The document discusses the central limit theorem and how it allows inferences to be made about a population based on a sample. It provides examples of using the central limit theorem to calculate probabilities related to sample means. It explains that as sample size increases, the sampling distribution of the sample mean will approach a normal distribution, regardless of the shape of the population. It emphasizes that the central limit theorem is important because it allows samples to be used to make inferences about populations.

1. Previous Lesson
What are the properties of
sampling distribution of the
sample means?

2. Refresher:
A. Revisiting the
A National Achievement Test
was given to a group of high
school graduating students.
The result shows that the mean
score is 95 with a standard
deviation of 15. Determine the
standard scores corresponding
to the following scores.
Score Standard
Score
110
100
85
70

3. B. Areas Under the Normal Curve
Find the area under the normal curve
given the following conditions.
Conditions Area
Between z=0.5 and z=1.5
To the left of z=0.27
To the right of z= 1.23

4. Answer:
A. B.
Score Standard
Score
110 z=1
100 z=0.33
85 z=-0.67
70 z=-1.67
Conditions Area
Between z=0.5 and z=1.5 0.2417 or
24.17%
To the left of z=0.27 0.6064 or
60.64%
To the right of z= 1.23 0.1093 or
10.93%

5. Central Limit Theorem
PREPARED BY:
AILEEN D. LOSITAÑO
SHS TEACHER

6. True or false?
1. If the population is normal, the sample size should be greater
than 30. _______
2. The Central Limit Theorem tells us that as the sample size
increases, the sampling distribution of the sample means
approaches a normal distribution regardless of the shape of the
population.__________
3. The sample mean can be standardized or converted to z-score
if the population is normal and any sample size can be use.
_____

7. The Central Limit Theorem
(for the sample mean x)
If a random sample of n observations is selected
from a population (any population), then when n is
sufficiently large, the sampling distribution of x will be
approximately normal.
(The larger the sample size, the better will be the
normal approximation to the sampling distribution of
x.)

8. The Central Limit Theorem
This theorem allows us to use a
sample to make inferences
about a population because it
states that if n is sufficiently
large, the sampling distribution
will be approximately normal no
matter what the population
distribution looks like.

9. To apply the Central Limit Theorem
-If the population distribution is normal,
any sample size can be used.
-If the population distribution is not
normal, then 𝑛 ≥ 30.
-The distribution can be converted to
standard scores provide that sample size
is large.

10. Using the Central Limit Theorem to Convert
to the Standard Normal Distribution

11. Example 1: Cholesterol Level
The average number of milligrams (mg) of cholesterol in a
cup of a certain brand of ice cream is 660mg, and the
standard deviation is 35mg. Assume the variable is
normally distributed.
A. If a cup of ice cream is selected, what is the probability
that the cholesterol content will be more than 670mg?
B. If a sample of 10 cups of ice cream is selected, what is
the probability that the mean of the sample will be larger
than 670mg?

12. Solution:
Given: µ=660; σ=35; X=670
a. Find P(X>670)
𝑧 =
𝑋 − 𝜇
𝜎
=
670 − 660
35
= 0.29
Find the area under the normal curve using the z-table:
The P(X>670) has the area of 0.3859 or 38.59%
b. Given: µ=660; σ=35; 𝑥 = 670; n=10
𝑧 =
𝑥 − 𝜇
𝜎
𝑛
=
670 − 660
35
10
= 0.90
Using the z-table: The P( 𝑥>670) is 0.1841 or 18.41%

13. Example 2: Time to complete
a Jingle
The average time it takes a group of
college students to complete a certain jingle
presentation in MAPEH is 46.2 minutes. The
standard deviation is 8 minutes. Assume that the
variable is normally distributed. If 50 randomly
selected college students create a jingle, what is
the probability that the mean time it takes the
group to complete the jingle will be less than 43
minutes?

