Topic 6: The distribution of
the sample mean and
linear combinations of
random variables
CEE 11 Spring 2002
Dr. Amelia Regan
These notes draw liberally from the class text, Probability and Statistics for
Engineering and the Sciences by Jay L. Devore, Duxbury 1995 (4th edition)
The distribution of the
sample mean
Let X1 , X 2 , X 3 ,..., Xn
be a random sample from a distribution with mean
value  and standard deviation  . Then
E(X)= x  
V (X)= x2   2 / n
x  / n
In addition, the sample Total To  X1  X 2  X 3  ...X n
E(To )=n
V (To )=n x2
 T  n
o
The Case of a Normal
Population Distribution
Let X1 , X 2 , X 3 ,..., Xn
be a random sample from a normal distribution with mean
value  and variance  2 . Then for any n,
X is normally distributed with  x  
and  x2   2 / n and To is also normally distributed
with To =n and  T2o  n 2 .
The Central Limit Theorem
Let X1 , X 2 , X 3 ,..., Xn
be a random sample from a distribution with mean
value  and variance  2 . Then if n is sufficiently large
X has approximately a normal distribution with  x  
and  x2   2 / n and To also has approximately a normal distribution
with To =n and  T2o  n 2 .
The central limit theorem

The larger value of n, the better the approximation.

What are the implications? Significant!


It means that we can use the normal distribution to
describe and analyze data from distributions about
which we know only the mean and the variance.
In n > 30 we can invoke the central limit theorem
Example



The nicotine content in a single cigarette of a particular brand is
a random variable with mean  = 0.8 mg and  = 0.10 mg.
If an individual smokes 5 packs of these cigarettes per week,
what is the probability that the total amount of nicotine consumed
in a week is at least 82 mg?
No matter what the underlying distribution, the total amount of
nicotine consumed is approximately normally distributed with
  (0.80)(20)(5)  80.0
  100(0.10)  1.0
Example

Therefore the probability that the total is more than 82
is using
z
x

82  80
P( x  82)  P( z 
)
1.0
P( z  2)  1  p( z  2)  0.0228
The distribution of a linear
combination
The sample mean and sample total are
special cases of a type of random variable
that arises very frequently in statistical
applications.
Given a collection of n rv's X1 , X 2 , X 3 ,..., X n
and n numerical constants a1 , a2 , a3 ,..., an

n
The rv Y  a1 X 1  a2 X 2  a3 X 3 ...  ...an X n   an X n
i 1

The random variable Y is called a linear
combination of random variables
The distribution of a linear
combination
Given a collection of n rv's X1 , X 2 , X 3 ,..., X n
and n numerical constants a1 , a2 , a3 ,..., an
n
The rv Y  a1 X 1  a2 X 2  a3 X 3 ...  ...an X n   an X n
i 1


The special case where all the coefficients are equal
to 1 gives us the sample total
The special case in which all of the coefficients are
equal to 1/n gives us the sample average
The distribution of a linear
combination
Let X 1 , X 2 , X 3 ,..., X n have mean values 1 , 2 , 3 ,..., n and
variances  12 ,  22 ,  32 ,... n2 respectively

Then – whether or not the random variables (X’s) are
independent
E  a1 X 1  a2 X 2  a3 X 3 ...  ...an X n  
a1E  X 1   a2 E  X 2   a3 E  X 3  ,...an E  X n 
The distribution of a linear
combination
Let X 1 , X 2 , X 3 ,..., X n have mean values 1 , 2 , 3 ,..., n and
variances  12 ,  22 ,  32 ,... n2 respectively

Further – if the random variables are independent
then
V  a1 X 1  a2 X 2  a3 X 3 ...  ...an X n  
a12V  X 1   a22V  X 2   a32V  X 3  ,...an2V  X n 