Lesson 6 : Distribution of the Sample Mean
Satya Mandal
Due Date:
See the Lecture Notes Site.
6.1 Introduction
The goal of this lesson is to establish the foundation for estimation of parameters that we will do in the next two lesson. This lesson can be viewed as a bridge between probability (the math) and estimation (statistics). We will mainly do the estimation of the population mean μ.
In this lesson, our final theorem would be that the sample mean
X = (X_{1}+X_{2}+ ... +X_{n})/n
has normal distribution, approximately.
Example.
Suppose we want to study the height
distribution of the U.S. population.
Let X represent the height of the whole
US population.
We may collect a data of size n = 713 as follows:
71 
62 
67 
73 
61 
58 
63 
58 
69 
68 
55 
57 
51 
57 
49 
63 
63 
64 
72 
59 
67 
59 
57 
69 
55 
56 
65 
66 
53 
53 
51 
66 
68 
71 
61 
63 
53 
76 
77 
67 
And so on
Our point of view is that the height
x_{1}, x_{2}, x_{3}, ... , x_{n}
(in our case 71, 62, 67, ... )
are, in fact, the observed values of random variables
X_{1}, X_{2}, X_{3}, ... , X_{n},
respectively.
Here X_{1} is the notation for height of the first member
of the sample, which could be the height of anybody from the whole
U.S. population, and in the case of our sample
the value of X_{1} is 71. Similarly, X_{2} is the
notation for height of the second member of the sample, which could
be the height of anybody from the whole U.S. population, and in the
case of our sample the value of X_{2}is
62.
In general, we collect sample of size n. The sample would be n numbers
x_{1}, x_{2}, …, x_{n}
representing the height of n individuals.
We shall consider the height x_{i} of the
i^{th} individual
as the observed value
of a random variable X_{i}.
Here X_{i} is the
notation for the height of the i^{th} member of the sample, which
could be the height of anybody from population.
Therefore, these n
measurements
x_{1}, x_{2}, …, x_{n}
are, respectively, the observed values of n random variables
X_{1}, X_{2}, …, X_{n}.
We (re)define the sample mean X
as the random
variable
X

= 
X_{1}+X_{2}+…+X_{n}
n

. 
So, the sample means that
we computed before in Lesson 2 are the values of the random variable
X.
We will consider the
sampling distribution of
the sample mean X.
Sampling Types
There are many ways to do sampling. Most commonly discussed
among them are
 Sampling without replacement,
 Sampling with replacement.
The Sampling without replacement is the
type of sampling where,
whenever a sample member is selected, the member is
excluded from the subsequent selections. It is analogous to
selecting n balls from a box of N balls.
Balls are selected one by one,
without replacing them back in the box before subsequent selections.
This type of sampling is meant to rule out the possibility of selecting
a member more than once. For small populations, possibility of selecting a
memeber twice may be significant. For such small
populations, sampling without replacement
would be appropriate.
The Sampling with replacement is the
type of sampling where each selection is done without any regard to previous selections.
In other words, each time a sample member is drawn,
it is placed back to the whole population
before the next selection is made. This way,
each selection is done from the same whole population.
A member could, therefore, be selected more than once. This
may seem unnatural.
But when working with large populations this
is not likely to happen and is most natural from
the statistical point of view. (How often would one receive
calls
twice for the same poll?)
We will only consider sampling with
replacement .
Properties
Let X (like height) be a random variable
with mean μ and
variance σ^{2}.
Let X_{1}, X_{2}, … , X_{n}
be a sample from the X  population.
We assume that the
sampling was done with replacement.
Thw following are comments and properties of such a sample:

X would be called the parent population
or the population random
variable. Also μ and σ^{2}
are called the population mean and variance.
 Each of the sample member X_{i}
has the same distribution
as X. So, mean of X_{i} is μ
and variance of X_{i} is σ^{2}.
 The sample members X_{1},X_{2}, …, X_{n}
are all mutually independent.
(In fact, one has to ensure that they
are drawn independently. This is a very difficult task in practice, because
of lack of accessibility
of the population members, human bias and other
reasons.)
 The distribution of X is called the sampling
distribution of X.
 Theorem. The mean of the sample mean
X is the population mean
μ,
that is
Mean(X)
= Mean(X) = μ
The variance of the sample mean X is
given by
Var(X)
= σ^{2}/n
So, the standard deviation of X, denoted
by σ _{X},
is given by
σ_{X}
= σ/√n.
 Definition. The standard deviation σ_{X}
is also called standard error.
6.2 Central Limit Theorem
The following theorem describes the sampling distribution of the Sample Mean. It is called the Central Limit Theorem (CLT)
Theorem (CLT).
Suppose X_{1},X_{2}, …,X_{n} is a sample
from a population X with mean μ and
variance σ^{2}.
Assume n is large.
Then the sample mean X, approximately,
has a normal
distribution with
mean(X) = μ
and standard deviation
σ_{X}=
σ/√n.
Therefore, approximately,
P(a < X
<b) ≈ P(L < Z < R)
where L=(aμ)/σ
_{X} and R=(bμ)/σ
_{X}
OR
P(a < 
X

