND PROBABILITY
1
3
2
Number of Tails (Z)
4
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Probability P(Y)
0.4
0.3
0.2
0.1
0
STATISTICS AND PROBABILITY
1
2
Number of Tails (Y)
3
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
n(S) = 5C3 = 10
S = {N1N2N3, D1N1N2, D1N1N3, D1N2N3, D2N1N2,
D2N1N3, D2N2N3, D1D2N1, D1D2N2, D1D2N3}
Number of Defective
Computer (X)
Probability P(x)
0
1
2
𝟏
𝟏𝟎
𝟑
𝟓
𝟑
𝟏𝟎
Probability P(X)
0.8
0.6
0.4
0.2
1
0
2
Number of Tails (X)
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Determine whether the table presents a probability
distribution. Explain your answer.
1)
X
P(X)
1
𝟏
𝟑
5
𝟏
𝟑
8
𝟏
𝟑
7
𝟏
𝟑
9
𝟏
𝟑
2)
X
P(X)
0
𝟏
𝟔
2
𝟏
𝟔
4
𝟏
𝟑
6
𝟏
𝟔
8
𝟏
𝟔
3)
X
P(X)
1
𝟏
𝟒
2
𝟏
𝟖
3
𝟏
𝟒
5
𝟏
𝟖
4)
X
P(X)
4
𝟏
𝟓
8
𝟏
𝟖
12
𝟏
𝟖
15
𝟏
𝟓
5)
X
P(X)
1
0.35
3
0.25
5
0.22
7
0.12
17
𝟏
𝟖
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
1
2
3
4
5
6
7
8
9
10
0.06
0.14
0.16
0.14
0.12
0.10
0.08
0.07
0.06
0.04
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
3. What is the probability that the demand
is between four and eight? 0.25
4. What is the probability that the demand
is less than nine? 0.93
5. What is the probability that the number
of demand is even number? 0.45
6. What is the probability that the demand
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,
y₁ = 2,
x₂ = 2,
y₂ = 1,
x₃ = 5,
y₃ = 0,
x₄ = 1
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
B. The following are the scores of 40
students in a test. Compute the mean score.
Score
42
50
53
38
46
Number of Students
8
12
9
7
4
Solution:
𝟒𝟐(𝟖) + 𝟓𝟎(𝟏𝟐) + 𝟓𝟑(𝟗) + 𝟑𝟖(𝟕) + 𝟒𝟔(𝟒) 𝟏𝟖𝟔𝟑
x=
=
= 46.575
𝟒𝟎
40
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
1
6
1
6
1
6
1
6
1
6
2
6
3
6
4
6
5
6
6
6
2
3
4
5
6
𝟐𝟏
Mean =
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:
𝟏𝟎 + 𝟖 +𝟕 +
x=
𝟕𝟔
𝟕
=
x = 10.8571
𝟏𝟓 + 𝟐𝟎 + 𝟖 + 𝟖
𝟕
𝑴𝒅 = 8
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:
𝟏𝟓𝟎+𝟖𝟎+𝟗𝟓+𝟏𝟏𝟓+𝟐𝟓𝟎+𝟑𝟎𝟎+𝟏𝟐𝟓+𝟏𝟑𝟎+𝟏𝟓𝟎+𝟏𝟓𝟎
x=
𝟏𝟎
=
𝟏𝟓𝟒𝟓
𝟏𝟎
𝑴𝒅 =
𝟏𝟑𝟎+𝟏𝟓𝟎
𝟏𝟎
=
= 154.5000 𝑴𝒐 = 150
𝟐𝟖𝟎
𝟐
= 140
II. Solve for x , 𝑴𝒅 and 𝑴𝒐 of the following set
of scores.
