Unit 6: Mean and Variance of Sampling Distribution
Espinola, Kristine Marian C.
March 26, 2021
STEM1-ROY
:
1.
Solve the mean of each sampling distribution for the sample mean.
a.
Sample
Probability
6
7
8
9
0.20
0.25
0.20
0.35
Solution: μ = 6 + 7 + 8 + 9 / 5
μ = 30 / 5
μ=6
b.
Sample
Probability
22
24
26
28
30
0.15
0.10
0.18
0.35
0.22
Solution: μ = 22 + 24 + 26 + 28 + 30/ 5
μ = 130/ 5
μ = 26
c.
Sample
Probability
10.5
13.8
15.6
19.1
20.2
0.11
0.31
0.25
0.18
0.15
Solution: μ = 10.5 + 13.8 + 15.6 + 19.1 + 20.2 / 5
μ = 79.2 / 5
μ =15.8
2. For each population, find the mean of the sampling distribution of the sample mean
with an indicated sample size, n.
a.
7, 10, 19, 20; n =2
Solution:
Sample
Mean
7, 10
7, 19
7, 20
10, 19
10, 20
19, 20
8.5
13
13.5
14.5
15
19.5
Sample
Mean
Frequency
Probability
8.5
13
13.5
14.5
15
19.5
1
1
1
1
1
1
1/6
1/6
1/6
1/6
1/6
1/6
= 8.5 (1) + 13(1) + 13.5(1) + 14.5(1) + 15 (1) +19.5(1) / 6
= 14
b.
46, 51, 55, 59, 61; n = 3
Solution:
Sample
Mean
46, 51, 55
46, 51, 59
46, 51, 61
46, 55, 59
46, 55, 61
46, 59, 61
51, 55, 59
51, 55, 61
51, 59, 61
51, 59, 61
50.7
52
52.7
53.3
54
55.3
55
55.7
57
58.3
Sample Mean
Frequency
Probability
50.7
52
52.7
53.3
54
55.3
55
55.7
57
58.3
1
1
1
1
1
1
1
1
1
1
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
0.10
= 50.7(1) + 52(1) + 52.7(1) + 53.3(1) + 54(1) + 55.3(1) + 55(1) + 55.7(1) + 57(1) +
58.3(1) / 10
= 54.4
c.
24, 26, 29, 32, 35; n =4
Solution:
Sample
Mean
24, 26, 29, 32
24, 26, 29, 35
24, 26, 32, 35
24, 29, 32, 35
27.75
28.5
29.25
30
Sample
Mean
Frequency
Probability
27.75
28.5
29.25
30
1
1
1
1
0.20
0.20
0.20
0.20
= 27.75(1) + 28.5(1) + 29.25 (1) + 30(1) / 4
= 28.87 or 28.88
:
1. For each of the following population, solve for the variance of the sampling
distribution of the sample mean whose sample size is indicated.
a.
6, 8, 9, 12; n = 2
Solution:
2=
b.
2=
6.5
2.166666667 or 2.2
54, 58, 68, 82, 91; n = 3
Solution:
2=
N= 5
2=
197.44
65.81333333 or 65.81
35, 36, 39, 40, 49; n = 4
Solution:
2=
N= 5
2=
e.
2.345 or 2.35
N= 4
2=
d.
4.69
15, 19, 20, 22; n = 3
Solution:
c.
2=
N= 4
24.56
6.14
65, 69, 74, 78, 79, 90; n = 4
Solution:
N= 6
2=
63.81
2=
15.95
2. The variance of a population is 3.26. If a sampling distribution of the sample mean is
constructed from the population with a sample size of 3, what is the variance of the
sampling distribution of the sample mean?
Cinema
1
2
3
4
5
Number of
people
99
95
112
130
150
SOLUTION:
2=
3.26
2=
n= 3
2/
n
3.26/ 3
2 = 1.086666667 or 1.09
2=
:
1. For each of the following population, solve for the standard error of the sampling
distribution of the sample mean whose sample size is indicated.
a.
5, 9, 11, 14; n = 2
Solution:
N= 4
2=
10.69
√
2=
√5.35
2=
13
√
2=
√4.33
197.4
√
2=
√6.14
√
2=
√16.64
√
2=
√5.62
= 2.31
b.
16, 20, 22, 26; n = 3
Solution:
N= 4
= 2.08
c.
54, 58, 68, 82, 91; n = 3
Solution:
N= 5
2=
= 2.4779 or 2.48
d.
38, 39, 41, 45, 49; n =4
Solution:
N= 5
2=
197.4
= 4.0792 or 4.07
e.
105, 109, 110, 112, 115, 120; n = 4
Solution:
N= 6
= 2.37
2=
22.47
2. The variance of a population is 5.18. If a sampling distribution of the sample mean is
constructed from the population with a sample size of 3, what is the standard error of
the sampling distribution of the sample mean?
Solution:
2=
2=
5.18 n= 3
2/
n
= 5.18/ 3
√1.73
=
= 1.32
3. The number of people in five ice cream stores of an ice cream company that wanted
to try the new ice cream flavor are reported as follows:
Stores
Number of
people
Solution:
2=
13876- (260)2 / 5 / 5
1
49
2
55
3
65
4
52
5
39
n= 5
= 13876- 13520/ 5
= 71.2/ 3
=
√23.73
= 4.87
If three stores are selected at random and a sampling distribution of the sample mean
for the number of people that wanted to try the new ice cream flavor is constructed,
what is the standard error of the sampling distribution?
