ProbStat - Central Limit Theorem.pptx

CENTRAL LIMIT
THEOREM
STATISTICS AND PROBABILITY

OBJECTIVES
At the end of this presentation, you are expected to:
• illustrate the Central Limit Theorem
(M11/12SPIIIe-2)
• define the sampling distribution of the sample mean
using the Central Limit Theorem (M11/12SPIIIe-3)
• solve problems involving sampling distributions of
the sample mean (M11SP-IIIe-f1)
2

3 Nagmahal, Nasaktan,
Nabaliw, Natuto
Puso'y tumibok,
Hindi hinimok.
Kusang sumibol,
Hindi naghabol.
Naramdaman ang pintig,
Binuhay ang pag-ibig.
Nagmahal ng buong puso,
Umibig ng totoo.
Ngunit…
Hindi sinuklian.
Hindi ipinaglaban.
Ipinagpalit sa iba,
Iniwang mag-isa.
Luhaan ang mata,
Damdami'y nagdusa.
Piniga't sinaktan,
Nawalang saysay ang
ipinaglaban.
Kaya...
Nagmukmok sa kuwarto,
Nagkulong, nagtago.
Sarili'y inilayo,
Nanlumo, tuliro.
Tulala't mga mata'y mugto,
Hindi kinakausap ibang tao.
Nabaliw, nagpakatanga.
Nagmahal na nga lang,
nasaktan pa.
Subalit…
Buhay mo'y mahalaga.
Pamilya mo'y naririyan pa.
Sarili mo'y nilikha,
Nang isang Diyos na
mapagpala.
Maging matatag ka,
Kalimutan ang sakit at
magsimula.
Matuto ka,
Magbalik loob sa Kanya...

1
What is Central Limit Theorem (CLT)?
5
CENTRAL LIMIT
THEOREM

CENTRAL LIMIT THEOREM
“The sum or average of an infinite sequence of
independent and identically distributed random
variables, when suitably rescaled, tends to a
normal distribution.”
1810
Pierre-Simon Laplace
“If the individual measurements could be
viewed as approximately independent and
identically distributed, then their mean could be
approximated by a normal distribution.”
1824
Siméon-Denis Poisson

7
“As the sample size (n) becomes
bigger, the sampling distribution of
the sample mean can be
approximated by a normal
probability distribution”.

8
“As the sample size becomes bigger, the sampling distribution
of the sample mean can be approximated by a normal

9

10

Example: A population consists of the numbers 2, 4, 5, 9 and 10. Take
three numbers as samples from the population.
SAMPLE MEAN 𝐱 FREQUENCY f(𝐱) PROBABILITY P(𝐱)
3.67 1 1/10 or 0.1
5.00 1 1/10 or 0.1
5.33 2 2/10 or 0.2
5.67 1 1/10 or 0.1
6.00 1 1/10 or 0.1
6.33 1 1/10 or 0.1
7.00 1 1/10 or 0.1
7.67 1 1/10 or 0.1
8.00 1 1/10 or 0.1
TOTAL 10 1

Example: A population consists of the numbers 2, 4, 5, 9 and 10. Take
three numbers as samples from the population.
𝐱 P(𝐱) 𝐱 ∙ P(𝐱)
3.67 0.1 0.367
5.00 0.1 0.500
5.33 0.2 1.066
5.67 0.1 0.567
6.00 0.1 0.600
6.33 0.1 0.633
7.00 0.1 0.700
7.67 0.1 0.767
8.00 0.1 0.800
TOTAL 1 𝝁𝐱 = 6.000
𝝁𝐱 = 𝐱 ∙ 𝑷(𝐱)
Mean of the sampling
distribution = 6.00
𝛍 =
𝐱
𝐍
Mean of the
population
𝛍 =
𝟐 + 𝟒 + 𝟓 + 𝟗 + 𝟏𝟎
𝟓
𝛍 =
𝟑𝟎
𝟓
𝛍 = 𝟔
So, 𝛍𝐱 = 𝛍 or 𝐱 = 𝛍

13
The mean of the sample distribution
of means (𝛍𝐱) is equal to the mean of
the population (𝛍).
𝛍𝐱 = 𝛍

14
The variance of the sampling
distribution is equal to the variance of
the population divided by the sample
size.
𝝈𝟐
𝐱 =
𝛔𝟐
𝐧

15
The standard deviation of the sampling
distribution is equal to the standard deviation
of the population divided by the square root of
the sample size.
𝛔𝐱 =
𝛔
𝐧

PROBLEM #1
The record of weights of the male
population follows the normal distribution.
Its mean and standard deviations are 70
kg and 15 kg respectively. If a researcher
considers the records of 50 males, then
what would be the (a) mean and (b)
standard deviation of the chosen sample?

SOLUTION
a. mean
Since 𝛍𝐱 = 𝛍 and 𝛍 = 70 kg, then:
𝛍𝐱 = 𝛍
𝛍𝐱 = 70 kg
Given:
𝛍 = 70 kg
𝛔 = 15 kg
n = 50 males

SOLUTION
b. standard deviation
Since 𝛔𝐱 =
𝛔
𝐧
, 𝛔 = 15 kg and n = 50 males,
then:
𝛔𝐱 =
𝛔
𝐧
𝛔𝐱 =
𝟏𝟓
𝟓𝟎
𝛔𝐱 = 𝟐. 𝟏𝟐
Given:
𝛍 = 70 kg
𝛔 = 15 kg
n = 50 males

19
The z-score formula is:
𝐳 =
𝐱 − 𝛍
𝛔

20
𝐳 =
𝐱 − 𝛍𝐱
𝛔𝐱 𝛔𝐱 =
𝛔
𝐧
𝛍𝐱 = 𝛍

21
𝐳 =
𝐱 − 𝛍
𝛔
𝐧
Large sample
size: n > 30
Small sample
size: n ≤ 30

PROBLEM #2
During this time of pandemic, a certain
group of welfare recipients receives cash
benefits of ₱1,100.00 per week with a
standard deviation of ₱200. If a random
sample of 25 is taken, what is the
probability that their mean benefit is
greater than ₱ 1,200.00 per week?

