STATISTICS and PROBABILITY

Chapter 3, Lesson 2: Finding the Mean and Variance of the Sampling Distribution of Sample
Means
I. Overview:
 This lesson will demonstrate an understanding of Sampling Distribution
II. Objectives:
 Finds the mean and variance of the sampling distribution of the sample mean.
III. Content Outline:
 Please take time to read and understand this part because all graded activities and
quizzes will be coming from these contents.

Mean
- Also known as the “average”
- Calculated by dividing the sum of the observations by total number of observations.

Variance
- The average of the squared deviations from the mean.
 POPULATION VARIANCE

2
N
( X i−μ )
σ =∑
2

i=1 N
Where:
σ is the variance of a population
2

μ is the population mean

X i are the values of observations in the population
N is the total number of observations in the
population

Standard Deviation (σ )
- Square root of the variance

SAMPLE VARIANCE

√∑
2
n
( x i−X )
s=
i=1 n−1
Where:
s is the variance of a sample
X is the sample mean
x i are the values of observations in the sample
n is the total number of observations in the samples

POPULATION VARIANCE

√∑
2
N
( X i−μ )
σ=
i=1 N
Where:
σ is the variance of a population
μ is the population mean
X i are the values of observations in the population
N is the total number of observations in the
population
EXAMPLE 1:
Consider a population consisting of 1, 2, 3, 4, and 5. Suppose sample size 2 are drawn
from this population. Describe the sampling distribution of the sample mean.
 What is the mean and variance of the sampling distribution of the sample means?

Compute the mean of the population mean.

N

∑ Xi 1+2+3+ 4+5 15
μ= i=1 = = =3.00
N 5 5
So ,the mean of the population is 3.00

Compute the variance of the population variance.

Xi X i −μ ¿
2
1 1−3=−2 (−2) =4
2
2 2−3=−1 (−1) =1
2
3 3−3=0 (0) =0
2
4 4−3=1 (1) =1
2
5 5−3=2 (2) =4
∑ ¿¿
2
N
( X i−μ ) 10
σ =∑
2
= =2
i=1 N 5

So, the variance of the population is 2.

Determine the number of possible samples of size n = 2.

Use the formula N C n , Where N = 5 and n = 2

N C n= 5 C 2=10
So, there are 10 possible samples of size n = 2 that can be drawn.

List all possible samples and their corresponding means.

SAMPLES MEAN
1, 2 1.50
1, 3 2.00
1, 4 2.50
1, 5 3.00
2, 3 2.50
2, 4 3.00
2, 5 3.50
3, 4 3.50
3, 5 4.00
4, 5 4.5
Construct the sampling distribution of the sample means

SAMPLING DISTRIBUTION of SAMPLE MEANS

SAMPLE MEAN PROBABILITY P( X
FREQUENCY
X )
1
1.50 1 =0.1
10
1
2.00 1 =0.1
10
2
2.50 2 =0.2
10
2
3.00 2 =0.2
10
2
3.50 2 =0.2
10
1
4.00 1 =0.1
10
1
4.50 1 =0.1
10
TOTAL 10 1.00

Compute the mean of the sampling distribution of the sample means ( μ X ¿ .

Follow these steps:
 Multiply the sample mean by the corresponding probability.
 Add the results.

SAMPLING DISTRIBUTION of SAMPLE MEANS

SAMPLE MEAN PROBABILITY
X ∙ P (X )
X P( X )
1
1.50 =0.1 0.15
10
1
2.00 =0.1 0.20
10
2
2.50 =0.2 0.50
10
2
3.00 =0.2 0.60
10
2
3.50 =0.2 0.70
10
1
4.00 =0.1 0.40
10
1
4.50 =0.1 0.45
10
TOTAL 1.00 3.00

μ X =∑ X ∙ P ( X )=3.00

So, the mean of the sampling distribution of the sample means is 3.00

Compute the variance (σ 2 X ¿ of the sampling distribution of the sample

means. Follow these steps:
 Subtract the population mean ( μ) from each sample mean ( X ).
Label this as X −μ.
2
 SAMPLING DISTRIBUTION of SAMPLE MEANS
P( X 2 2
X X −μ (X −μ) P( X)∙(X −μ)
)
1 2
1.50 1.50 – 3.00 = -1.50 (−1.50) =2.25 0.225
10
1 2
2.00 2.00 – 3.00 = -1.00 (−1.00) =1.00 0.100
10
2 2
2.50 2.50 – 3.00 = -0.50 (−0.50) =0.25 0.050
10
2 2
3.00 3.00 – 3.00 = 0.00 (0.00) =0.00 0.000
10
2 2
3.50 3.50 – 3.00 = 1.50 (1.50) =0.25 0.050
10
1 2
4.00 4.00 – 3.00 = 1.00 (1.00) =1.00 0.100
10
1 2
4.50 4.50 – 3.00 = 1.50 (1.50) =2.25 0.225
10
TOTAL 1.00 0.750

=∑ P ( X ) ∙ ( X−μ )
2 2
σ X
= 0.75

So, the variance of the sampling distribution of the sample means is 0.75

Construct the histogram for the sampling distribution of the sample means.
Probability X

0.25
0.2
0.15
0.1
0.05
0
1.5 2 2.5 3 3.5 4 4.5
EXAMPLE 2:
Sample Mean X
 Considering a population consisting of 1, 2, 3, 4, and 5. Suppose samples of
size 3 are drawn from this population. Describe the sample means. What is
the mean and variance of the sampling distribution of the sample means?

