POLYTECHNIC UNIVERSITY OF THE PHILIPPINES

A. MABINI CAMPUS, ANONAS ST., STA. MESA,

MANILA SHS – STATISTICS AND PROBABILITY

Bondoc, Shayne F. May 1, 2021

STEM 11-1 Ma’am Grace Pediongco

Consider all samples of size 5 from this population:

2 5 6 8 10 12 13
1. Compute the mean (μ) and standard deviation (σ) of the population.
N=7

∑𝑥
𝜇=
𝑁
2 + 5 + 6 + 8 + 10 + 12 + 13
= = 8.00
7
𝑇ℎ𝑒 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 8.00.
𝑿 𝑿 − 𝝁 (𝑿 − 𝝁)𝟐
2 −6 36
5 −3 9
6 −2 4
8 0 0
10 2 4
12 4 16
13 5 25
∑(𝑋 − 𝜇)2 = 94

2
∑(𝑋 − 𝜇)2
𝜎 =
𝑁

94
=
7
= 13. ̅̅̅̅̅̅̅̅̅̅
428571
The variance of the population is 13. ̅̅̅̅̅̅̅̅̅̅
428571.
2. List all the samples of the sample size 5 and compute the mean (X̅) for each
sample.

N = 7, n = 5

𝑁!
𝐶(𝑁, 𝑛) =
(𝑁 − 𝑛)! 𝑛!

7!
𝐶(7, 5) =
(7 − 5)! 5!

7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
=
2 ∙ 1 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

5040
= = 21
240

There are 21 possible samples of size 5 that can be drawn

Samples Mean
2, 5, 6, 8, 10 6.2
2, 5, 6, 8, 12 6.6
2, 5, 6, 8, 13 6.8
2, 5, 6, 10, 12 7
2, 5, 6, 10, 13 7.2
2, 5, 6, 12, 13 7.6
2, 5, 8, 10, 12 7.4
2, 5, 8, 10, 13 7.6
2, 5, 8, 12, 13 8
2, 5, 10, 12, 13 8.4
2, 6, 8, 10, 12 7.6
2, 6, 8, 10, 13 7.8
2, 6, 8, 12, 13 8.2
2, 6, 10, 12, 13 8.6
2, 8, 10, 12, 13 9
5, 6, 8, 10, 12 8.2
5, 6, 8, 10, 13 8.4
5, 6, 8, 12, 13 8.8
5, 6, 10, 12, 13 9.2
5, 8, 10, 12, 13 9.6
6, 8, 10, 12, 13 9.8

The means of the samples vary from sample to sample. The mean of the population
𝜇 = 8, while the means of the samples may be less than, greater than, or equal to 8
 2. Construct the sampling distribution of the sample means.
Sample Mean Frequency Probability
̅
𝑿 𝑷(𝑿̅)
6.2 1 1
21
6.6 1 1
21
6.8 1 1
21
7 1 1
21
7.2 1 1
21
7.4 1 1
21
7.6 3 1
7
7.8 1 1
21
8 1 1
21
8.2 2 2
21
8.4 2 2
21
8.6 1 1
21
8.8 1 1
21
9 1 1
21
9.2 1 1
21
9.6 1 1
21
9.8 1 1
21
Total 21 1.00

4. Calculate the mean (μX̅ ) of the sampling distribution of the sample means.
Compare this to mean of the population.

Sample Mean Probability ̅ ∙ 𝑷(𝑿

𝑿 ̅)
̅
𝑿 𝑷(𝑿̅)
6.2 1 31
̅̅̅̅̅̅̅̅̅̅
≈ 0.2952380
21 105
 6.6 1 11
̅̅̅̅̅̅̅̅̅̅
≈ 0.3142857
21 35
6.8 1 34
̅̅̅̅̅̅̅̅̅̅
≈ 0.3238095
21 105
7 1 1
≈ 0. 3̅
21 3
7.2 1 12
̅̅̅̅̅̅̅̅̅̅
≈ 0.3428571
21 35
7.4 1 37
̅̅̅̅̅̅̅̅̅̅
≈ 0.3523809
21 105
7.6 1 38
̅̅̅̅̅̅̅̅̅̅
≈ 1.0857142
7 35
7.8 1 13
̅̅̅̅̅̅̅̅̅̅
≈ 0.3714285
21 35
8 1 8
≈ 0. ̅̅̅̅̅̅̅̅̅̅
380952
21 21
8.2 2 82
̅̅̅̅̅̅̅̅̅̅
≈ 0.7809523
21 105
8.4 2 4
≈ 0.8
21 5
8.6 1 43
̅̅̅̅̅̅̅̅̅̅
≈ 0.4095238
21 105
8.8 1 44
̅̅̅̅̅̅̅̅̅̅
≈ 0.4190476
21 105
9 1 3
̅̅̅̅̅̅̅̅̅̅
≈ 0.4285714
21 7
9.2 1 46
̅̅̅̅̅̅̅̅̅̅
≈ 0.4380952
21 105
9.6 1 16
̅̅̅̅̅̅̅̅̅̅
≈ 0.4571428
21 35
9.8 1 7
≈ 0.46̅
21 15
Total 1.00 8.00

𝝁𝑿̅ = ∑[𝑋̅ ∙ 𝑃(𝑋̅)]

= 8.00
The mean of the sampling distribution of the sample means(𝜇𝑋̅ ) is 8.00, while the mean of the
population is 8.00 therefore shows that the mean of the sampling distribution of the sample
means is is equal to the mean (𝜇) of the population.
5. Calculate the standard deviation (σx̅) of the sampling distribution of the sample means.
Compare this to the standard deviation of the population.

