Sampling Distribution
If We have information on all member of population and we use this
information to prepare a probability distribution
The properties of the population:
N→ Number of population
μ→ Mean
ẟ→ Standard Deviation
The properties of the Sampling Distribution:
n→ Number of Sampling
𝜇𝑥̅ → 𝑀𝑒𝑎𝑛 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑖𝑛𝑔𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
ẟ𝑥̅ → 𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓𝑠𝑎𝑚𝑝𝑙𝑖𝑛𝑔𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
𝑛
𝑁
𝑛
> 0.05
𝑁
𝑛
≤ 0.05
𝑁
μ𝑥̅ =μ
μ𝑥̅ = μ
ẟ
ẟ𝑥̅ =
𝑛
ẟ𝑥̅ =
ẟ 𝑁−𝑛
𝑛 𝑁−1
1. A population consists of the following four values: 12, 12, 14, and 16.
a. List all samples of size 2, and compute the mean of each sample.
b. Compute the mean of the distribution of the sample mean and the
population mean.
c. Compare the dispersion in the population with that of the sample
mean.
12 , 12 , 14 , 16
↓
↓
↓
↓
A , B , C , D
sample
A,B
A,C
A,D
B,C
B,D
C,D
score
12,12
12,14
12,16
12,14
12,16
14,16
𝜇=
12+12+14+16
4
ẟ𝑥̅ = √
𝑛=2
̅−μ
𝒙
-1.5
-0.5
0.5
-0.5
0.5
1.5
Mean
12
13
14
13
14
15
N=4
ẟ𝑥̅ =
=13.5
(𝑥̅ − 𝜇)2
= 0.95
𝑛
μ=13.5
ẟ√𝑁 − 𝑛
√𝑛√𝑁 − 1
𝑛
𝑁
> 0.05
= 0.95
̅ − 𝝁)𝟐
(𝒙
2.25
0.25
0.25
0.25
0.25
2.25
5.5
The mean wage per hour for all 5000 employees who work at a large
company is $27.50 and the standard deviation is $3.70 . Let𝑥̅ be the
mean wage per hour for a random sample of certain employees selected
from this company . Find the mean and standard deviation of𝑥̅ for a
sample size of
(a) 30
(b) 200 .
μ=27.50
N=5000
(a)
𝑛
n=30
𝑁
=
μ=27.50
𝜇𝑥̅ = μ
(b)
ẟ=3.7
𝑛
n=200
𝑁
𝜇𝑥̅ = μ = 27.5
30
5000
= 6 × 10−3 < 0.05
ẟ𝑥̅ =
=
200
5000
ẟ
√𝑛
=
3.7
√30
= 0.67
= 0.04 < 0.05
ẟ𝑥̅ =
ẟ
√𝑛
=
3.7
√200
= 0.2616
A normal population has a mean of 60 and a standard deviation of 12.
You select a random sample of 9. Compute the probability the sample
mean is:
a. Greater than 63.
b. Less than 56.
c. Between 56 and 63.
(a). μ=60
ẟ=12
n=9
𝑛
< 0.05
𝑁
P(x> 63) →
𝑧=
𝑥−𝜇𝑥̅
ẟ𝑥̅
ẟ
ẟ𝑥̅ =
P(z>
√𝑛
=4
63−60
4
) = 𝑃(𝑧 > 0.75)
=1-P(z< 0.75) = 1 − 0.7734 = 0.2266 (form table p.p 126)
(B) P(x< 56) →
𝑃(𝑧 <
𝑃(𝑧 < −1) = 0.1587
56−60
4
)
(form table p.p126)
( c) 𝑃(56 < 𝑥 < 63) = 𝑃(−1 < 𝑧 < 0.75)
𝑝(𝑧 < 0.75) − 𝑃(𝑧 < −1) = 0.7734 − 0.1587 = 0.6156
μ=P
ẟ= pq
when
𝑛
𝑁
> 0.05
𝑛
𝑁
≤ 0.05
ẟ𝑥̅ =
ẟ𝑥̅ =
𝑝𝑞
√𝑛
𝑝𝑞√𝑁−𝑛
√𝑛√𝑁−1
Suppose that 75% of adults said that college education has become too
expensive for most people. Let Ộ be the proportion in a random sample
of 1400 adult Americans who will hold the said opinion. Find the
probability that 76.5 % to 78% of adults in this sample will hold this
opinion.
P=0.75
q=1-0.75=0.25
𝜇 = 𝑃 = 0.75
ẟ𝑥̅ =
𝑝𝑞
√𝑛
=
n=1400
0.75 × 0.25
√1400
= 5.011 × 10−3
𝑝(0.765 < 𝑥 < 0.78)
np=1400× 0.75 = 1050
𝑛𝑝 > 5
𝑛𝑞 = 1400 × 0.25 = 350
𝑛𝑞 > 5
0.765 − 0.75
0.78 − 075
𝑃(
<
𝑧
<
)
5.011 × 10−3
5.011 × 10−3
P(2.99< 𝑧 < 5.98) = 𝑃(𝑧 < 5.98) − 𝑃(𝑧 > 2.99)
= 0.9999 - 0.9986 = 0.0013