Answer:

Compute the mean of the population mean.

N

∑ Xi 1+2+3+ 4+5 15
μ= i=1 = = =3.00
N 5 5
So ,the mean of the population is 3.00

Compute the variance of the population variance.

Xi X i −μ ¿
2
1 1−3=−2 (−2) =4
2
2 2−3=−1 (−1) =1
2
3 3−3=0 (0) =0
2
4 4−3=1 (1) =1
2
5 5−3=2 (2) =4
∑ ¿¿
2
N
( X i−μ ) 10
σ =∑
2
= =2
i=1 N 5

So, the variance of the population is 2.

Determine the number of possible samples of size n = 3.

Use the formula N C n , Where N = 5 and n = 3

N C n= 5 C 3=10
So, there are 10 possible samples of size n = 3 that can be drawn.

List all possible samples and their corresponding means.

SAMPLES MEAN
1, 2, 3 2.00
1, 2, 4 2.33
1, 2, 5 2.67
1, 3, 4 2.67
1, 3, 5 3.00
1, 4, 5 3.33
2, 3, 4 3.00
2, 3, 5 3.33
2, 4, 5 3.67
3, 4, 5 4.00

Construct the sampling distribution of the sample means

 SAMPLING DISTRIBUTION of SAMPLE MEANS
SAMPLE MEAN PROBABILITY P( X
FREQUENCY
X )
1
2.00 1 =0.1
10
1
2.33 1 =0.1
10
2
2.67 2 =0.2
10
2
3.00 2 =0.2
10
2
3.33 2 =0.2
10
1
3.67 1 =0.1
10
1
4.00 1 =0.1
10
TOTAL 10 1.00

Compute the mean of the sampling distribution of the sample means ( μ X ¿ .

Follow these steps:
 Multiply the sample mean by the corresponding probability.
 Add the results.

SAMPLING DISTRIBUTION of SAMPLE MEANS

SAMPLE MEAN PROBABILITY
X ∙ P (X )
X P( X )
1 1
2.00
10 5
1 7
2.33
10 30
2 8
2.67
10 15
2 3
3.00
10 5
2 2
3.33
10 3
1 11
3.67
10 30
1 2
4.00
10 5
TOTAL 1.00 3.00

μ X =∑ X ∙ P ( X )=3.00

So, the mean of the sampling distribution of the sample means is 3.00

Compute the variance (σ 2 X ¿ of the sampling distribution of the sample

means. Follow these steps:
 Subtract the population mean ( μ) from each sample mean ( X ).
Label this as X −μ.
 Square the difference. Label this as(X −μ)2.
 Multiply the results by the corresponding probability. Label this as
2
P( X)∙(X −μ)
 SAMPLING DISTRIBUTION of SAMPLE MEANS
P( X 2 2
X X −μ (X −μ) P( X)∙(X −μ)
)
1 2
2.00 2.00 – 3.00 = -1 (−1) =1.0000 0.10000
10
1 2
2.33 2.33 – 3.00 = -0.67 (−0.67) =0 . 4489 0.04489
10
2 2
2.67 2.67 – 3.00 = -0.33 (−0. 33) =0. 1089 0.02178
10
2 2
3.00 3.00 – 3.00 = 0.00 (0.00) =0.00 00 0.00000
10
2 2
3.33 3.33 – 3.00 = 0.33 (0.33) =0. 1089 0.02178
10
1 2
3.67 3.67 – 3.00 = 0.67 (0.67) =0 . 4489 0.04489
10
1 2
4.00 4.00 – 3.00 = 1.00 (1. 00) =1.0000 0.10000
10
TOTAL 1.00 0.33334

=∑ P ( X ) ∙ ( X−μ )
2 2
σ X
= 0.33334
≈ 0.33

So, the variance of the sampling distribution of the sample means is 0.33

Probability
Construct the
X histogram for the sampling distribution of the sample means.

0.25
0.2
0.15
0.1
0.05
2.00 2.33 2.67 3.00 3.33 3.67 4.00
Sample Mean X

PROPERTIES of the SAMPLING DISTRIBUTION of SAMPLE MEAN

If all possible samples of size n are drawn from a population of size N with mean μ
and variance σ 2 then the sampling distribution of the sample means has the
following properties:

1. The mean of the sampling distribution of the sample means is equal

to the population mean μ. That is μ X =μ
2. The variance of the sampling distribution of the sample means σ is
given by
2
σ N −n
2
σ X= ∙  for finite population
n N −n
2
σ
2
σ X=  for infinite population
n

3. The standard deviation of the sampling distribution of the sample

means is given by

√ n N √
σ 2 N −n
σ X= ∙
−n
population correction factor, and
√
 finite population where N −n is the finite
N −n

σ2
σ X=  for infinite population
√n
The standard deviation( σ X ) of the sampling distribution of the sample
means is also known as standard error of the mean.

The first and second example are all finite population. They have
always have equal mean of population.
Finite Population
- One that consists of a finite or fixed number of elements
measurements or observations
Infinite Population
- One that consists hypothetically at least, infinitely many
elements.

IV. Activities:

6
Consider all samples of size 5 from this population:
2 5 6 8 10 12 13
 Compute the mean and standard deviation of the population
 List all sample of size 5 and compute the mean for each sample
 Construct the sampling distribution of the sample means

V. Assignment:

Rest and study your modules. Be ready for your SUMMATIVE TEST 