Probability A. The number of pets at home was defined as : Last Name Response Rasales 2 Lasrera 2 Launto 0 Cantila 3 Kho 3 Magrina 3 Sugabo 3 Mirafelix 2 Dela Cruz 4 Monteurgen 0 Martinez 6 Luma-as 2 Meja 4 Malonzo 0 Magcalas 3 Palangdao 8 Sagcal 0 Salazar 4 Pineda 8 Mina 3 From the data given, discrete probability distribution as : x Frequency f Px=fn x×Px 0 4 0.20 0.0000 2 4 0.20 0.4000 3 6 0.30 0.9000 4 3 0.15 0.6000 6 1 0.05 0.3000 8 2 0.10 0.8000 Total n=20 1 3 From the data given in the table,  ∑x×Px=3μ=3 Thus, the mean is 3     B. Given : From the information given as : Mean value is given as 3.  x Frequency P(x) x×P(x)x×Px (x−μ)x-μ (x−μ)2x-μ2 (x−μ)2×P(x)x-μ2×Px 0 4 0.20 0.0000 -3 9 1.8 2 4 0.20 0.4000 -1 1 0.2 3 6 0.30 0.9000 0 0 0 4 3 0.15 0.6000 1 1 0.15 6 1 0.05 0.3000 3 9 0.45 8 2 0.10 0.8000 5 25 2.5 Total 20 1 3     5.1  From the table value,  Variance : V(x)==∑(x−μ)2×P(x)5.1Vx=∑x-μ2×Px=5.1 Thus, variance is 5.1 Standard deviation : σ===V(x)⎯⎯⎯⎯⎯⎯⎯⎯√5.1⎯⎯⎯⎯⎯⎯√2.26σ=Vx=5.1=2.26 Thus, the standard deviation is 2.26   What is the interpretation of this two? please provide the interpretation

A First Course in Probability (10th Edition)
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Chapter1: Combinatorial Analysis
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Probability

A. The number of pets at home was defined as :

Last Name Response
Rasales 2
Lasrera 2
Launto 0
Cantila 3
Kho 3
Magrina 3
Sugabo 3
Mirafelix 2
Dela Cruz 4
Monteurgen 0
Martinez 6
Luma-as 2
Meja 4
Malonzo 0
Magcalas 3
Palangdao 8
Sagcal 0
Salazar 4
Pineda 8
Mina 3

From the data given, discrete probability distribution as :

x Frequency f Px=fn x×Px
0 4 0.20 0.0000
2 4 0.20 0.4000
3 6 0.30 0.9000
4 3 0.15 0.6000
6 1 0.05 0.3000
8 2 0.10 0.8000
Total n=20 1 3

From the data given in the table, 

∑x×Px=3μ=3

Thus, the mean is 3

 

 

B. Given :

From the information given as :

Mean value is given as 3. 

x Frequency P(x) x×P(x)x×Px (x−μ)x-μ (x−μ)2x-μ2 (x−μ)2×P(x)x-μ2×Px
0 4 0.20 0.0000 -3 9 1.8
2 4 0.20 0.4000 -1 1 0.2
3 6 0.30 0.9000 0 0 0
4 3 0.15 0.6000 1 1 0.15
6 1 0.05 0.3000 3 9 0.45
8 2 0.10 0.8000 5 25 2.5
Total 20 1 3     5.1

 From the table value, 

Variance :

V(x)==∑(x−μ)2×P(x)5.1Vx=∑x-μ2×Px=5.1

Thus, variance is 5.1

Standard deviation :

σ===V(x)⎯⎯⎯⎯⎯⎯⎯⎯√5.1⎯⎯⎯⎯⎯⎯√2.26σ=Vx=5.1=2.26

Thus, the standard deviation is 2.26

 

What is the interpretation of this two? please provide the interpretation

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