̅
𝑿 ̅)
𝑷(𝑿 ̅− 𝝁
𝑿 ̅ − 𝝁)𝟐
(𝑿 ̅ ) ∙ (𝑿
𝑷(𝑿 ̅ − 𝝁)𝟐
1
6.2 21 −1.8 3.24 ̅̅̅̅̅̅̅̅̅̅
0.15428571
1
6.6 21 −1.4 1.96 0.093̅
1
6.8 21 −1.2 1.44 ̅̅̅̅̅̅̅̅̅̅
0.06857142
1
7 21 −1 1 0. ̅̅̅̅̅̅̅̅̅̅
047619
1
7.2 21 −0.8 0.64 ̅̅̅̅̅̅̅̅̅̅
0.03047619
1
7.4 21 −0.6 0.36 ̅̅̅̅̅̅̅̅̅̅
0.01714285
1
7.6 7 −0.4 0.16 ̅̅̅̅̅̅̅̅̅̅
0.02285714
1
7.8 21 −0.2 0.04 ̅̅̅̅̅̅̅̅̅̅
0.00190476
1
8 21 0 0 0
2
8.2 21 0.2 0.04 ̅̅̅̅̅̅̅̅̅̅
0.00380952
2
8.4 21 0.4 0.16 ̅̅̅̅̅̅̅̅̅̅
0.01523809
1
8.6 21 0.6 0.36 ̅̅̅̅̅̅̅̅̅̅
0.01714285
1
8.8 21 0.8 0.64 ̅̅̅̅̅̅̅̅̅̅
0.03047619
1
9 21 1 1 0. ̅̅̅̅̅̅̅̅̅̅
047619
1
9.2 21 1.2 1.44 ̅̅̅̅̅̅̅̅̅̅
0.06857142
1
9.6 21 1.6 2.56 ̅̅̅̅̅̅̅̅̅̅
0.12190476
1
9.8 21 1.8 3.24 ̅̅̅̅̅̅̅̅̅̅
0.15428571
Total 1.00 ̅ ) − (𝑿
∑ 𝑷(𝑿 ̅ − 𝝁)𝟐
̅̅̅̅̅̅̅̅̅̅̅
= 0.8𝟗𝟓𝟐𝟑𝟖𝟎

̅ = ∑ 𝑷(𝑿
𝝈𝟐 𝑿 ̅ ) − (𝑿
̅ − 𝝁)𝟐
 ̅̅̅̅̅̅̅̅̅̅̅
= 0.8𝟗𝟓𝟐𝟑𝟖𝟎

̅̅̅̅̅̅̅̅̅̅ while the variance of

The variance of the sampling distribution of the sample means is 0.8952380
the population is 13. ̅̅̅̅̅̅̅̅̅̅
428571 therefore shows that it’s not equal and the variance of the sampling
distribution of the sample means is less than the variance of the population.

1
7
̅)
Probability 𝑷(𝑿

2
21

1
21

6.2 6.6 6.8 7 7.2 7.4 7.6 7.8 8 8.2 8.4 8.6 8.8 9 9.2 9.6 9.8

̅
Sample Mean 𝑿

Assessment 7: Look Back and Reflect

Form a group of five students in your class. Determine the general weighted average of
the members of the group last semester. List them.

Sample General Weighted Average

Ajero 97
Alimboyogen 98
Biating 96
Bondoc 97
Cortez 97
Tomaquin 95
Vicenta 97
Villon 95
∑𝑥
𝜇=
𝑁
97 + 98 + 96 + 97 + 97 + 95 + 97 + 95
=
8
= 96.5
 The mean of the population is 96.5.
𝑿 𝑿 − 𝝁 (𝑿 − 𝝁)𝟐
97 0.5 0.25
98 1.5 2.25
96 -0.5 0.25
97 0.5 0.25
97 0.5 0.25
95 -1.5 2.25
97 0.5 0.25
95 -1.5 2.25
∑(𝑋 − 𝜇)2 = 8

∑(𝑋 − 𝜇)2
𝜎2 =
𝑁
8
= =1
8
The variance of the population is 1.
1. List all possible samples of size 2 and their corresponding means.
N = 8, n = 2

𝑁!
𝐶(𝑁, 𝑛) =
(𝑁 − 𝑛)! 𝑛!