4.87 is the standard error of the sampling distribution.
Unit 7: The Central Limit Theorem
1. Solve the following.
a. Given a normally distributed set of data with μ =32 and σ =10, what is the mean and
standard deviation of the sampling distribution of the sample mean with a sample size of
25?
Solution:
μ =32
σ =10
n= 25
= σ/ √ n
= 10/ √ 25
= 10/ 5
μ x̅ = 32
=2
b. Given a normally distributed set of data with μ = 50 and σ = 5, what is the mean and
standard deviation of the sampling distribution of the sample mean with a sample size of
4?
Solution:
μ =50
σ =5
n= 4
= σ/ √ n
= 5/ √ 4
= 5/ 2
= 2.5
μ x̅ = 50
c. Given a normally distributed set of data with μ = 60 and 𝜎2 = 12, what is the mean
and standard deviation of the sampling distribution of the sample mean with a sample
size of 4?
μ =60
Solution:
√
σ =12
n= 4
= σ/ n
= 12/ 4
= √3
= 1.73
μ x̅ = 60
2. Determine the population mean and population standard deviation given the
following information about the sampling distribution of the sample mean of a normally
distributed population.
a.
=
Solution:
= 10, n = 90
=
/ √ n = 10
/ √ 90 = 10
= (10) (9.49)
= 105
= 94.9
/ 9.49
b.
=
Solution:
= 0.1, n = 4
=
/ √ n = 0.1
/ √ 4 = 0.1
= (0.1) (2)
=1
= 0.2
/2
c.
Solution:
=
= 2, n = 50
=
/√n=2
/ √ 50 = 2
/ 7.07
= (2) (7.07)
= 14.14
= 20
3. Given a normally distributed population with standard deviation
, determine
the sample size such that the sampling distribution of the sample mean has a standard
deviation
= 9.
Solution:
2=
2/
2=
( )2
= (9) 2 = 81
2 = ( ) 2
= (135) 2 = 18225
n= 81
= 18225/ n
n = 225
1. Solve the following.
a. Given a normally distributed set of data with mean μ = 10 and a sample standard
deviation 𝑠 = 4, what is the mean and standard deviation of the sampling distribution of
the sample mean with a sample size of 36?
Solution: μ = 10
= 𝑠/ √ n
= 4/ √ 36
= 4/ 6 = 2/3
= 0.66
𝑠=4
n= 36
= 10
b. Given a normally distributed set of data with mean μ = 50 and a sample standard
deviation 𝑠 = 20, what is the mean and standard deviation of the sampling distribution of
the sample mean with a sample size of 49?
Solution: μ = 50
= 𝑠/ √ n
= 20/ √ 49
= 20/ 7
= 2.86
𝑠 = 20
= 50
n= 49
c. Given a normally distributed set of data with mean μ = 45 and a sample standard
deviation 𝑠 = 50, what is the mean and standard deviation of the sampling distribution of
the sample mean with a sample size of 121?
Solution: μ = 45
= 𝑠/ √ n
= 50/ √ 121
= 50/ 11
= 4.55
𝑠 = 50
n= 121
= 45
2. Determine the population mean and sample standard deviation used given the
following information about the sampling distribution of the sample mean of a normally
distributed population whose population standard deviation is unknown.
105,
a.
= 10,
SOLUTION:
= 𝑠/ √ n= 10
= 𝑠 / √ 90
1,
b.
𝑠= (10) (9.49)
𝑠= 94.9
= 105
= 0.1,
SOLUTION:
= 𝑠/ √ n= 0.1
= 𝑠 / √ 4 = 0.1
=𝑠/2
𝑠= (0.1) (2)
𝑠= (0.2)
=1
,
c.
= 2,
SOLUTION:
= 𝑠/ √ n= 2
= 𝑠 / √ 50 = 2
= 𝑠 / 7.07
𝑠= (2) (7.07)
𝑠= (14.14)
= 20
1. Determine the mean and standard deviation of the sampling distribution of the
sample mean with a sample size of and whose population mean and population
standard deviation are as follows:
a.
𝑛 = 50; 𝜇 = 13.15; 𝜎 = 3.74
SOLUTION:
= 𝜎/ √n
= 3.74/ √50 • √50/ √50
= 3.74 √50/ 50 ÷ 3.74/ 3.74
= √50/ 13.37
= 1.93
b.
= 13.15
𝑛 = 20; 𝜇 = 26.14; 𝜎 = 5.14
SOLUTION:
= 𝜎/ √n
= 5.14/ √20 • √20/ √20
= 5.14 √20/ 20 ÷ 5.14/ 5.14
= √20/ 3.89
= 1.15
= 26.14
c.
𝑛 = 30; 𝜇 = 31.15; 𝜎 = 1.25
SOLUTION:
= 𝜎/ √n
= 1.25/ √30 • √30/ √30
= 1.25 √30/ 30 ÷ 1.25/ 1.25
= √30/ 24
= 0.23
d.
= 31.15
𝑛 = 35; 𝜇 = 16.17; 𝜎 = 0.72
SOLUTION:
= 𝜎/ √n
= 0.72/ √35 • √35/ √35
= 0.72 √35/ 35 ÷ 0.72/ 0.72
= √35/ 48.61
= 0.12
= 16.17