Scores Frequency
25
10
23
5
20
4
15
11
Solution:
𝟐𝟓(𝟏𝟎)+𝟐𝟑(𝟓)+𝟐𝟎(𝟒)+𝟏𝟓(𝟏𝟏)
x=
𝟑𝟎
=
𝟔𝟏𝟎
𝟑𝟎
= 20.3333
𝑴𝒅 =
𝟐𝟑+𝟐𝟎
𝟐
=
𝑴𝒐 = 15
𝟒𝟑
𝟐
= 21.5000
Formula for the Mean of the Probability
Distribution
µ = Ʃx · P(x)
Examples:
1. The probabilities that a customer will buy 1, 2, 3,
𝟑 𝟏 𝟏 𝟐 𝟑
, , , , .
𝟏𝟎 𝟏𝟎 𝟏𝟎 𝟏𝟎 𝟏𝟎
4, or 5 items in a grocery store are
What
is the average number of items that customer will buy?
Solution:
𝟑
𝟏
𝟏
𝟐
𝟑
µ = 1( ) + 2( ) + 3( ) + 4( ) + 5( )
𝟏𝟎
𝟏𝟎
𝟏𝟎
𝟑
𝟐
𝟑
𝟖
𝟏𝟓
= + + + +
𝟏𝟎
𝟏𝟎
𝟏𝟎
𝟏𝟎
𝟏𝟎
µ = 3.1
𝟏𝟎
𝟏𝟎
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
Probability
-$2.00
0.30
STATISTICS AND PROBABILITY
-$1.00
0.40
$2.00
0.20
$3.00
0.10
SAMSUDIN N. ABDULLAH, Ph.D.
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)
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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)
1
0.10
2
0.18
5
0.22
6
0.19
7
0.15
11
0.16
x.P(x)
Ʃx.P(x) =
STATISTICS AND PROBABILITY
x-µ
(x - µ)²
(x - µ)².P(x)
Ʃ(x - µ)².P(x) =
SAMSUDIN N. ABDULLAH, Ph.D.
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)
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
CHAPTER II
NORMAL
DISTRIBUTION
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Skewed to
the Left
Skewness is less
than zero (negative).
STATISTICS AND PROBABILITY
Skewed to
the Right
Skewness is greater
than zero (positive).
SAMSUDIN N. ABDULLAH, Ph.D.
Types of Kurtosis
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
The skewness of
a normal curve is 0
and its kurtosis is
3.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
C. Determine the area of the region indicated
by the following. Draw a normal curve for
each.
1. -1 < z < 1
2.
3.
4.
5.
6.
7.
8.
9.
10.
-2 < z < 2
-1.5 < z < 2.5
0.18 < z < 3.2
-3 < z < 1.65
-0.1 < z < 1.47
-2.33 < z < 1.64
-2.88 < z < 3
-1.96 < z < 1.96
-2.96 < z < -0.01
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Find the area of the shaded region of the normal curve.
1.
A = 0.3413 or 34.13%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
2.
A = 2(0.4938)
= 0.9876 or 98.76%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
3.
2.
-1.25
A = 0.5 – 0.3944
= 0.1056 or 10.56%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
4.
A = 0.4938 + 0.2734
= 0.7672 or 76.72%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
5.
A = (0.50 – 0.3944) + (0.4772 – 0.1915)
= 0.1056 + 0.2857
= 0.3913 or 39.13%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
A = 0.5 – 0.3944
= 0.1056 or 10.56%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
A = 0.5 – 0.3944 + 0.4772 – 0.3159
= 0.1056 + 0.1613
= 0.2669 or 26.69%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
A = 0.5 – 0.3944 + 0.3413 + 0.5 – 0.3159
= 0.1056 + 0.3413 + 0.1841
= 0.6310 or 63.10%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
-2.75
A = 0.5 – 0.4970
= 0.003 or 0.30%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
-2.75
A = 0.5 – 0.4970 + 0.3944
= 0.003 + 0.3944
= 0.3974 or 39.74%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
-2.75
A = 0.5 – 0.4970 + 0.3944 + 0.5 – 0.4394
= 0.003 + 0.3944 + 0.0606
= 0.458 or 45.80%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
A = 1 – 2(0.4750)
= 1 – 0.95
= 0.05 or 5%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Applications of
Normal Curve
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
The following formula is used when
sample size is not given:
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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 zscores corresponding to each of the following.
