The document discusses the Central Limit Theorem, which defines the sampling distribution of the sample mean and states that as sample size increases, the sampling distribution approaches a normal distribution. It provides formulas for calculating the mean, variance, and standard deviation of sampling distributions, along with practical examples to illustrate these concepts. Additionally, it covers the computation of probabilities using the z-score transformation in various scenarios.

1. Statistics and Probability
Central Limit Theorem

2. Illustrates the central limit
theorem.
2
Defines the sampling
distribution of the sample
mean using the central limit
theorem.
Learning
Competencies:
👩
📖
M11/12SP-IIIe-2
M11/12SP-IIIe-3

3. The mean of the sampling distribution of
the sample mean is equal to the population
mean.
3
Theorem 3-1
𝝁𝒙 - mean of the sampling distribution of sample means
𝝁 – population mean
𝝁𝒙 = 𝝁

4. The variance and standard deviation of a
sampling distribution of the sample means are
as follows:
4
Theorem 3-2
𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
(𝝈𝒙)𝟐 =
𝝈𝟐
𝒏
𝝈𝒙 =
𝝈
𝒏

5. EXAMPLE
Find the sample variance and the sample
standard deviation of a population, given 𝜎 = 5 and
𝑛 = 10
5
Given:
=
52
10
𝝈 = 𝟓 𝒏 = 𝟏𝟎
(𝝈𝒙)𝟐
=
𝝈𝟐
𝒏
= 𝟐. 𝟓
Solution:
=
25
10
𝝈𝒙 =
𝝈
𝒏
=
5
10
=
5
3.162
≈ 𝟏. 𝟓𝟖

6. If random samples of size n are drawn
from a population, then as n becomes
larger, the sampling distribution of the
mean approaches the normal
distribution, regardless of the shape of
the population distribution.
6
Central Limit Theorem

7. The standard error of the mean measures the
degree of accuracy of the sample mean as an
estimate of the population mean. It is also
known as the standard deviation of the
sampling distribution of the sample mean,
denoted by sample mean.
7
Standard Error of the Mean

8. Given a die, it has 6 faces in which each
faces have either dot/s of x = 1, 2, 3, 4, 5, and 6.
8
Example:
1. Compute the population mean
2. Compute the population variance
3. Compute the population standard deviation
4. Illustrate the probability histogram of the
sampling distribution of the mean.

9. 9
1. Compute the population mean.
µ =
∑x
N
where:
𝜇 − population mean
𝛴 x − sum of all data in terms of x
N − population size
µ =
∑𝒙
𝑵
=
1+2+3+4+5+6
6
=
21
6
= 3.5

10. 10
2. Compute the population variance.
X (X-𝜇)
1
2
3
4
5
6
N=6
1-3.5 = -2.5
2-3.5 = -1.5
5-3.5 = 1.5
4-3.5 = 0.5
6-3.5 = 2.5
3-3.5 = -0.5

11. 11
3. Compute the population standard deviation
X (X-𝜇) (X- 𝜇)²
1 1-3.5 = -2.5
2 2-3.5 = -1.5
3 3-3.5 = -0.5
4 4-3.5 = 0.5
5 5-3.5 = 1.5
6 6-3.5 = 2.5
N=6
(2.5)² = 6.25
(0.5)² = 0.25
(1.5)² = 2.25
(2.5)² = 6.25
(-1.5)² = 2.25
(-0.5)² = 0.25
𝛴 (x -𝜇)² = 17.5

12. 12
X (X-𝜇) (X- 𝜇)² P(X)
1 1-3.5 = -2.5 (2.5)² = 6.25
2 2-3.5 = -1.5 (-1.5)² = 2.25
3 3-3.5 = -0.5 (-0.5)²= 0.25
4 4-3.5 = 0.5 (0.5)² = 0.25
5 5-3.5 = 1.5 (1.5)² = 2.25
6 6-3.5 = 2.5 (2.5)²= 6.25
N=6 𝛴 (x -𝜇)² = 17.5
1/6
1/6
1/6
1/6
1/6
1/6

13. 13
𝛔𝟐
=
∑ 𝐗 − µ 𝟐
𝒏
Population variance
=
17.5
6
≈ 𝟐. 𝟗𝟐
𝛔 = 𝛔𝟐
Population standard deviation
= 2.92 ≈ 𝟏. 𝟕𝟏

14. 14
The population mean and
sampling distribution
means are both equal
which is 3.5. It has a
variance of approximately
2.92 and a standard
deviation of approximately
1.71. Since all samples
have the same probability
of 1/6. The trend of the
histogram is like a flat line
horizontally.

