Mean and Variance of Sampling Distribution of Sample Means

1. Mean and Variance of
Sampling Distribution of
Sample Means
Lesson 2
Page 1
Statistics and Probability – Week 5

2. Page 2
In the previous lesson, it is explained that the parameter has a fixed value, while the
value of the statistic varies. Since a statistic is derived from a sample, the value of a
statistic will differ from one sample to the next. This is why statisticians are interested
not only in explaining the variance of individual data about the population mean, but
also in understanding how the means of similar-sized samples obtained from the same
population differ about the population means.
Illustrative Example 1. A population consists of five numbers 1, 2, 3, 4, and 5. Suppose
samples of size 2 are drawn from this population.
(a) Find the mean and variance of the population
(b) Describe the sampling distribution of the sample means.
(c) Find the mean and variance of the sampling distribution of the sample means.
Solution.
(a) In computing the mean of the population, we have to get the sum of all the
values and divide the result by the number of values involved. That is,
𝜇 =
𝑥
𝑁
=
1 + 2 + 3 + 4 + 5
5
=
15
5
= 3
𝜇 = 3
✓ The mean of the population is 3.

3. Page 3
In computing the variance of the population we must construct a table
consists of three columns. The first column will be for the values given which are
1, 2, 3, 4, and 5. In the second column, subtract the mean from each value given.
Lastly, in the third column, square each value on the second column. After
completing the values, get the sum of the values on the third column.
𝒙 𝒙 − 𝝁 (𝒙 − 𝝁)𝟐
1 -2 4
2 -1 1
3 0 0
4 1 1
5 2 4
(𝒙 − 𝝁)𝟐 = 𝟏𝟎
Computing for the variance of the
population, apply the formula, that is,
𝜎2 =
(𝒙 − 𝝁)𝟐
𝑁
=
10
5
= 2
𝜎2 = 2
For the standard deviation, extract the
square root of the variance, that is,
𝜎 = 2
𝜎 = 1.41
✓ The variance of the population is 2.
✓ The standard deviation of the population is 1.41

4. Page 4
(a) In describing the sampling distribution of the sample means, construct a
table. The first column is for the sample or the possible outcomes when a
sample size of two will be drawn. In the second column, compute the
corresponding means of each sample. That is,
1+2
2
,
1+3
2
,
1+4
2
, and so on.
Sample Mean
1, 2 1.5
1, 3 2
1, 4 2.5
1, 5 3
2, 3 2.5
2, 4 3
2, 5 3.5
3, 4 3.5
3, 5 4
4, 5 4.5
In constructing the sampling distribution of the sample means, create a
table consists of three columns. In the first column, arrange the sample
means from lowest to highest. In the second column, write the frequency or
the number of times the mean occurs. In the third column, compute the

5. Page 5

6. Page 6
(a) In finding the mean of the sampling distribution of the sample means,
construct a table of values wherein the first two columns are the sample mean
and its corresponding probability. In the third column, compute the product
of sample mean and probability. As in, 1.5 0.1 , 2 0.1 , 2.5 0.2 , 3 0.2 ,
and so on. Then, get the sum of the results. The sum of the values in the third
column is the mean of the sampling distribution.
(𝒙) 𝑃(𝒙) (𝒙) ∙ 𝑷(𝒙)
1.5 0.1 0.15
2 0.1 0.20
2.5 0.2 0.50
3 0.2 0.60
3.5 0.2 0.70
4 0.1 0.40
4.5 0.1 0.45
Total 1 3.00
𝜇𝑥 = (𝒙) ∙ 𝑷(𝒙)
𝜇𝑥 = 3
The mean of the sampling distribution of the sample means is 3

7. Page 7
In finding the variance of the sampling distribution of the sample
means, construct a table of values wherein the first two columns are the
sample mean and its corresponding probability. In the third column,
subtract the population mean from each sample mean. That is 1.5 − 3, 2 −
3, 2.5 − 3, 3 − 3, and so on. In the next column, square each result in the
third column. In the last column, multiply the results to the corresponding
probability. Then get the sum of the values.
(𝒙) 𝑃(𝒙) 𝒙 − 𝝁 𝒙 − 𝝁 2 𝑃(𝒙) ∙ 𝒙 − 𝝁 2
1.5 0.1 -1.5 2.25 0.225
2 0.1 -1 1 0.1
2.5 0.2 -0.5 0.25 0.05
3 0.2 0 0 0
3.5 0.2 0.5 0.25 0.05
4 0.1 1 1 0.1
4.5 0.1 1.5 2.25 0.225
Total 1 0.750
Computing for the variance of the
sampling distribution of the sample
means, that is,
𝜎2
𝑥 = 𝑷(𝒙) ∙ 𝒙 − 𝝁 2
𝜎2
𝑥 = 0.75
For the standard deviation, extract the
square root of the variance, that is,
𝜎𝑥 = 0.75
𝜎𝑥 = 0.87
✓ The variance of the sampling distribution of the sample means is
0.75.
✓ The standard deviation of the sampling distribution of the sample
means is 0.87

8. Page 8
Illustrative Example 2. A population consists of five numbers 1, 2, 3, 4, and 5. Suppose
samples of size 3 are drawn from this population.
(a) Describe the sampling distribution of the sample means.
(b) Find the mean and variance of the sampling distribution of the sample means.
Solution.
Since the given numbers are the same in illustrative example 1, the mean and
variance of the population will also be 𝜇 = 3 and 𝜎2 = 2.
(a) In describing the sampling distribution of the sample means, the same
process as illustrated in example 1, construct a table. The first column is for
the sample or the possible outcomes when a sample size of three will be
drawn. In the second column, compute the corresponding means of each
sample. That is, (1+2+3)/3, (1+2+4)/3, (1+2+5)/3, and so on.
Sample Mean
1, 2, 3 2
1, 2, 4 2.33
1, 2, 5 2.66
1, 3, 4 2.66
1, 3, 5 3
1, 4, 5 3.33
2, 3, 4 3
2, 3, 5 3.33
2, 4, 5 3.66
3, 4, 5 4

