The household size in a certain neighborhood varies from 1 to 6 The probability distribution of the size of a random household, X, is as follows: 2 P(X = x₁)| Ti 2 3 4 5 1 6 0.13 0.19 0.13 0.31 The expected value is computed, E[X] = 3.12. To find the variance and standard deviation the following table was set up: P(X = x₁) 0.2 0.03 3 0.31 SD[X] 4 0.2 5 = 0.14 (x₁ - E[X])² (1-3.12)² = (-2.12)² = 4.4944 (23.12)² = (-1.12)² = 1.2544 (3-3.12)² = (-0.12)² = (4- 3.12)² = (0.88)² = 0.7744 (53.12)² = (1.88)² = 3.5344 6 0.03 (x₁ — E[X])²P(X = x₁) - Total 4.4944 0.13 = 0.5843 1 1.2544 0.19 = 0.2383 0.7744 0.2 0.0144 0.31 = 0.0045 3.5344 0.14 = EX] = 3.12 Total: Fill out the blanks (round to 4 decimals) in the table above and compute the VAR[X] and SD[X]: VAR[X] = (Round the answer to 4 decimals) (Round the answer to 2 decimals) 0.4948 8.2944 0.03= 0.2488
The household size in a certain neighborhood varies from 1 to 6 The probability distribution of the size of a random household, X, is as follows: 2 P(X = x₁)| Ti 2 3 4 5 1 6 0.13 0.19 0.13 0.31 The expected value is computed, E[X] = 3.12. To find the variance and standard deviation the following table was set up: P(X = x₁) 0.2 0.03 3 0.31 SD[X] 4 0.2 5 = 0.14 (x₁ - E[X])² (1-3.12)² = (-2.12)² = 4.4944 (23.12)² = (-1.12)² = 1.2544 (3-3.12)² = (-0.12)² = (4- 3.12)² = (0.88)² = 0.7744 (53.12)² = (1.88)² = 3.5344 6 0.03 (x₁ — E[X])²P(X = x₁) - Total 4.4944 0.13 = 0.5843 1 1.2544 0.19 = 0.2383 0.7744 0.2 0.0144 0.31 = 0.0045 3.5344 0.14 = EX] = 3.12 Total: Fill out the blanks (round to 4 decimals) in the table above and compute the VAR[X] and SD[X]: VAR[X] = (Round the answer to 4 decimals) (Round the answer to 2 decimals) 0.4948 8.2944 0.03= 0.2488
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![Question 3
The household size in a certain neighborhood varies from 1 to 6 The probability distribution of the size of a
random household, X, is as follows:
1
2
0.13
0.19
xi
P(X = x₁)
Xi
1
2
3
4
5
6
0.13
0.31
The expected value is computed, E[X] = 3.12. To find the variance and standard deviation the following
table was set up:
P(X = x₂)
0.2
0.03
3
0.31
4
0.2
SD[X] =
5
0.14
(x₁ - E[X])²
(1-3.12)² = (-2.12)² = 4.4944
(23.12)² = (-1.12)² = 1.2544
(3-3.12)² = (-0.12)² =
(4- 3.12)² = (0.88)² = 0.7744
(53.12)² = (1.88)² = 3.5344
6
0.03
4.4944 -0.13
(x₁ — E[X])²P(X = x₂)
1.2544 -0.19
0.0144 - 0.31
0.7744 -0.2
=
(Round the answer to 2 decimals)
Total
1
=
0.5843
0.2383
3.5344 0.14 = 0.4948
EX] =3.12
Total:
Fill out the blanks (round to 4 decimals) in the table above and compute the VAR[X] and SD[X]:
VAR[X] =
(Round the answer to 4 decimals)
0.0045
8.2944 0.03 0.2488](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc61e6f9-1283-4c33-8aac-5d95a690c4e7%2F76426620-e20a-43fd-abf7-ade0bde3c550%2Fn4dql3d_processed.png&w=3840&q=75)
Transcribed Image Text:Question 3
The household size in a certain neighborhood varies from 1 to 6 The probability distribution of the size of a
random household, X, is as follows:
1
2
0.13
0.19
xi
P(X = x₁)
Xi
1
2
3
4
5
6
0.13
0.31
The expected value is computed, E[X] = 3.12. To find the variance and standard deviation the following
table was set up:
P(X = x₂)
0.2
0.03
3
0.31
4
0.2
SD[X] =
5
0.14
(x₁ - E[X])²
(1-3.12)² = (-2.12)² = 4.4944
(23.12)² = (-1.12)² = 1.2544
(3-3.12)² = (-0.12)² =
(4- 3.12)² = (0.88)² = 0.7744
(53.12)² = (1.88)² = 3.5344
6
0.03
4.4944 -0.13
(x₁ — E[X])²P(X = x₂)
1.2544 -0.19
0.0144 - 0.31
0.7744 -0.2
=
(Round the answer to 2 decimals)
Total
1
=
0.5843
0.2383
3.5344 0.14 = 0.4948
EX] =3.12
Total:
Fill out the blanks (round to 4 decimals) in the table above and compute the VAR[X] and SD[X]:
VAR[X] =
(Round the answer to 4 decimals)
0.0045
8.2944 0.03 0.2488
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