15. Consider a population of Senior High School
consisting of the values 2, 4, 6, 8, 10 and 12.
15
Problem:
1. Compute the population mean
2. Compute the population variance
3. Compute the population standard deviation
4. Illustrate the probability histogram of the
sampling distribution of the mean.

16. 16
1. Compute the population mean.
µ =
∑x
N
where:
𝜇 − population mean
𝛴 x − sum of all data in terms of x
N − population size
µ =
∑𝒙
𝑵
=
2+3+6+9+10+12
6
=
42
6
= 7

17. 17
2. Compute the population variance.
X (X-𝜇)
2 2 - 7 = - 5
4 4 - 7 = - 3
6 6 - 7 = - 1
8 8 - 7 = 1
10 10 - 7= 3
12 12 - 7 = 5
N=42

18. 18
3. Compute the population standard deviation
X (X-𝜇) (X- 𝜇)²
2 2 - 7 = - 5 (-5)² = 25
4 4 - 7 = - 3 (-3)² = 9
6 6 - 7 = - 1 (- 1)²= 1
8 8 - 7 = 1 (1)² = 1
10 10 - 7= 3 (3)² = 9
12 12 - 7 = 5 (5)²= 25
N=42 𝛴 (x -𝜇)² = 70

19. 19
X (X-𝜇) (X- 𝜇)² P(X)
2 2 - 7 = - 5 (-5)² = 25
4 4 - 7 = - 3 (-3)² = 9
6 6 - 7 = - 1 (- 1)²= 1
8 8 - 7 = 1 (1)² = 1
10 10 - 7= 3 (3)² = 9
12 12 - 7 = 5 (5)²= 25
N=42 𝛴 (x -𝜇)² = 70
1/6
1/6
1/6
1/6
1/6
1/6

20. 20
𝛔𝟐
=
∑ 𝐗 − µ 𝟐
𝒏
Population variance
=
70
6
≈ 𝟏𝟏. 𝟔𝟕
𝛔 = 𝛔𝟐
Population standard deviation
= 11.67 ≈ 𝟑. 𝟒𝟐

21. 21
The population mean and
sampling distribution
means are both equal
which is 7. It has a
variance of approximately
11.67 and a standard
deviation of approximately
3.42. Since all samples
have the same probability
of 1/6. The trend of the
histogram is like a flat line
horizontally.

22. Consider a population of Delfin Albano consisting of
the values 5, 10, 15, 20, 25, 30, 35 and 40.
22
Problem:
1. Compute the population mean
2. Compute the population variance
3. Compute the population standard deviation
4. Illustrate the probability histogram of the
sampling distribution of the mean.

23. 23
1. Compute the population mean.
µ =
∑x
N
where:
𝜇 − population mean
𝛴 x − sum of all data in terms of x
N − population size
µ =
∑𝒙
𝑵
=
5+10+15+20+25+30+35+40
6
=
180
10
= 18

24. 24
X (X-𝜇) (X- 𝜇)² P(X)
5 5 - 8 = - 3 (-3)² = 9
10 10 - 8 = 2 (2)² = 4
15 15 - 8 = 7 (-7)²= 49
20 20 - 8 = 12 (12)² = 144
25 25 - 8= 17 (17)² = 289
30 30 - 8 = 22 (22)²= 484
35 35 – 8 = 27 (27)² = 729
40 40 – 8 =32 (32)² =1,024
N=180
𝛴 (x -𝜇)² = 2732
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8

25. 25
𝛔𝟐
=
∑ 𝐗 − µ 𝟐
𝒏
Population variance
=
2732
8
≈ 𝟑𝟒𝟏. 𝟓
𝛔 = 𝛔𝟐
Population standard deviation
= 341.5 ≈ 𝟏𝟖. 𝟒𝟖

26. 26
X (X-𝜇) (X- 𝜇)² P(X)
10
15
25
34
43
56
ASSIGNMENT:

27. 27

28. Statistics and Probability
Central Limit Theorem

29. If the standard deviation is used as an
estimator of population parameter instead of
the mean, then the standard error (SE) is:
29
Standard Error
𝑺𝑬 = 𝝈𝒙
𝟐 =
𝝈
𝒏

30. The transformation of the z – score
considering the standard error is:
30
z – transformation
𝒛 =
𝒙 − 𝝁
𝝈
𝒏

31. The formula for the z – score when
working with the sample mean is given by
31
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
Theorem 3-3
𝒏 = sample size
𝒙 = sample mean
𝝁= population mean
𝝈 = population standard deviation

32. 32
REMEMBER
The distribution is approximately normal if
𝒏 ≥ 𝟑𝟎 regardless of the shape of distribution.
If 𝐧 ≤ 𝟑𝟎 , the sample mean 𝒙 is also
approximately normal as long as the
population is normally distributed.

