Download 5-1 Random Variables and Probability Distributions

7-1 and 7-2
Sampling Distribution
Central Limit Theorem
Let’s construct a sampling distribution (with
replacement) of size 2 from the sample set
{1, 2, 3, 4, 5, 6}
1, 1
2, 1
3, 1
4, 1
5, 1
6, 1
1, 2
2, 2
3, 2
4, 2
5, 2
6, 2
1, 3
2, 3
3, 3
4, 3
5, 3
6, 3
1, 4
2, 4
3, 4
4, 4
5, 4
6, 4
1, 5
2, 5
3, 5
4, 5
5, 5
6, 5
1, 6
2, 6
3, 6
4, 6
5, 6
6, 6
Mean = 
Probability
1
1/36
1.5
2/36
2
3/36
2.5
4/36
3
5/36
3.5
6/36
4
5/36
4.5
4/36
6
3/36
5.5
2/36
6
1/36
Theorem 7-1
Some variable x has a normal distribution with
mean = μ and standard deviation = σ
For a corresponding random sample of size n
from the x distribution
- the  distribution will be normal,
- the mean of the  distribution is μ
- the standard deviation is σ
n
What does this mean?
If you have a population and have the
luxury of measuring a lot of sample
means, those means (called xbar) will
have a normal distribution and those
means have a mean (i.e. average
value) of mu.
For the sample size 2
What is the mean of {1, 2, 3, 4, 5, 6}?
What appear to be the mean of the
distribution of 2 out of 6?
Theorem 7-1 (Formula)
μx = μ
σx 
σ
n
x  μx
x μ
z

σ
σx
n
Why doesn’t the SD
stay the same?
Because the
sample size is
smaller… you will
see a smaller
deviation than you
would expect for
the whole
population
Central Limit Theory
Allows us to deal with not knowing about
original x distribution
(Central = fundamental)
The Mean of a random sample has a sampling
distribution whose shape can be
approximated y the Normal Model as the
value of n increases.
Larger Sample = Bigger Approximation
The standard is that n ≥ 30.
Example
Coal is carried from a mine in West
Virginia to a power plant in NY in
hopper cars on a long train. The
automatic hoper car loader is set to
put 75 tons in each car. The actual
weights of coal loaded into each car
are normally distributed with μ = 75
tons and σ = 0.8 tons.
What is the probability that one car chosen at
random will have less than 74.5 tons of coal?
This is a basic probability – last chapter
P(r  74.5)
74.5  75
z
 .625
.8
P(z  .625)
 .2657
What is the probability that 20 cars chosen at
random will have a mean load weight of less
than 74.5 tons?
The question here is that the sample of
20 cars will have  (xbar) ≤ 74.5
x  74.5
σx 
.8
μx =75
 .1789
20
74.5  75
z
 2.795
.1789
P(z  2.795)  .0024
Another Example
Invesco High Yield is a mutual fund that
specializes in high yield bonds. It has
approximately 80 or more bonds at the
B or below rating (junk bonds). Let x
be a random variable that represents
the annual percentage return for the
Invesco High Yield Fund. Based on
information, x has a mean μ = 10.8%
and σ = 4.9%
Why would it be reasonable to assume
that x (the average annual return of all
bonds in the fund) has a distribution
that is approximately normal?
80 is large enough for the Central Limit
Theorem to apply
Compute the probability that
after 5 years  is less than 6%
(Would that seem to indicate that μ is
less than 10.8% and that the junk bond
market is not strong?)
x  .06
σx 
.049
μx =.108
 .0219
5
.06  .108
z
 2.19
.0219
P(z  2.19)  .0140
N = 5 because we are looking over 5
years
Yes. The probability that it is less
than 6% is approx. 1%. If it is
actually returning only 6%, then it
looks like the market is weak.
Compute the probability that after 5
years  is greater than 16%
x  .16
σx 
.049
μx =.108
 .0219
5
.16  .108
z
 2.37
.0219
P(z  2.37)  1 P(z  2.37)  .0087
Note
The Normal model applies to quantitative
data…