Given:
𝛍 = ₱1,100.00
𝛔 = ₱200
n = 25
P(𝐱 >₱1,200.00)
𝛍 = ₱1,100.00 ₱1,200.00
greater than, above ₱1,200.00
Convert to z-score.
to the right of
z=n:
0.5000 - area of
normal curve
with the z-value
P(𝐱 >₱1,200.00)

SOLUTION
𝐳 =
𝐱 − 𝛍
𝛔
𝐧
𝐳 =
₱1,200.00 − ₱1,100.00
₱200
𝟐𝟓
𝐳 =
₱100.00
₱200
𝟐𝟓
Given:
𝛍 = ₱1,100.00
𝛔 = ₱200
n = 25
P(𝐱 >₱1,200.00)
𝐳 =
₱100.00
₱200
𝟓
𝐳 =
₱100.00
₱𝟒𝟎
𝐳 = 𝟐. 𝟓𝟎

SOLUTION
𝐳 = 𝟐. 𝟓𝟎
𝐳𝟐.𝟓𝟎 = 𝟎. 𝟒𝟗𝟑𝟖
𝐅𝐢𝐧𝐝 𝐳𝟐.𝟓𝟎.

𝛍 = ₱1,100.00 ₱1,200.00
SOLUTION
to the right of
z=n:
0.5000 - area of
normal curve
with the z-value
P(𝐱 >₱1,200.00)
𝐳𝟐.𝟓𝟎 = 𝟎. 𝟒𝟗𝟑𝟖

𝛍 = ₱1,100.00 ₱1,200.00
SOLUTION
to the right of
z=n:
0.5000 - area of
normal curve
with the z-value
P(𝐱 >₱1,200.00)
𝐳𝟐.𝟓𝟎 = 𝟎. 𝟒𝟗𝟑𝟖
0.5 – 0.4938 = 0.0062
∴ The probability that their mean benefit is greater than
₱1,200.00 per week is 0.0062 or 0.62%.

PROBLEM #3
There are 64 pawikan hatchlings in a marine
sanctuary in Batangas which can creep their
way to the sea from the shore at an average
speed of 0.025 meter per second with a standard
deviation of 0.012 meter per second. Assuming
that the variable is normally distributed and 16
pawikan hatchlings are chosen at random, what
is the probability that they have an average
speed of less than 0.03 meter per second?

Given:
𝛍 = 0.025
𝛔 = 0.012
n = 16
P(𝐱 < 0.03)
𝛍 = 0.025 0.03
less than, below 0.03
Convert to z-score.
to the left of z=n:
0.5000 + area of
normal curve
with the z-value
P(𝐱 < 0.03)

SOLUTION
𝐳 =
𝐱 − 𝛍
𝛔
𝐧
𝐳 =
0.03 − 0.025
0.012
𝟏𝟔
𝐳 =
0.005
0.012
𝟏𝟔
𝐳 =
0.005
0.012
𝟒
𝐳 =
0.005
0.003
𝐳 = 𝟏. 𝟔𝟕
Given:
𝛍 = 0.025
𝛔 = 0.012
n = 16
P(𝐱 < 0.03)

SOLUTION
𝐳 = 𝟏. 𝟔𝟕
𝐳𝟏.𝟔𝟕 = 𝟎. 𝟒𝟓𝟐𝟓
𝐅𝐢𝐧𝐝 𝐳𝟏.𝟔𝟕.

𝛍 = 0.025 0.03
P(𝐱 < 0.03)
SOLUTION
𝐳𝟐.𝟓𝟎 = 𝟎. 𝟒𝟗𝟑𝟖
to the left of z=n:
0.5000 + area of
normal curve
with the z-value

𝛍 = 0.025 0.03
SOLUTION
to the right of
z=n:
0.5000 - area of
normal curve
with the z-value
𝐳𝟏.𝟔𝟕 = 𝟎. 𝟒𝟓𝟐𝟓
0.5 + 0.4525 = 0.9525
∴ The probability that they have an average speed of
less than 0.03 meter per second is 0.9525 or 95.25%.
P(𝐱 < 0.03)

34
1. ILAW Manufacturing company
produces bulbs that last a mean of
900 hours with a standard deviation of
110 hours. What is the probability that
the mean lifetime of a random sample
of 15 of these bulbs is less than 850
hours?

35
2. A school principal claims that grade 11
students have a mean grade of 86 with a
standard deviation of 4. Suppose that the
distribution is approximately normal. There
are 20 students that were gathered for the
data. What is the probability that a
randomly selected grade will be greater
than 82?

37
𝐳 =
𝐱 − 𝛍
𝛔
𝐧
Large sample
size: n > 30
Small sample
size: n ≤ 30

38
1. The IQs of Grade 11 students in
MAKATAO NATIONAL HIGH SCHOOL
were measured and found to be normally
distributed with a mean of 98 and a
standard deviation of 8. What is the
probability that a random sample of 4
students will have an average of above
110?
SEATWORK

39
2. The mean annual salary of all the frontlines
(nurses, medical technologists, radiologic
technologists, phlebotomists) in the Philippines
is Php 42,500. Assume that this is normally
distributed with standard deviation Php 5,600. A
random sample of 25 health workers is drawn
from this population, find the probability that the
mean salary of the sample is less than Php
40,500?
SEATWORK

ProbStat - Central Limit Theorem.pptx

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