Let’s construct a sampling distribution (with replacement) of size 2 from the sample set
{1, 2, 3, 4, 5, 6}
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6
2, 1 2, 2 2, 3 2, 4 2, 5 2, 6
3, 1 3, 2 3, 3 3, 4 3, 5 3, 6
4, 1 4, 2 4, 3 4, 4 4, 5 4, 6
5, 1 5, 2 5, 3 5, 4 5, 5 5, 6
6, 1 6, 2 6, 3 6, 4 6, 5 6, 6

Mean = 
1
1.5
2
4
4.5
6
2.5
3
3.5
5.5
6
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36

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

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.

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?

σ x z 

σ n

σ x

σ
 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

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.

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 z 
.8
P(z   .625)
 .2657
  .625

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
σ x

.8
20
 .1789
z 
.1789
 
P(z   
2.795

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


(Would that seem to indicate that μ is less than 10.8% and that the junk bond market is not strong?)
σ x

.049
5
 .0219
z 
.0219
P(z   
  2.19
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%
σ x

.049
5
 .0219
z 
.0219
 2.37

The Normal model applies to quantitative data…