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Section 6 – 6:
Sampling Distribution of the Sample Means ( X )
The last section developed the probabilities for data that came from a normal distribution of x
values. These questions looked like P(x > 1) or P(x < 1) or P( 1< x < 4). These probability
questions ask the question if I select 1 z value from the entire population of z values what is the
probability that the selected x value will be within the stated range of x values.
There are many problems where the concern is not the values of x but the average value for a
sample of several x values taken as a group. The concern in this type of problem is about the
value of the average of the entire sample. We call the average of a sample the sample mean.
Examples of sample means:
1. A Ski Lift Chair has room for three riders. A sign on the chair lift says the mean (average) weight
for the three riders should be less than 200 pounds. The lift may fail if the average, or mean,
weight for the three riders is more than 200 pounds. There are many people in line waiting to ride
the chair. Some samples of 3 riders may have riders that weighs as much as 250. Some
samples will contain lighter riders. The question we want to ask is “what is the probability that any
sample of three riders selected at random will have a mean of more than 200 pounds?
2. An elevator manufacture makes a 10 passenger elevator. They limit the total weight to 2000
pounds for the 10 passengers. If a sample of 10 people total 2000 pounds then the average or
mean weight for the sample is 200 pounds. The manufacturer would need to know what the
probability is that any random sample of 10 people would have an mean weight of more than
200 pounds.
3. An airplane company is not concerned about the weight of a single passenger on a 200 passerger
plane. They are concerned about the average weight for the 200 passengers that may be on
the plane. They need the mean of any 200 passengers to be below a given value.
These are different questions than in the past section. The questions in this section are based on the
probability of selecting a sample of n values and having the mean of the n values selected being
within a given range. Many different samples of size n can be taken and the value for each sample
mean can be recorded.
If we had a distribution (collection) of the values of all the possible sample means for samples of
n values taken from a population of N values, we could easily find how many of these means x n are
within a given range. A distribution of the sample means looks like this X = {x1 ,x 2 ,x 3 , x 4 , x 5 ,....x n }
We will use the symbol X to represent the distribution of all the possible sample means. The
following examples develop the procedure for creating the distribution of all the possible sample
means X and using that distribution to answer the probability of any mean x n being within a given
range.
6 – 6 Central Limit Theorem Lecture
Page 1 of 10
© 2012 Eitel
An example of the Distribution of the Sample Means ( X )
A FLC Child Center swing has room for two riders. A sign on the swing says the mean (average)
weight for the pair of riders cannot be more than 56 pounds. We do not care about the
weight of a single rider. The question they want to know is “what is the probability that any sample of
two riders selected at random will have a sample mean of more than 56 pounds?” One way
to answer this question is to select every possible sample of two riders and find the mean for each
sample. This procedure would produce a collection of all the possible sample means for two riders.
One day the FLC Child Center has five children that want to ride the swing. The weights of the five
children are 42 pounds, 52 pounds, 54 pounds, 58 pounds and 60 pounds. None of the children are
too heavy for the swing alone. The question the Center wants answered is “what is the probability
that the mean weight for any pair of the five children will be more than 56 pounds?”
The table below shows every possible sample of two weights taken from the five weights. The
mean of each sample is calculated and each mean is labeled. The mean of the first sample is x 1 = 47,
the mean of the second sample is x 2 = 48 , the mean of the third sample is x 3 = 50 ect.
42
52
54
58
60
x:
The weights of
the 5 children
42 + 52
= x 1 = 47
2
52 + 58
= x 6 = 55
2
42 + 54
= x 2 = 48
2
52 + 60
= x 7 = 56
2
42 + 58
= x 3 = 50
2
54 + 58
= x 8 = 56
2
42 + 60
= x 4 = 51
2
54 + 60
= x 9 = 57
2
52 + 54
= x 5 = 53
2
58 + 60
= x10 = 59
2
the calculation of the sample
mean for every possible
sample of the 5 weights
taken two at a time
x 1 = 47
x 6 = 55
x 2 = 48
x 7 = 56
x 3 = 50
x 8 = 56
x 4 = 51
x 9 = 57
x 5 = 53
x 10 = 59
X
the distribution of all the
sample means of size 2
There are 10 possible sample means of the five children taken 2 at a time. All 10 of these means
form a Distribution of the Sample Means of size n = 2. We label this distribution X . This distribution
can be used to answer the question the FLC Child Center wants to know, “what is the probability that
the mean weight for any pair of the four children will be more than 56 pounds?”
From the Distribution X we can see that 2 out of the 10 possible sample means are more than 56
pounds. There is a 20% chance that two of the five children listed will get on the swing and their
average weight will be more than 56 pounds.
