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DISTRIBUTION OF THE SAMPLE MEAN
X1 , X 2 ,
, X n sample from a distribution/population with mean μ
and standard deviation σ.
1 n
X   Xi
n i 1

We know:
can take different values for different
samples – sampling distribution.

FACT1. The mean and standard deviation for the distribution of
are given by:

 X   and  X 
X
n
The mean of X is the same as the population mean μ. So,
unbiased for μ.
The standard deviation of
standard deviation.

X is
X is
n times smaller than the population
Averages have smaller variability than single observations!
Law of Large Numbers
Closer look at the standard deviation of
: X 
X

n
As n=sample size increases,  X
0; i.e. as n increases, spread
of the sample mean decreases to zero.
What random variable has spread zero? Constant!
X
comes arbitrarily close to μ
1.5
1.5
2.0
2.0
Conclusion- Law of Large Numbers:
for large enough n.
Distribution with mean=10,
st.dv.= 2/100^0.5=0.2
1.0
1.0
Distribution with mean=10,
st.dv.= 2/10^0.5=0.63
0.0
0.0
0.5
0.5
Distribution with
mean=10, st.dv.=2
5
10
5
15
n=10, n=100
10
15
DISTRIBUTION OF THE SAMPLE MEAN – NORMAL DATA
X1 , X 2 ,
, Xn
sample from a Normal distribution, N(μ, σ ).

From FACT 1, We know:

FACT 2: If X 1 , X 2 ,
distribution.
X  
and
X 

n
, X n are from N(μ, σ ), then X has a N(μ, σ/ √n )
NOTE: Since X is normally distributed, with μX = μ and
standardize it:
X 
X 
n( X  )
Z

.

/ n

n
, then we may
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 X is the average weight
for this sample, find P( X > 160).

Solution: Since the sample came from a normal distribution, by Fact 2, the
sample mean has a normal distribution as well.
X ~N(μ, σ/ √n )=N(150, 20/ √16)=N(150, 5). Thus,
P( X  160)  P(
X  150 160  150

)  P( Z  2)  1  P( Z  2)  1  0.9772  0.0228.
5
5
EXAMPLE, CONTD.

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:
X ~N(μ, σ/ √n )=N(150, 20/ √9)=N(150, 6.67). Also,
P( Total weight < 1500)=P(
X
<1500/9)=P(
X <166.67).
So,
X  150 166.67  150
P( X  166.67)  P(

)  P(Z  2.5)  0.9938.
6.67
6.67
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 X their average
score. Find P( 550 < X ≤ 600).
Solution: Since the data come from a normal distribution, by Fact 2, X has a
normal distribution with mean
 X    500 and  X   / n  100 / 16  25.
Thus, P(550 <
X ≤ 600) =
P(
550  500 X  500 600  500


)
25
25
25
= P(2 < Z ≤ 4) = P(Z ≤ 4) - P(Z ≤ 2 ) =
= 1 – 0.9772 = 0.0228.
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 X , 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 X .
Fact 3 holds for samples from any distribution, but gives an approximate
distribution for X .
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( X ≤ 702), where X 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:
X
So,
has approx. a N(μ, σ/√n) = N(700, 10/√25) = N(700, 2) distr.
X  700 702  700
P( X  702)  P(

)  P(Z  1)  0.8413.
2
2
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