1.
37
2.
22
3.
33
4.
28
5.
40
6.
27
7.
34 17
22
27
32
37
42 47
8.
30
9.
32
10. 25
Solutions:
1.
2.
3.
4.
5.
𝒙 − 𝝁 𝟑𝟕 −𝟑𝟐 𝟓
z=
=
= =1
σ
𝟓
𝟓
z=
𝒙−𝝁
σ
z=
𝒙−𝝁
σ
z=
𝒙−𝝁
σ
z=
𝒙−𝝁
σ
STATISTICS AND PROBABILITY
=
𝟐𝟐 −𝟑𝟐
𝟓
=
𝟑𝟑 −𝟑𝟐
𝟓
=
𝟐𝟖 −𝟑𝟐
𝟓
=
𝟒𝟎 −𝟑𝟐
𝟓
=
−𝟏𝟎
𝟓
= -2
𝟏
𝟓
= = 0.2
=
−𝟒
𝟓
= -0.8
𝟖
𝟓
= = 1.6
SAMSUDIN N. ABDULLAH, Ph.D.
6.
7.
8.
9.
10.
𝒙 − 𝝁 𝟐𝟕 −𝟑𝟐 −𝟓
z=
=
= = -1
σ
𝟓
𝟓
z=
𝒙−𝝁
σ
z=
𝒙−𝝁
σ
z=
𝒙−𝝁
σ
z=
𝒙−𝝁
σ
STATISTICS AND PROBABILITY
=
𝟑𝟒 −𝟑𝟐
𝟓
𝟐
𝟓
=
𝟑𝟎 −𝟑𝟐
𝟓
=
𝟑𝟐 −𝟑𝟐
𝟓
= =0
=
𝟐𝟓 −𝟑𝟐
𝟓
−𝟕
𝟓
= = 0.4
=
−𝟐
𝟓
= -0.4
𝟎
𝟓
=
= -1.4
SAMSUDIN N. ABDULLAH, Ph.D.
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:
𝒙 − 𝝁 𝟕𝟐 −𝟔𝟎 𝟏𝟐
1. z =
=
= = 1.5
σ
𝟖
𝟖
Referring to the z-table, the area
below z = 1.5 is 0.9332. Therefore, about
93.32% of the group got below 72.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
2. z =
𝒙−𝝁
σ
z=
𝒙−𝝁
σ
=
𝟓𝟖 −𝟔𝟎
𝟖
=
𝟕𝟔 −𝟔𝟎
𝟖
=
−𝟐
𝟖
=
𝟏𝟔
𝟖
= -0.25
=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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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:
𝒙 −𝟐𝟖𝟓
𝟏𝟐
z=
A = (1 – 0.05 ) – 0.5
= 0.95 – 0.5
= 0.45
𝒙 −𝟐𝟖𝟓
1.65 =
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 zscore 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)
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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) =
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
a) Given:
x = 43 minutes
μ = 46.2 minutes
σ=8
Solution:
P(x < 43) = ?
𝒙−𝝁
z= σ
=
43 − 46.2
8
=
−3.2
8
Thus, the probability that a
randomly selected college student
will complete the test in less than
43 minutes is 34.46%.
= -0.40
P(x < 43) = P(z < -0.40)
= 0.500 – 0.1554
= 0.3446 or 34.46%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
b) Given:
x = 43 minutes
μ = 46.2 minutes
σ=8
n = 50
Solution:
P(x > 43) = ?
𝒙−𝝁
z= 𝓸
𝒏
=
=
=
43 − 46
= 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%.
𝟖
𝟓𝟎
−3.2
𝟖
𝟕.𝟎𝟕
−𝟑.𝟐
𝟏.𝟏𝟑
= -2.83
P(x > 43) = P(z > -0.2.83)
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
a) Given:
x = 78
μ = 78
σ = 10
Solution:
P(x < 78) = ?