6 – 6 Central Limit Theorem Lecture
Page 2 of 10
© 2012 Eitel
Example 1
A Distribution of Sample Means of size n= 2 (with replacement)
for the Normal population of { 4, 6, 8 }
4
6
8
x:
A population
of 3 values
4+ 4
= x1 = 4
2
6+8
= x6 = 7
2
4+6
= x2 = 5
2
8+ 4
= x7 = 6
2
4+8
= x3 = 6
2
8+6
= x8 = 7
2
6+ 4
= x4 = 5
2
8+8
= x9 = 8
2
x1 = 4
x6 = 7
x2 = 5
x7 = 6
x3 = 6
x8 = 7
x4 = 5
x9 = 8
x5 = 6
6+6
= x5 = 6
2
the calculation of the sample means
for every possible sample of 3 values
taken two at a time n = 2
(with replacement)
X
the distribution of all the
sample means of size 2
from a population of size 3
x
f( x )
4
the mean of the x values = µ x
the SD of the x values = σ x
4
1
6
Input the x data into the Calculator
5
2
8
µx = 6
6
3
σ x = 1.633
7
2
input the X data into the calculator
8
1
ux = 6
x
σ x 1.633
=
= 1.15
n
2
sample size = n = 2
the mean of all the sample means = µ x
the SD of all the sample means = σ x
sample size = n = 2
σ x = 1.15
The mean of the population of x values µ x is equal to the mean of the distribution of sample means
µx
We will make the claim (without proof at this time) that µ x = µx
The standard deviation of the population of x values σ x is NOT equal to the standard deviation of the
distribution of sample means σ x
σ
We will make the claim (without proof at this time) that σ x = x
n
Notice that the distribution of the sample means X is normal in this example
6 – 6 Central Limit Theorem Lecture
Page 3 of 10
© 2012 Eitel
Example 2
A Distribution of Sample Means of size n= 2 (with replacement)
for the Normal population of { 2, 4, 6, 8}
2
4
6
8
x:
A population
of 4 values
2+2
= x1 = 2
2
6+2
= x9 = 4
2
2+ 4
= x2 = 3
2
6+ 4
= x10 = 5
2
2+6
= x3 = 4
2
6+6
= x11 = 6
2
2+8
= x4 = 5
2
6+8
= x12 = 7
2
4+2
= x5 = 3
2
8+2
= x13 = 5
2
4+ 4
= x6 = 4
2
8+ 4
= x14 = 6
2
4+6
= x7 = 5
2
8+6
= x15 = 7
2
4+8
= x8 = 6
2
8+8
= x16 = 8
2
the mean of the x values = µ x
the SD of the x values = σ x
x
f( x )
2
1
Input the x data into the Calculator
3
2
6
µx = 5
4
3
8
σ x = 2.236
5
4
6
3
7
2
ux = 5
8
1
σ x = 1.58
2
4
σ x 2.236
=
= 1.58
n
2
x9 = 4
x2 = 3
x 10 = 5
x3 = 4
x 11 = 6
x4 = 5
x 12 = 7
x5 = 3
x 13 = 5
x6 = 4
x 14 = 6
x7 = 5
x 15 = 7
x8 = 6
x 16 = 8
X
the distribution of all the
sample means of size 2
from a population of size 4
the calculation of the sample means
for every possible sample of 4 values
taken two at a time n = 2
(with replacement)
x
x1 = 2
sample size = n = 2
the mean of all the sample means = µ x
the SD of all the sample means = σ x
sample size = n = 2
input the X data into the calculator
We will make the claim (without proof ) that µ x = µx
We will make the claim (without proof ) that σ x =
σx
n
Notice that the distribution of the sample means X is normal
6 – 6 Central Limit Theorem Lecture
Page 4 of 10
© 2012 Eitel
X population
of x values
x
Example 3
All the Samples of
size n = 3 (with Replacement)
samp
samp
0
0 ,0 ,0
6, 6, 0
0/ 3 = 0
12/ 3 = 4
3
0, 0, 3
6, 6, 3
3/ 3 = 1
15/ 3 = 5
6
0, 0, 6
6, 6, 6
6/ 3 = 2
18/ 3 = 6
9
0, 0, 9
6, 6 ,9
9/ 3 = 3
21/ 3 = 7
0, 3, 0
6, 0, 0
3/ 3 = 1
6/3 = 2
0, 3, 3
6, 0, 3
6/ 3 = 2
9/ 3 = 3
0, 3, 6
6, 0, 6
9/ 3 = 3
12/ 3 = 4
0, 3, 9
6, 0, 9
12/ 3 = 4
15/ 3 = 5
0, 6, 0
6, 3, 0
6/ 3 = 2
9/ 3 = 3
0, 6, 3
6, 3, 3
9/3=3
12/ 3 = 4
0, 6, 6
6, 3, 6
12/ 3 = 4
15/ 3 = 5
0, 6, 9
6, 3, 9
15/ 3 = 5
18/ 3 = 6
0, 9, 0
6, 9, 0
9/ 3 = 3
15/ 3 = 5
0, 9, 3
6, 9, 3
12/ 3 = 4
18/ 3 = 6
0, 9, 6
6, 9, 6
15/ 3 = 5
21/ 3 = 7
0, 9, 9
6, 9, 9
18/ 3 = 6
24 3 = 8
3, 3, 0
9, 9, 0
6/ 3 = 2
18/ 3 = 6
3, 3, 3
9, 9, 3
9/ 3 = 3
21/ 3 = 7
3, 3, 6
9, 9, 6
12/ 3 = 4
24/ 3 = 8
3, 3, 9
9, 9, 9
15/ 3 = 5