< b)
≈ P 

a μ
σ/√n

< Z < 
bμ
σ/√n


. 
Further,
if the parent population X is Normal, then the above equations are exact.
Standard Error and Precision
The standard error
σ
_{X}=
σ
/√n
decreases to zero, as the sample size n increases. Because of this,
while estimating the mean μ by the sample mean X,
by increasing the
sample size n,
the sample mean X can
simultaneously achieve precision and level of confidence
(i.e. probability of a give precision). The Animation 6.1.1
below demonstrates the same.
The following animation demonstration of the same.
Problem Solving: The Central Limit Theorem (CLT) would be used to compute approximate probability for the sample mean
X. This would be similar to normal approximation to Binomial (Lesson 5).
Problems on 6.2: Central Limit Theorem
Exercise 6.2.1. It is known that the tuition
paid per semester by students in a university has a distribution with
mean $2,050 and standard deviation $310. If 64 students are interviewed,
what is the approximate probability that the sample mean tuition paid
will be above $2,060?
Solution by TI84:
Here the population mean μ = 2,050 and the population standard deviation σ = 310. The sample size n = 64.
First step is to compute the mean
μ_{X} =μ = 2050
and the standard deviation
σ_{X}
=
σ/
√n =
310/√64=38.75
Let X = Tuition paid by the students.
Then, the distribution of X
is, approximately, N(2050, 38.75)
Now " X will be above 2060" means
" X > 2060".
P(2060 < X)
= P([2060  μ]/σ_{X}
<
[X  μ]/
σ_{X} )
≈
P([2060  μ]/σ
_{X} < Z )
= P([2060  2050]/38.75 < Z )
= P (.2580, < Z ) = normalcdf(.2580, 5)= .3982
Exercise 6.2.2.
The monthly water consumption X per household in a subdivision in Kansas
City has normal distribution with mean 15000 gallons and standard deviation
3000 gallons. What is the probability that the mean consumption of the
44 households in the subdivision will exceed 16000 gallons?
Solution by TI84:
Here the population mean μ = 15000 and the population standard deviation σ = 3000. The sample size n = 44.
First step is to compute the mean
μ_{X}
=μ = 15000.
and the standard deviation
σ_{X}
=
σ/
√n =
3000/√44=452.2670
Let X = monthly water consumption by the households.
Then, the distribution of X
is, approximately, N(15000, 452.2670)
Now " X will exceed 16000" means
" X > 16000".
P(16000 < X)
= P([16000  μ]/σ
_{X}
<
[X  μ]/
σ_{X} )
≈ P([16000  μ]/σ
_{X}
< Z )
= P([1600015000]/452.2670 < Z )
= P (2.2111, < Z ) = normalcdf(2.2111, 5)= .0135
Exercise 6.2.3.
In a class of more than thousand students, the instructor announced after a test that the mean score was μ = 77 point and standard deviation σ = 24 points. You took a sample of 81 students. What would be the approximate probability that the sample mean would be less than 80?
Solution by TI84:
Here the population mean μ = 77 and the population standard deviation σ = 24. The sample size n = 81.
First step is to compute the mean
μ_{X}
=μ = 77.
and the standard deviation
σ_{X}
=
σ/
√n =
24/√81= 2.6667
Let X = Points scored by students.
Then, the distribution of X
is, approximately, N(77, 2.6667)
Now "the sample mean would be less than 80" means
" X < 80".
P(X < 80)
= P([X  μ]/
σ_{X}
<
[80  μ]/σ
_{X})
≈ P(Z < [80  μ]/σ_{X}
< Z )
= P(Z < [80 77]/2.6667)
= P (Z < 1.1250) = normalcdf(5, 1.1250)= .8697
Exercise 6.2.4.
The mean salary X of the university professors in a state is μ = $65,000 and standard deviation
σ = $14,000. You collect a sample of 75 professors. What is the probability that sample mean salary of these 75 professors would be above $60,000.
Solution by TI84:
Here the population mean μ = 65000 and the population standard deviation σ = 14000. The sample size n = 75.
First step is to compute the mean
μ_{X}
=μ = 65000.
and the standard deviation
