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
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
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 30E
Related questions
Question
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 :
| 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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