𝒙−𝝁
z= σ
=
Thus, the probability of a randomly
selected student to have a score of
less than 78 is 50%.
78 − 78
10
0
= 10
=0
P(x < 78) = P(z < 0)
= 0.50 or 50%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
b) Given:
x = 76
μ = 78
σ = 10
Solution:
P(x < 76) = ?
𝒙−𝝁
z= σ
=
Thus, the probability of a randomly
selected student to have a score
less than 76 is 7.93%.
76 − 78
10
−2
= 10
= -0.2
P(x < 78) = P(z < -0.2)
= 0.0793 or 7.93%
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
c) Given:
x₁ = 75
x₂ = 80
μ = 78
σ = 10
z=
80 − 78
10
2
= 10
= 0.2
Solution:
P(75 < x < 80) = ?
z=
=
=
𝒙−𝝁
σ
75 − 78
10
P(75 < x < 80) = 0.1179 + 0.0793
= 0.1972 or 19.72%
Thus, the probability of a randomly
selected student to have a score
between 75 and 80 is 19.72%.
−3
10
= -0.3
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
d) Given:
x = 95
μ = 78
σ = 10
P(x > 95) = P(x > 1.7)
= 0.500 – 0.4554
Solution:
P(x > 95) = ?
z=
= 0.0446 or 4.46%
Thus, the probability of a randomly selected
student to have a score above 95 4.46%.
𝒙−𝝁
σ
=
95 − 78
10
=
17
10
= 1.7
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
e) Given:
x = 76
μ = 78
σ = 10
n = 45
−𝟐
= 𝟏.𝟒𝟗
= -1.34
P(x > 76) = P(x > 1.34)
Solution:
P(x > 95) = ?
𝒙−𝝁
z= ℴ
= 0.4099 + 0.5000
= 0.9099 or 90.99%
𝒏
=
=
76 − 78
𝟏𝟎
𝟒𝟓
Thus, the probability that the 45 randomly
selected freshmen can have a mean of greater
than 76 is 90.99%.
−2
𝟏𝟎
𝟔.𝟕𝟏
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
a) Given:
x = 75
μ = 80
ℴ = 12
Solution:
P(x > 75)
z=
0.6628 (1000) = 662.8 or 663
Thus, there were 663 freshmen who passed the
entrance examination..
𝒙−𝝁
σ
=
=
𝟕𝟓 − 𝟖𝟎
𝟏𝟐
−𝟓
𝟏𝟐
= -0.42
P(x > 75) = P(z > -0.42)
= 0.1628 + 0.5000
= 0.6628
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
CHAPTER III
SAMPLING AND
SAMPLING
DISTRIBUTION
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
Sample Mean
10
8
40
29
17
32.7
38
26
80
Figure 1
24
55
34 34
33
32
35 32.7
32
33
31
33 30
Figure 2
Descriptive Statistics of the two given sets of sample data
Figure 1
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Figure 2
32.7
6.92989
27.5
None
21.9142
480.233
1.3037
1.13241
72
8
80
327
10
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
32.7
0.4726
33
33
1.4944
2.2333
-0.1518
-0.3595
5
30
35
327
10
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Probability Sampling Methods
1. Simple Random Sampling
- Fishbowl method
- Lottery Method
2. Systematic Sampling
3. Stratified Sampling
4. Cluster Sampling
5. Multistage Sampling
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Non-probability Sampling
Methods
1. Quota Sampling
2. Convenience Sampling
3. Purposive Sampling
4. Self-Selection Sampling
5. Snowball Sampling
6. Judgemental Sampling
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
Grade 7
Grade 8
Grade 9
Grade 10
Grade 11
Grade 12
n=
– 1209
– 1083
– 985
– 889
– 1087
– 780
6033
𝑵
𝟏 + 𝑵𝒆𝟐
=
𝟔𝟎𝟑𝟑
𝟏 + 𝟔𝟎𝟑𝟑(𝟎.𝟎𝟓)𝟐
=
𝟔𝟎𝟑𝟑
𝟏 + 𝟔𝟎𝟑𝟑(𝟎.𝟎𝟎𝟐𝟓)
STATISTICS AND PROBABILITY
=
𝟔𝟎𝟑𝟑
𝟏 + 𝟏𝟓.𝟎𝟖𝟐𝟓
=
𝟔𝟎𝟑𝟑
𝟏𝟔.𝟎𝟖𝟐𝟓
n = 375
Proportional Percentage:
𝟑𝟕𝟓
𝟔𝟎𝟎𝟑
= 0.0622
SAMSUDIN N. ABDULLAH, Ph.D.