σ_{X}
=
σ/
√n =
14000/√75=1616.5808
Let X = monthly water consumption by the households.
Then, the distribution of X
is, approximately, N(65000, 1616.5808)
Now " X would be above 60,000" means
" X > 60000".
P(16000 < X)
= P([60000  μ]/σ_{X}
<
[X  μ]/
σ_{X} )
≈ P([16000  μ]/σ_{X} ; < Z )
= P([6000065000]/1616.5808 < Z )
= P (3.0929, < Z ) = normalcdf(3.0929, 5)= .9990
Exercise 6.2.5.The time X that a child spends watching TV on week ends has a normal distribution with mean μ = 330 minutes and standard deviation σ = 95 minutes. You sample 50 kids in a school. What is the probability that the sample time
X
that these kids watch TV on a weekend will be less than 300 minutes.
Exercise 6.2.6. The weight X of fish in a lake has mean μ = 12 pounds
and standard deviation σ = 4.5 pounds. Suppose you catch 150 fish.
What is the probability that the mean weight of of the catch
will be less than 1900/150 pounds?
Solution by TI84:
Here the population mean μ = 12 and the population standard deviation σ = 4.5. The sample size n = 150.
First step is to compute the mean
μ_{X}
=μ = 12.
and the standard deviation
σ_{X}
=
σ/
√n =
4.5/√150= .3674
Let X = Weight of the fish population.
Then, the distribution of X
is, approximately, normally distributed.
P(X < 1900/150)
= P(X < 12.6667)
= P([X  μ]/
σ_{X}
<
[12.6667  μ]/σ
_{X})
≈ P(Z < [12.6667  μ]/σ_{X})
= P(Z < [12.6667  12]/.3674)
= P (Z < 1.8146) = normalcdf(5, 1.8146)= .9652
Exercise 6.2.7.
The amount X of water used when a person takes a shower has a
mean μ = 30 gallons and standard deviation σ = 16 gallons.
Suppose 36 people take a shower in a swimming pool facility.
What is the probability that average of more than 25 gallons of
water will be used by these 36 people.
Solution by TI84:
Here the population mean μ = 30 and the population standard deviation σ = 16. The sample size n = 36.
First step is to compute the mean
μ_{X}
=μ = 30.
and the standard deviation
σ_{X}
=
σ/
√n =
16/√36=2.6667
Let X = water used when a person takes a shower.
Then, the distribution of X
is, approximately, normally distributed.
P(25 < X)
= P([25  μ]/σ_{X}
<
[X  μ]/
σ_{X} )
≈ P([25  μ]/σ
_{X} ; < Z )
= P([25  30]/2.6667 < Z )
= P (1.8750, < Z ) = normalcdf(1.8750, 5)= .9696
Exercise 6.2.8. The waiting time for the campus
bus has a mean μ= 7 minutes and the standard deviation σ = 2 minutes.
A student used the bus 120 times in a month.
What is the probability that the student would have waited more than
7.5 minutes, on an average, during the whole month?
Solution by TI84:
Here the population mean μ =7 and the population standard deviation σ = 2. The sample size n = 120.
First step is to compute the mean
μ_{X}
=μ =7.
and the standard deviation
σ_{X}
=
σ/
√n =
2/√120=
.1826
Let X = waiting time for the bus.
Then, the distribution of X
is, approximately, Normal.
P(7.5 < X)
= P([7.5  μ]/σ_{X}
<
[X  μ]/
σ_{X} )
≈ P([7.5  μ]/σ
_{X} ; < Z )
= P([7.5  7]/.1826 < Z )
= P (2.7382, < Z ) = normalcdf(2.7382, 5)= .0031
Exercise 6.2.9. According to some data, the
annual Kansas wheat export X has a mean 733 million dollars and standard
deviation 163 million dollars. What is the probability that over the
next 10 years mean wheat exports will exceed 804 million dollars?
Solution by TI84:
Here the population mean μ = 733 and the population standard deviation σ = 163. The sample size n = 10.
First step is to compute the mean
μ_{X}
=μ = 733.
and the standard deviation
σ_{X}
=
σ/
√n =
163/√10= 51.5451
Let X = Kansas wheat export annually.
Then, the distribution of X
is, approximately, Normal.
P(804 < X)
= P([804  μ]/σ_{X}
<
[X  μ]/
σ_{X} )
≈ P([804  μ]/σ < Z )
= P([804  733]/51.5451 < Z )
= P (1.3774, < Z ) = normalcdf(1.3774, 5)= .0842
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