27/ 3 = 9
3, 0, 0
9, 0, 0
3/ 3 = 1
9/3 = 3
3, 0, 3
9, 0, 3
6/ 3 = 2
12/3 = 4
3, 0, 6
9, 0, 6
9/ 3 = 3
15/ 3 = 5
3, 0, 9
9, 0, 9
12/ 3 = 4
18/ 3 = 6
3, 6, 0
9, 3, 0
9/ 3 = 3
12/ 3 =4
3, 6, 3
9, 3, 3
12/ 3 = 4
15/ 3 = 5
3, 6, 6
9, 3, 6
15/ 3 = 5
18/ 3 = 6
3, 6, 9
9, 3, 9
18/ 3 = 6
21/ 3 = 7
3, 9, 0
9, 6, 0
12/ 3 = 4
15/ 3 = 5
3, 9, 3
9, 6, 3
15/ 3 = 5
18/ 3 = 6
3, 9, 6
9, 6, 6
18/ 3 = 6
21/ 3 = 7
3, 9, 9
9, 6, 9
21/ 3 = 7
24/ 3 = 8
6 – 6 Central Limit Theorem Lecture
xn
Page 5 of 10
X All the Sample Means of
size n = 3 (with Replacement)
xn
© 2012 Eitel
Example 3 Continued
A Distribution of Sample Means of size n= 3 (with replacement)
for the Normal population of { 0, 3, 6, 9}
The Distribution of the
population of x values
The Distribution of all the possible Sample
Means X of size n = 3 (with Replacement)
the mean of the x values = µ x
the SD of the x values = σ x
x
f( x )
0
1
Input the x data into the Calculator
1
3
6
µ x = 4.5
2
6
3
10
input the X data into the calculator
9
σ x = 3.354
4
12
ux = 4.5
5
12
6
10
σ x = 1.94
7
6
8
3
9
1
x
0
3
σ x 3.354
=
= 1.94
n
3
sample size = n = 3
the mean of all the sample means = µ x
the SD of all the sample means = σ x
sample size = n = 3
The mean of the population of x values µ x
is equal
to the mean of the distribution of sample means µ x
µ x = µx = 4.5
We will make the claim (without proof at this time) that µ x = µx
The SD of the population of x values σ x
is equal to the SD of the distribution of sample means σ x
σx =
σx
= 1.94
n
We will make the claim (without proof at this time) that σ x =
σx
n
Notice that the distribution of the sample means X is normal
6 – 6 Central Limit Theorem Lecture
Page 6 of 10
© 2012 Eitel
Extra Example 1
A Distribution of Sample Means of size n= 2 (without replacement)
for the Normal population of { 3, 5, 7, 9}
3
5
7
9
x:
A population
of 4 values
3+ 5
= x1 = 4
2
7+3
= x7 = 5
2
3+ 7
= x2 = 5
2
7+5
= x8 = 6
2
3+ 9
= x3 = 6
2
7+9
= x9 = 8
2
5+ 3
= x4 = 4
2
9+ 3
= x10 = 6
2
5+ 7
= x5 = 6
2
9+5
= x11 = 7
2
9+ 7
5+9
= x 12 = 8
= x6 = 7
2
2
the calculation of the sample means
for every possible sample of 4 values
taken two at a time n = 2
(without replacement)
x1 = 4
x7 = 5
x2 = 5
x8 = 6
x3 = 6
x9 = 8
x4 = 4
x 10 = 6
x5 = 6
x 11 = 7
x6 = 7
x 12 = 8
X
the distribution of all the
sample means of size 2
from a population of size 4
the mean of the x values = µ x
the SD of the x values = σ x
x
f( x )
4
2
Input the x data into the Calculator
5
2
7
µx = 6
6
4
7
2
input the X data into the calculator
9
σ x = 2.236
8
2
ux = 6
x
3
5
σ x 2.236
=
= 1.58
n
2
sample size = n = 2
the mean of all the sample means = µ x
the SD of all the sample means = σ x
sample size = n = 2
σ x = 1.58
The mean of the population of x values µ x is equal to the mean of the distribution of sample means
µ x We will make the claim (without proof at this time) that µ x = µx
The standard deviation of the population of x values σ x is NOT equal to the standard deviation of the
distribution of sample means σ x
σ
We will make the claim (without proof at this time) that σ x = x
n
Notice that the distribution of the sample means X is normal in this example
6 – 6 Central Limit Theorem Lecture
Page 7 of 10
© 2012 Eitel
Extra Example Example 2
A Distribution of Sample Means of size n= 2 (with replacement)
for the population of { 2, 4, 8)
2
4
8
x:
A population
of 3 values
2+2
= x1 = 2
2
4+8
= x6 = 6
2
2+4
= x2 = 3
2
8+2
= x7 = 5
2
2+8
= x3 = 5
2
8+4
= x8 = 6
2
4+2
= x4 = 3
2
8+8
= x9 = 8
2
x6 = 6
x2 = 3
x7 = 5
x3 = 5
x8 = 6
x4 = 3
x9 = 8
x5 = 4
4+4
= x5 = 4
2
the calculation of the sample means