Grade 7
Grade 8
Grade 9
Grade 10
Grade 11
Grade 12
– 1209 x 0.0622 = 75
– 1083 x 0.0622 = 67
– 985 x 0.0622 = 61
– 889 x 0.0622 = 55
– 1087 x 0.0622 = 68
– 780 x 0.0622 = 49
375
Then, apply the simple random
sampling technique in choosing the
individual respondent per group.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
Teachers
Parents
Principals
–
–
–
–
3050
550
320
150
Compute for the total number of sample as well
as the sample per group.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
Students
Teachers
Parents
Principals
– 3050
– 550
– 320
– 150
4070
=
𝟒𝟎𝟕𝟎
𝟏 + 𝟏𝟎.𝟏𝟕𝟓
=
𝟒𝟎𝟕𝟎
𝟏𝟏.𝟏𝟕𝟓
n = 364
n=
𝑵
𝟏 + 𝑵𝒆𝟐
=
𝟒𝟎𝟕𝟎
𝟏 + 𝟒𝟎𝟕𝟎(𝟎.𝟎𝟓)𝟐
=
𝟒𝟎𝟕𝟎
𝟏 + 𝟒𝟎𝟕𝟎(𝟎.𝟎𝟎𝟐𝟓)
STATISTICS AND PROBABILITY
Proportional Percentage:
𝟑𝟔𝟒
𝟒𝟎𝟕𝟎
= 0.0894
SAMSUDIN N. ABDULLAH, Ph.D.
Students
Teachers
Grade 9
Grade 10
– 3050 x 0.0894 = 273
– 550 x 0.0894 = 49
– 320 x 0.0894 = 29
– 150 x 0.0894 = 13
364
Then, apply the simple random
sampling technique in choosing the
individual respondent per group.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Population
USM - Kabacan
MSU - Maguindanao
CCSPC
SKSU
STATISTICS AND PROBABILITY
– 1580
– 1398
– 1409
– 1216
SAMSUDIN N. ABDULLAH, Ph.D.
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)
STATISTICS AND PROBABILITY
=
𝟓𝟔𝟎𝟑
𝟏 + 𝟏𝟒.𝟎𝟎𝟕𝟓
=
𝟓𝟔𝟎𝟑
𝟏𝟓.𝟎𝟎𝟕𝟓
n = 373
Proportional Percentage:
𝟑𝟕𝟑
𝟓𝟔𝟎𝟑
= 0.0666
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
Koronadal City
Cotabato City
Tacurong City
Kidapawan City
–
–
–
–
–
4050
2890
3060
2079
1980
Compute for the total number of sample as well
as the sample per group.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Select your answers from the following:
1.
Mean
13. t-distribution
2.
Median
curve
3.
Mode
14.Normal Curve
4.
Range
15. Statistics
5.
Standard Deviation
16. Zero
6.
Variance
17. Bell-Shaped
7.
Coefficient of Variation
18. Research
8.
Kurtosis
19. Statistics and
9.
Skewness
Probability
10. Scatteredness
20. Simple Random
11. Frequency
Sampling
12. Percentage
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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 ztest, t-test, analysis of variance (ANOVA),
simple linear correlation (Pearson r),
Spearman’s Rho, chi-square (x²) and
regression.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Testing a Hypothesis About a Single Mean
Using Large Samples (z-test)
z=
STATISTICS AND PROBABILITY
𝑥− 𝜇
ℴ
𝑛
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
=
I.