33. Population Variance is given
The formula for the sampling
distribution of a normal population if the
population variance is given is
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
33

34. EXAMPLE 1: The average time it takes a group
of CCSASHS students to complete a Statistics
test is 54.8 minutes. The standard deviation is
5 minutes. We shall assume that the data are
normally distributed.
If there are 50 randomly selected students.
What is the probability that the mean time it
takes the group to finish the test will be less
than 53 minutes?

35. 35
𝑛 = 50 𝒙 = 53
𝝁 = 54.8 𝝈 = 5
1. Identify the given

36. 36
2. Identify what is asked
P( 𝑥˂ 53)

37. 37
3. Identify the formula
𝒛 =
𝒙 − 𝝁
𝝈
𝒏

38. 38
4. Solution to the problem
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
=
53 − 54.8
5
50
= −𝟐. 𝟓𝟓
=
−1.8
0.707
or
.0054 or 54%

39. 39
.0054
-2.5
.05

40. 40
5. Illustrate

41. 41
6. State your final answer
The probability that 50
randomly selected students will
finish the test in less than 53
minutes is 0.54%

42. EXAMPLE 2: Suppose the average age of
the family members is 34 with a standard
deviation of 4. If 100 members of the
community decided to have a summer
outing bonding and relaxation. Find the
probability that the average of these
member/s is less than 35?

43. 43
𝑛 = 100 𝒙 = 35
𝝁 = 34 𝝈 = 4
1. Identify the given

44. 44
2. Identify what is asked
P( 𝑥˂ 35)

45. 45
3. Identify the formula
𝒛 =
𝒙 − 𝝁
𝝈
𝒏

46. 46
4. Solution to the problem
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
=
35 − 34
4
100
=
1
0.4
= 2.5
or
0.9938 or 99.38 %

47. 47
2.5
.00
.9938

48. 48
5. Illustrate

49. 49
6. State your final answer
The probability that the random
sample of 100 persons has an
average of fewer than 35 years
old is 0.9938 0r 99.38%

50. EXAMPLE 3: The mean time it takes a group of
senior high students to complete a
certain examination is 50.6 minutes. The
standard deviation is 6 minutes.
Assume that the variable is normally distributed.
If 49 randomly selected senior high school
students take the examination, what is the
probability that the mean time it takes the group
to complete the test is between 47.8 and 53

51. 51
𝒏 = 49 𝒙 = 47.8 𝑎𝑛𝑑 53
𝝁 = 50.6 𝝈 = 6
1. Identify the given

52. 52
2. Identify what is asked
P( 47.8 < 𝑥 < 53)

53. 53
3. Identify the formula
𝒛 =
𝒙 − 𝝁
𝝈
𝒏

54. 54
4. Solution to the problem
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
=
47.8 − 50.6
6
49
= -3.27
=
−2.8
0.857
P(𝑥 >47.8) = P(𝑧 > - 3.27)
=0.0005
4.a

55. 55
4. Solution to the problem
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
=
53 − 50.6
6
49
= 2.8
=
2.4
0.857
P(𝑥 <53) = P(z > 2.8)
=0.9974
4.b

56. 56
3.2
.07
.0005
2.8
.00
.9974
0.9974-0.0005=0.9969 or 99.69%

57. 57
5. Illustrate

58. 58
6. State your final answer
The probability that 49 randomly
selected senior high school
students will complete the test
between 47.8 and 53 minutes is
99.69%

59. PROBLEM: The average time it takes a group of
Senior High School students to complete a
certain examination is 48.2 minutes. The
standard deviation is 7 minutes. Assume that
the variable is normally distributed.
1. If 40 randomly selected SHS student take the
examination, what is the probability that the mean it
takes the group to complete the test will be less than
45 minutes??

60. 60
𝒏 = 40, 𝒙 = 45, 𝝁 = 48.2,𝝈 = 7
1. Identify the given
2. Identify what is asked P( 𝑥˂ 45)
3. Identify the formula
𝒛 =
𝒙 − 𝝁
𝝈
𝒏

61. 𝒛 =
𝒙 − 𝝁
𝝈
𝒏
61
4. Solution to the problem
=
45 − 48.2
7
40
= - 2.89
=
−3.2
7
40
2.9 .0019
.00
.0019 or
19% |

62. 62
5. Illustrate
6. State your final answer
The probability that 40 randomly selected SHS
students will complete the test in less than 45
minutes is 0.0019 or 0.19%

63. PROBLEM: An investigator of a case of food
poisoning found that the amount of salmonella
in every serving of food is normally distributed
with an average of 3.7 colony forming units per
(cfu/g) and a standard deviation of 1.19 cfu/g.
1. What is the probability that a selected
serving has at least 4.2 cfu/g of
salmonella?