14. Solution:
Given: µ=46.2 ; σ=8 ; 𝑥=43 ;n=50 find P( 𝑥<43)
𝑧 =
𝑥−𝜇
𝜎
𝑛
= 𝑧 =
43−46.2
8
50
=
−3.2
1.131
= −2.83
Using z table, the area of z=-2.83 to the left is 0.0023
or 0.23%
So, the probability that 50 randomly selected college
student will complete the jingle in less than 43 minutes
is 0.23%.

15. Board Work:
1. The average cholesterol content of a certain canned
goods is 215 milligrams and the standard deviation is 15
milligrams. Assume the variable is normally distributed. If a
sample of 25 canned goods is selected, what is the
probability that the mean of the sample will be larger than
220 milligrams?
2. The average public high school has 468 students with a
standard deviation of 87. If a random sample of 38 public
elementary schools is selected, what is the probability that
the mean number of students enrolled is between 445 and
485?

16. Solution:
1. Given: µ=215; σ=15; 𝑥 =220; n=25
Find P( 𝑥 >220)
𝑧 =
𝑥 − 𝜇
𝜎
𝑛
=
220 − 215
15
25
= 1.67
2. Given: µ=468; σ=87; 𝑥 = 445 𝑎𝑛𝑑 485; n=38
Find P(445 ≤ 𝑥 ≤ 485)
𝑧1 =
𝑥 − 𝜇
𝜎
𝑛
=
445 − 468
87
38
=
−23
14.113
= −1.63 𝑧2 =
𝑥 − 𝜇
𝜎
𝑛
=
485 − 468
87
38
=
17
14.113
= 1.20
Using the z-table: The PP(445 ≤ 𝑥 ≤ 485) is 0.8849-0.0516=0.8333 or 83.33%
Find the area under
the normal curve
using the z-table:
The P( 𝑥 >220) has the
area of 0.0475 or
4.75%

17. What do you think is the importance
of Central Limit Theorem?

19. Seatwork:
 Try this!
1. An electrical company claims that the average life of the
bulbs it manufactures is 1200 hours with a standard
deviation of 250 hours. If a random sample of 100 bulbs
is chosen, what is the probability that the sample mean
will be:
a. Greater than 1,150 hours?
b. Less than 1,250 hours?
c. Between 1,150 and 1,250 hours?

20. Solution:
Given: µ=1200; σ=250; n=100
a. Find P( 𝑥 >1,150)
𝑧 =
𝑥 − 𝜇
𝜎
𝑛
=
1,150 − 1,200
250
100
= −2
Find the area under the normal curve using the z-table:
The P( 𝑥 >1,150) has the area of 0.9772 or 97.72%
b. Find P( 𝑥 <1,250)
𝑧 =
𝑥 − 𝜇
𝜎
𝑛
=
1,250 − 1,200
250
100
= 2
Find the area under the normal curve using the z-table:
The P( 𝑥 <1,250) has the area of 0.9772 or 97.72%

21. c. Find P(1,150 ≤ 𝑥 ≤1,250)
(Answer in A and B)
The P( 𝑥 >1,150) has the area of 0.9772 or 97.72%
The P( 𝑥 <1,250) has the area of 0.9772 or 97.72%
Solution:
.9972-.9972= 0
P(1,150 ≤ 𝑥 ≤1,250)=0 %

22. Evaluation:
1. Manufacturer of ball bearings claims that
this product has a mean weight 502g and
standard deviation of 30g. What is the
probability that a random sample of 100
ball bearings will have combined weight:
a. Between 496g and 500g?
b. More than 510g?

23. Answer:
Given: µ=502; σ=30; n=100
a. Find P(496 ≤ 𝑥 ≤ 500)
𝑧 =
𝑥 − 𝜇
𝜎
𝑛
=
496 − 502
30
100
= −2
𝑧 =
𝑥 − 𝜇
𝜎
𝑛
=
500 − 502
30
100
= −0.67
Find the area under the normal curve using the z-table:
The 0.2514 – 0.0228=0.2286 or 22.86%

24. Thank you and
God bless!!!