Ho: x = ₱8,048.25
Ha: x > ₱8,048.25
1520.15
124.38
z = 12.22
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
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.
1563.42
12.57
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
=
I.
Ho: x = 250 mL
Ha: x < 250 mL
−2
0.5
z = -4
II. α = 0.01
III. z-test (left-tailed)
IV. The z-critical value = -2.33
V. Computation:
z=
𝑥− 𝜇
=
=
ℴ
𝑛
VI. Decision Making/Conclusion
248 − 250
5
100
−2
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.
5
10
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
=
I.
Ho: x = 38.25
Ha: x ≠ 38.25
−1.45
1.17
z = -1.24
II. α = 0.05
III. z-test (two-tailed)
IV. The z-critical value = 1.96
V. Computation:
z=
𝑥− 𝜇
=
=
ℴ
𝑛
36.80− 38.25
10.5
81
−1.45
10.5
9
VI. Decision Making/Conclusion
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)
Types of Test/Significant Level 0.01 0.05 0.10
One-Tailed/One-Sided Test 2.33 1.65 1.29
Two-Tailed/Two-Sided Test 2.58 1.96 1.65
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Direction: Fill-in the boxes with the correct answers regarding hypothesis
testing. Second row serves as your example.
zcomp value
-5.256
1
2
3
4
5
6
7
8
9
10
Inequality
Symbol
<
15.783
-1.678
-2.05
0.247
-4.097
7.89
-5.079
-2.32
1.98
40.235
STATISTICS AND PROBABILITY
Zcritical value
-2.33
Decision
Reject Ho.
Interpretation
There is a significant
difference between the
sample mean and
population mean.
1.65
-1.96
-2.33
1.65
-2.33
1.96
-1.65
-2.33
1.96
1.96
SAMSUDIN N. ABDULLAH, Ph.D.
Direction: Fill-in the boxes with the correct answer regarding hypothesis testing.
Second row serves as your example.
1
2
zcomp value Inequality Zcritical value Decision
Interpretation
Symbol
-5.256
<
-2.33
Reject Ho. There is a significant
difference between the
sample mean and
population mean.
15.783
>
1.65
Reject Ho. There is a significant
difference between the
sample mean and
population mean.
-1.678
>
-1.96
Accept Ho. There is no significant
difference between the
sample mean and
population mean.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
3
-2.05
>
-2.33
4
0.247
<
1.65
5
-4.097
<
-2.33
STATISTICS AND PROBABILITY
Accept Ho. There is no significant
difference between the
sample mean and
population mean.
Accept Ho. There is no significant
difference between the
sample mean and
population mean.
Reject Ho. There is a significant
difference between the
sample mean and
population mean.
SAMSUDIN N. ABDULLAH, Ph.D.
6
7.89
>
1.96
7
-5.079
<
-1.65
8
-2.32
>
-2.33
STATISTICS AND PROBABILITY
Reject Ho. There is a significant
difference between the
sample mean and
population mean.
Reject Ho. There is a significant
difference between the
sample mean and
population mean.
Accept Ho. There is no significant
difference between the
sample mean and
population mean.
SAMSUDIN N. ABDULLAH, Ph.D.
9 1.98
10 40.235
>
>
STATISTICS AND PROBABILITY
1.96 Reject Ho. Significant
1.96 Reject Ho. There is a significant
difference between the
sample mean and
population mean.
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
=
I.
Ho: x = 24,500.00
Ha: x < 24,500
−9,500
365.8042
z = -25.9702
II. α = 0.05
III. z-test (Left-tailed)
IV. The z-critical value = -1.65
V. Computation:
z=
=
𝑥− 𝜇
ℴ
𝑛
15,000−24,500
4,480.15
150
VI. Decision Making/Conclusion
Since that z-computed value of -25.9702 is
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.
−9,500
= 4,480.15
12.2474
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
V. Computation:
Solution:
I.
