SAMPLING
DISTRIBUTION
OF THE SAMPLE MEAN + PROBABILITIES BASED ON
NORMAL DISTRIBUTION
Consider a population consisting of 4 values
Population mean
𝜇=5
Now suppose we select all possible samples of size 2 with replacement Standard deviation
Population
mean:
from
this population
𝝈 = 10.25
2 2
2 +24 +
2 6 4+ 8
4
4
4
6
6
6
6
8
8
8
8
𝜇 2= 4 6 8 2 4 =6 5 8 2 4 6 8 2 4 6 8
4 3 4 5 6 4 5 6 7 5 6 7 8
2
3
4 0.13
𝑥
Population standard
This nowdeviation:
is the distribution of sample mean
𝝈=
𝟐+𝟓
𝟐
+ 𝟒+𝟓
𝟐
+ 𝟔+𝟓
𝟒
𝟐
+ 𝟖+𝟓
𝟐
= 10.25
𝒏=𝟐
𝜇𝑥 = 𝜇 = 5
Mean of sampling distribution of the sample mean will
always be equal to the population mean
𝜎𝑥 = 7.25 =
10.25
2
=
σ
n
Standard deviation of sampling distribution of the
sample mean ( sample error) will always be equal to
standard deviation divided by square root of n
Central Limit Theorem
As n increases, the sampling distribution of sample means approaches a normal distribution
• If Population is normal, then distribution of
sample mean is also normal.
• If population is not normal, the distribution of
sample means becomes approximately normal
when n is large (by large we often say n is at
least 30).
with
Standard Score
To standardize the mean of a sample
of size n, we replace the x by x-bar,
and change the standard deviation to
the standard error of the mean
Example
Given a normal population with a mean of 63 and a standard deviation of 12, find the probability that a
random sample of size 16 has a mean greater than 61.5
Because we are finding the probability of sample mean we can
apply Z- formula for sampling distribution of the mean
𝜎𝑥 =
12
16
=3
𝑧=
𝑥−𝜇
𝜎𝑥
=
61.5−63
= -0.5
3
µ = 63
σ = 12
n = 16
𝑥 > 61.5
P(x > 61.5) = P ( z > -0.5)
= 1-0.3085
= 0.6915