1. DISTRIBUTION OF THE SAMPLE MEAN
sample from a distribution/population with mean μ
and standard deviation σ.
We know: can take different values for different
samples – sampling distribution.
FACT1. The mean and standard deviation for the distribution of are given by:
and
The mean of is the same as the population mean μ. So, is unbiased for μ.
The standard deviation of is times smaller than the population standard deviation.
Averages have smaller variability than single observations!

2. Law of Large Numbers Closer look at the standard deviation of :
As n=sample size increases, ; i.e. as n increases, spread of the sample mean decreases to zero.
What random variable has spread zero? Constant!
Conclusion- Law of Large Numbers: comes arbitrarily close to μ for large enough n.
Distribution with mean=10, st.dv.= 2/100^0.5=0.2
Distribution with mean=10, st.dv.= 2/10^0.5=0.63
Distribution with mean=10, st.dv.=2
n=10, n=100

3. DISTRIBUTION OF THE SAMPLE MEAN – NORMAL DATA
sample from a Normal distribution, N(μ, σ ).
From FACT 1, We know: and
FACT 2: If are from N(μ, σ ), then has a N(μ, σ/ √n ) distribution.
NOTE: Since is normally distributed, with μX = μ and , then we may standardize it:

4. SAMPLING DISTRIBUTION OF THE SAMPLE MEAN
EXAMPLE: Students in an university have a weight distribution that is known to be N(150, 20). Let X1, X2, …, X16 represent the weights of 16 randomly selected students from this university. If is the average weight for this sample, find P( > 160).
Solution: Since the sample came from a normal distribution, by Fact 2, the sample mean has a normal distribution as well.
~N(μ, σ/ √n )=N(150, 20/ √16)=N(150, 5). Thus,

5. EXAMPLE, CONTD. ~N(μ, σ/ √n )=N(150, 20/ √9)=N(150, 6.67). Also,
An elevator at this university has a capacity of 1500 pounds. What is the probability that 9 students who enter the elevator will have a safe ride, i.e. their total weight is less than 1,500 lb?
Solution: Again, by Fact 2, the sample mean has a normal distribution:
~N(μ, σ/ √n )=N(150, 20/ √9)=N(150, 6.67). Also,
P( Total weight < 1500)=P( <1500/9)=P( <166.67). So,

6. DISTRIBUTION OF THE SAMPLE MEAN, CONTD.
EXAMPLE. Suppose X is the score on a test and X~N(500, 100). Let X1, X2, …X16 be a sample of scores for 16 individuals and their average score. Find P( 550 < ≤ 600).
Solution: Since the data come from a normal distribution, by Fact 2, has a normal distribution with mean
Thus, P(550 < ≤ 600) =
= P(2 < Z ≤ 4) = P(Z ≤ 4) - P(Z ≤ 2 ) =
= 1 – =

7. The CENTRAL LIMIT THEOREM (CLT)
What if the data does not come from the normal distribution?
FACT 3. (CLT): If X1, X2, …Xn are any set of observations with mean μ
and standard deviation σ, their sample mean , has approximately
normal N(μ, σ/√n) distribution, if n is sufficiently large.
How large is sufficiently large? Depends on the distribution the data
comes from. Definitely n should be at least 20 before we use this
approximation.
Difference between Fact 2 and Fact 3: Fact 2 holds only for samples from
Normal distribution and gives exact distribution of
Fact 3 holds for samples from any distribution, but gives an approximate distribution for

8. The Central Limit Theorem contd.
Example. Suppose X1, X2, …, X25 are lifetimes of electronic components, with μ=700 hours and σ=10 hours. Find P( ≤ 702), where is the sample mean of the lifetimes of 25 components.
Solution. Usually lifetime data is skewed to the right, so not normal (Why?)
Since n=25 (reasonably large), we will use CLT and the normal approximation of the distribution of the sample mean:
has approx. a N(μ, σ/√n) = N(700, 10/√25) = N(700, 2) distr.
So,