9. Page 9
Now, arrange the sample means from lowest to highest. Then, write the frequency
or the number of times the mean occurs. Lastly, compute the probability or the quotient
of frequency and the total number of frequencies.
Sample Mean (𝒙) Frequency Probability 𝑃(𝒙)
2 1 0.1
2.33 1 0.1
2.66 2 0.2
3 2 0.2
3.33 2 0.2
3.66 1 0.1
4 1 0.1
Total 10 1

10. Page 10
(a) As illustrated in example 1, to find the mean of the sampling distribution of
the sample means, construct a table of values wherein the first two columns
are the sample mean and its corresponding probability. Then, compute the
product of sample mean and probability.
(𝒙) 𝑃(𝒙) (𝒙) ∙ 𝑷(𝒙)
2 0.1 0.2
2.33 0.1 0.233
2.66 0.2 0.532
3 0.2 0.60
3.33 0.2 0.666
3.66 0.1 0.366
4 0.1 0.4
Total 1 3.00
𝜇𝑥 = (𝒙) ∙ 𝑷(𝒙)
𝜇𝑥 = 3
✓ The mean of the sampling distribution of the sample means is 3.

11. Page 11
To find the variance of the sampling distribution of the sample
means, construct a table of values wherein the first two columns are the
sample mean and its corresponding probability. Then, subtract the
population mean from each sample mean. Next, square each result in the
third column. Lastly, multiply the results to the corresponding probability.
Then get the sum of the values.
(𝒙) 𝑃(𝒙) 𝒙 − 𝝁 𝒙 − 𝝁 2
𝑃(𝒙) ∙ 𝒙 − 𝝁 2
2 0.1 -1 1 0.1
2.33 0.1 -0.67 0.4489 0.0448
2.66 0.2 -0.34 0.1156 0.0231
3 0.2 0 0 0
3.33 0.2 0.33 0.1089 0.0217
3.66 0.1 0.66 0.4356 0.0435
4 0.1 1 1 0.1
Total 1 0.3331
Computing for the variance of the
sampling distribution of the sample
means,
𝜎2
𝑥 = 𝑷(𝒙) ∙ 𝒙 − 𝝁 2
𝜎2
𝑥 = 0.33
For the standard deviation,
𝜎𝑥 = 0.33
𝜎𝑥 = 0.57
✓ The variance of the sampling distribution of the sample means is
0.33.
✓ The standard deviation is of the sampling distribution of the
sample means 0.57

12. Page 12
Now, compare the values obtained from the two examples illustrated above.
Complete the table below.
Illustrative Example 1 Illustrative Example 2
Population
(𝑵 = 𝟓)
Sampling
Distribution
of the Sample
Means
(𝒏 = 𝟐)
Population
(𝑵 = 𝟓)
Sampling
Distribution
of the Sample
Means
(𝒏 = 𝟑)
Mean 𝜇 = 3 𝜇𝑥 = 3 𝜇 = 3 𝜇𝑥 = 3
Variance 𝜎2
= 2 𝜎2
𝑥 = 0.75 𝜎2
= 2 𝜎2
𝑥 = 0.33
Standard
Deviation
𝜎 = 1.41 𝜎2
𝑥 = 0.87 𝜎 = 1.41 𝜎𝑥 = 0.57
Comparing these computed values of mean, variance, and standard deviation, we
may derive and summarize the properties of the sampling distribution.

13. Page 13
Properties of the Sampling Distribution of Sample Means
If all possible samples of size n are drawn from a population of size N with mean 𝜇 and
variance 𝜎2, 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, 𝜇𝑥 = 𝜇
2. The variance of the sampling distribution of the sample means 𝜎2
𝑥 is given by
• 𝜎2
𝑥 =
𝜎2
𝑛
∙
𝑁−𝑛
𝑁−1
for finite population; and
• 𝜎2
𝑥 =
𝜎2
𝑛
for infinite population
3. The standard deviation of the sampling distribution of the sample means is
given by
• 𝜎𝑥 =
𝜎
𝑛
∙
𝑁−𝑛
𝑁−1
for a finite population where
𝑁−𝑛
𝑁−1
is the finite population
correction factor
• 𝜎𝑥 =
𝜎
𝑛
for infinite population
The first two examples illustrated are all finite populations. A finite population is
made up of a collection of elements, measurements, or observations with a specified
number. An infinite population, on the other hand, includes, at the very least, an infinite
number of elements. To understand how the sampling distribution of the sample means
from an infinite population is described, study the example illustrated below.

14. Page 14
Illustrative Example 3. A population has a mean of 60 and a standard deviation of 5.
A random sample of 16 measurements is drawn from this population. Describe the
sampling distribution of the sample means by computing its mean and standard
deviation.
Solution. Since the number of population is not given, assume that the population is
infinite.
Given: 𝜇 = 60, 𝜎 = 5 and 𝑛 = 16
Using the first property, 𝜇𝑥 = 𝜇, we can conclude that 𝜇𝑥 = 60.
Using the third property, 𝜎𝑥 =
𝜎
𝑛
𝜎𝑥 =
5
16
𝜎𝑥 =
5
4
𝜎𝑥 = 1.25
✓ The mean sampling distribution of the sample means is 60.
✓ The standard deviation of the sampling distribution of the sample means is
1.25