Ho: x = 594.41
Ha: x ≠ 594.41
578+599+605+589+790+554+615+568+498+598+625+618+608+589
15
9014
= 15
x=
= 600.9333
II. α = 0.05
z=
III. z-test (Two-tailed)
IV. The z-critical value = 1.96
=
600.9333 − 594.41
87.16
15
=
6.5233
87.16
3.8730
6.5233
22.5046
= 0.2899
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
V. . Since that the z-comp = 0.2899 is less than zcritical = 1.96, we must reject the null hypothesis.
Thus, the sample mean is not significantly different
from the population mean.
I
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
=
I.
Ho: x = 48.25
Ha: x ≠ 48.25
−9,75
1.1739
z = -8.30565
II. α = 0.01
III. z-test (Two-tailed)
IV. The z-critical value = -2.58
V. Computation:
z=
𝑥− 𝜇
=
=
ℴ
𝑛
38.50−48.25
5.25
20
VI. Decision Making/Conclusion
Since that z-computed value of -8.30565 is
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.
−9.75
5.25
4.4721
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Solution:
=
I.
Ho: x = 70.08
Ha: x ≠ 70.08
5.5287
2.6815
z = 2.0618
II. α = 0.05
III. z-test (Two-tailed)
IV. The z-critical value = 1.96
V. Computation:
z=
𝑥− 𝜇
=
=
ℴ
𝑛
75.6087−70.08
12.86
23
5.5287
12.86
4.7958
STATISTICS AND PROBABILITY
VI. Decision Making/Conclusion
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.
SAMSUDIN N. ABDULLAH, Ph.D.
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 tdistribution change depending upon to
the degrees of freedom.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
Testing a Hypothesis About a Single Mean
Using Small Samples (t-test)
t=
STATISTICS AND PROBABILITY
𝑥− 𝜇
𝑠
𝑛
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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
t=
500.9 − 500
0.9
6.61
10
= 2.09
z = 0.4306
V. Computation:
495+503+507+498+490+505+510+502+493+506
10
5009
=
10
x=
x = 500.9
t=
𝑥− 𝜇
=
=
𝑠
𝑛
− 250
5
100
−2
5
10
VI. Decision Making/Conclusion
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)
t=
𝒙₁−𝒙₂
𝒔₁² 𝒔₂²
+
𝒏₁ 𝒏₂
; 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
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
t=
I. Hₒ: x₁ = x₂
Hₐ: x₁ ≠ x₂
II. α = 0.05
𝒙₁−𝒙₂
𝒔₁² 𝒔₂²
+
𝒏₁ 𝒏₂
=
III. t-test (two-tailed)
=
III. df = 8 + 9 – 2 = 15
tcritical = 2.1315
=
V. Computation:
25+20+24+25+26+28+20+18
x₁ =
= 23.25
=
8
x₂ =
23+21 23+26+25+27+19+17+19
9
STATISTICS AND PROBABILITY
23.25−22.2222
12.2142 11.9444
+
8
9
1.0278
1.5268 +1.3272
1.0278
2.854
1.0278
1.6894
tcomp = 0.6084
= 22.22
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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.
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
IDENTIFYING THE
STATISTICAL TOOL
APPLICABLE FOR THE
GIVEN STATEMENT OF
THE PROBLEM
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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 ActivityOriented?
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
17.
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?
18.
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?
19.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
20.
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?
21.
5. Does the academic
performance of struggling
students in the tool subjects
significantly differ from each
other?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
22.
23.
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?
STATISTICS AND PROBABILITY
SAMSUDIN N. ABDULLAH, Ph.D.
24.
STATISTICS AND PROBABILITY
8. What intervention
programs can be
proposed
to
minimize, if not
totally
eradicate
absenteeism among
the low performing
students?
SAMSUDIN N. ABDULLAH, Ph.D.
25.
STATISTICS AND PROBABILITY
9. Is there a
significant difference
between
the
academic
performance of TVL
and
HUMSS
students?
SAMSUDIN N. ABDULLAH, Ph.D.
Thank you so much
From
SAMSUDIN N. ABDULLAH, Ph.D.
Master Teacher II