8!
𝐶(8, 2) =
(8 − 2)! 2!

8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
=
6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 ∙ 2 ∙ 1

40320
= = 𝟐𝟖
1440

Samples General Weighted Mean

Average (GWA)
Ajero, Alimboyogen 97, 98 97.5
Ajero, Biating 97, 96 96.5
Ajero, Bondoc 97, 97 97
Ajero, Cortez 97, 97 97
Ajero, Tomaquin 97, 95 96
 Ajero, Vicenta 97, 97 97
Ajero, Villon 97, 95 96
Alimboyogen, Biating 98, 96 97
Alimboyogen, Bondoc 98, 97 97.5
Alimboyogen, Cortez 98, 97 97.5
Alimboyogen, Tomaquin 98, 95 96.5
Alimboyogen, Vicenta 98, 97 97.5
Alimboyogen, Villon 98, 95 96.5
Biating, Bondoc 96, 97 96.5
Biating, Cortez 96, 97 96.5
Biating, Tomaquin 96, 95 95.5
Biating, Vicenta 96, 97 96.5
Biating, Villon 96, 95 95.5
Bondoc, Cortez 97, 97 97
Bondoc, Tomaquin 97, 95 96
Bondoc, Vicenta 97, 97 97
Bondoc, Villon 97, 95 96
Cortez, Tomaquin 97, 95 96
Cortez, Vicenta 97, 97 97
Cortez, Villon 97, 95 96
Tomaquin, Vicenta 95,97 96
Tomaquin, Villon 95, 95 95
Vicenta, Villon 97, 95 96

2. Construct the sampling distribution of the sample means.

Sample Mean Frequency Probability
̅
𝑿 𝑷(𝑿̅)
95 1 1
28
95.5 2 1
14
96 8 2
7
96.5 6 3
14
97 7 1
4
97.5 4 1
7
Total 28 1.00
 3. Draw the histogram of the sampling distribution of the sample means.

2
General Weighted Average
76

1
5
̅)

4
Probability 𝑷(𝑿

34
14

13
7
1
2
14

11
28

0
95 95.5 96 96.5 97 97.5

̅
Sample Mean 𝑿
4. Calculate the mean μX̅ of the sampling distribution of the sample means. Compare
this to the mean of the population.

Sample Mean Probability ̅ ∙ 𝑷(𝑿

𝑿 ̅)
̅
𝑿 𝑷(𝑿̅)
95 1 95
̅̅̅̅̅̅̅̅̅̅
≈ 3.39285714
28 28
95.5 1 191
̅̅̅̅̅̅̅̅̅̅
≈ 6.82142857
14 28
96 2 192
≈ 27. ̅̅̅̅̅̅̅̅̅̅
428571
7 7
96.5 3 579
̅̅̅̅̅̅̅̅̅̅
≈ 20.67857142
14 28
97 1 97
≈ 24.25
4 4
97.5 1 195
̅̅̅̅̅̅̅̅̅̅
≈ 13.9285714
7 14
Total 1.00 96.5

𝝁𝑿̅ = ∑[𝑋̅ ∙ 𝑃(𝑋̅)]

= 96.5
The mean of the sampling distribution of the sample mean is 96.5 and the mean of the
population is also 96.5 therefore they are equal.
5. Calculate the standard deviation (σX̅) of the sampling distribution of the sample
means. Compare this to the standard deviation of the population.

̅
𝑿 ̅)
𝑷(𝑿 ̅− 𝝁
𝑿 ̅ − 𝝁)𝟐
(𝑿 ̅ ) ∙ (𝑿
𝑷(𝑿 ̅ − 𝝁)𝟐

95 1 -1.5 2.25 ̅̅̅̅̅̅̅̅̅̅

0.0803571428
28

95.5 1 -1 1 ̅̅̅̅̅̅̅̅̅̅
0.0714285
14

96 2 -0.5 0.25 ̅̅̅̅̅̅̅̅̅̅

0.0714285
7

96.5 3 0 0 0
14

97 1 0.5 0.25 0.0625

4

97.5 1 1 1 ̅̅̅̅̅̅̅̅̅̅
0.142857
7
Total 1.00 ̅ ) − (𝑿
∑ 𝑷(𝑿 ̅ − 𝝁)𝟐
̅̅̅̅̅̅̅̅̅̅̅
= 0.𝟒𝟐𝟖𝟓𝟕𝟏

̅ = ∑ 𝑷(𝑿
𝝈𝟐 𝑿 ̅ ) − (𝑿
̅ − 𝝁)𝟐

̅̅̅̅̅̅̅̅̅̅̅
= 0.𝟒𝟐𝟖𝟓𝟕𝟏

̅̅̅̅̅̅̅̅̅̅ while the variance of

The variance of the sampling distribution of the sample means is 0.428571
the population is 1, therefore shows that it’s not equal and the variance of the sampling distribution
of the sample means is less than the variance of the population.