for every possible sample of 3 values
taken two at a time n = 2
(with replacement)
The Distribution of the population
of x values
2
4
Input the x data into the Calculator
8
µ x = 4.6667
4
1
σ x = 2.4944
5
2
6
2
8
1
σ x 2.4944
=
= 1.763
n
2
X
the distribution of all the
sample means of size 2
from a population of size 3
The Distribution of all the possible Sample
Means X of size n = 2 (with Replacement)
x freq. the mean of all the sample means = µ x
the SD of all the sample means = σ x
2
1
input the X data into the calculator
3
2
the mean of the x values = µ x
the SD of the x values = σ x
x
x1 = 2
ux = 4.6667
σ x = 1.7638
The mean of the population of x values µ x is equal to the mean of the distribution of sample means
µ x We will make the claim (without proof at this time) that µ x = µx
The standard deviation of the population of x values σ x is NOT equal to the standard deviation of the
distribution of sample means σ x
σ
We will make the claim (without proof at this time) that σ x = x
n
Notice that the distribution of the sample means X is NOT normal in this example
6 – 6 Central Limit Theorem Lecture
Page 8 of 10
© 2012 Eitel
Extra Example 2B
A Distribution of Sample Means of size n= 2 (without replacement)
for the uniform population of { 2, 4, 8}
2
4
8
x:
A population
of 3 values
2+4
= x1 = 3
2
4+8
= x4 = 6
2
2+8
= x2 = 5
2
8+2
= x5 = 5
2
4+2
= x3 = 3
2
8+4
= x6 = 6
2
the calculation of the sample means
for every possible sample of 3 values
taken two at a time n = 2
(without replacement)
x
f( x )
2
the mean of the x values = µ x
the SD of the x values = σ x
3
2
4
Input the x data into the Calculator
5
2
8
µ x = 4.6667
6
2
x
x1 = 3
x4 = 6
x2 = 5
x5 = 5
x3 = 3
x6 = 6
X
the distribution of all the
sample means of size 2
from a population of size 3
sample size = n = 2
the mean of all the sample means = µ x
the SD of all the sample means = σ x
sample size = n = 2
input the X data into the calculator
σ x = 2.4944
ux = 4.6667
σ x 2.4944
=
= 1.763
n
2
σ x = 1.2472
The mean of the population of x values µ x is equal to the mean of the distribution of sample means
µ x We will make the claim (without proof at this time) that µ x = µx
The standard deviation of the population of x values σ x is NOT equal to the standard deviation of the
distribution of sample means σ x . Also notice that σ x =
σx
is NOT true as is has been in past
n
examples. The original population was NOT normal and the samples were created WITHOUT
replacement.
6 – 6 Central Limit Theorem Lecture
Page 9 of 10
© 2012 Eitel
Central Limit Theorem:
Every possible sample of size n is taken from a population of values x 1 , x 2 , x 3 ....
and the average of each sample is recorded. This creates a population of sample means.
This population is called the Distribution of Sample Means and is labeled X
The Distribution of Sample Means X is Normal
if the distribution of x values is normal or n > 30
σ
µ x = µx and σ x = x
n
We will make the claim (without proof at this time) that µ x = µx is true for all populations of x that are
sampled for any sample size n.
The standard deviation of the Distribution of Sample Means σ x
is not equal to the standard deviation of the original distribution of x data σ x
σx
is true for all populations
n
of x that are sampled (with replacement) for any sample size n. We will also make the claim
σ
(without proof at this time) that σ x = x is true for all populations of x that are normal or the sample
n
size is greater than 30.
However: We will make the claim (without proof at this time) that σ x =
6 – 6 Central Limit Theorem Lecture
Page 10 of 10
© 2012 Eitel