64. 64
𝒏 = 1,
𝒙 = 42
𝑐𝑓𝑢
𝑔
, 𝝁 = 3.7
𝑐𝑓𝑢
𝑔
𝝈 = 1.19
𝑐𝑓𝑢
𝑔
1. Identify the given
2. Identify what is asked P( 𝑥 ≥ 4.2)
3. Identify the formula
𝒛 =
𝒙 − 𝝁
𝝈
𝒏

65. 𝒛 =
𝒙 − 𝝁
𝝈
𝒏
65
4. Solution to the problem
=
4.2 − 3.7
1.19
1
= 𝟎. 𝟒𝟐
=
0.5
1.19
1
0.4
.02
.6628
.6628 or
66.28%

66. 66
5. Illustrate
6. State your final answer
The probability that a selected serving has at least
4.2 cfu/g of salmonella is 33.72%.

67. PROBLEM: An investigator of a case of food poisoning
found that the amount of salmonella in every serving of
food is normally distributed with an average of 3.7
colony forming units per (cfu/g) and a standard
deviation of 1.19 cfu/g.
1. What is the probability that a selected serving
has at least 4.2 cfu/g of salmonella?
2. What is the probability that the mean of 10
randomly selected servings is at least 4.2 cfu/g?
ASSIGNMENT:

68. 𝝁 = 3.7
𝑐𝑓𝑢
𝑔
68
Given: 𝒙 = 4.2
𝑐𝑓𝑢
𝑔 𝝈 = 1.19
𝑐𝑓𝑢
𝑔
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
=
4.2 − 3.7
1.19
10
= 1.33
= 𝟎. 𝟎𝟗𝟏𝟖 𝒐𝒓 𝟗. 𝟏𝟖%
𝑛 = 10,
=
0.5
0.3763
𝐴1.33 = 0.9082
∴ 𝑃 𝑋 ≥ 1.33 = 1 − 0.9082
The probability that ten serving has at least 4.2 cfu/g of salmonella is 9.18%
𝟏. 𝟑𝟑
1.3
.03
.9082

69. 69
𝝈𝒙 =
𝝈
𝒏 𝝈𝒙 =
𝟏. 𝟏𝟗
𝟏𝟎 = 0.376
𝝈𝒙 =
𝟏. 𝟏𝟗
𝟏
= 1.19
Lower standard deviation
Higher standard
deviation

70. 70
Urphone Company claims that the life of their battery products
is approximately normally distributed with a mean length of 8
hours of battery life when used in DC motors, with a standard
deviation of 3.8 hours. What is the probability that a 10 batteries
will have a mean lifespan of less than 6 hours in DC motors?
8
3.8 10
6

71. The probability that the sample of 10 batteries will have a mean lifespan of
less than 6 hours is 4.85%
71
Given: 𝑛 = 10, 𝝈 = 3.8
𝝁 =8,
𝒙 = 𝟔,
𝒛 =
𝒙 − 𝝁
𝝈
𝒏
=
6 − 8
3.8
10
=
−2
1.2017
= −1.66
.0485
=.0485
-1.6
.06
0.0485 𝑜𝑟 4.85%
−𝟏. 𝟔𝟔

72. 72
Activity 1 Solve the following problems
1. The average
production of all the
machines in a fabrication
is 15 finished materials in
one hour with a standard
deviation of 3 materials
per hour. What is the
probability that the
sample of 5 machines will
produce more than 17
materials per hour?
1. Identify the given
2. Identify what is asked
4. Solution to the problem
3. Identify the formula
5. Illustrate
6. State your final answer

73. 73
2. The average
rating of a public
high school
students in a
certain division of
region II is 85%
with a standard
deviation of 3.
a) If a student is
selected, what is the
probability that his
average rating is
greater than 80%
b) If a random sample
of 20 senior high
school students is
selected, what is the
probability that the
average rating is less
than 80%?
1. Identify the given
2. Identify what
is asked
4. Solution to the
problem
3. Identify the
formula
5. Illustrate
6. State your final
answer

74. 74
Source: Statistics & Probability Book
DepEd Module
Thank you!
Disclaimer:
This video lesson is created for educational
purposes only.
No copyright claim
Keep safe everyone!

75. 75

76. Z - Table
( Negative )

77. Z - Table
( Positive )