An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, X, is as follows: 1 2 3 0.18 Ti P(X=2₁) 0.11 Ti 1 2 3 To find the expected value the following table was set up: P(X = x₁) 0.11 4 5 0.34 6 0.34 0.2 4 0.05 0.2 3. T₁ - P(X = I₁) 1.0.11 0.11 2-0.18 0.36 4.0.2 5 0.12 Total: Fill out the blanks in the table above and compute the E[X]: E[X] = = 1.02 5.0.12 = 0.6 (Round the answer to 2 decimals) 6 0.05 Total 1
An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, X, is as follows: 1 2 3 0.18 Ti P(X=2₁) 0.11 Ti 1 2 3 To find the expected value the following table was set up: P(X = x₁) 0.11 4 5 0.34 6 0.34 0.2 4 0.05 0.2 3. T₁ - P(X = I₁) 1.0.11 0.11 2-0.18 0.36 4.0.2 5 0.12 Total: Fill out the blanks in the table above and compute the E[X]: E[X] = = 1.02 5.0.12 = 0.6 (Round the answer to 2 decimals) 6 0.05 Total 1
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 1
An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability
distribution of the face value, X, is as follows:
1
2
3
0.34
Xi
P(X = x₁)
0.11
0.18
Xi
1
To find the expected value the following table was set up:
P(X = x₂)
0.11
2
3
4
5
6
E[X]
0.34
=
0.2
4
0.05
0.2
3.
Total:
Fill out the blanks in the table above and compute the E[X]:
x₁ · P(X = x₁)
1.0.11 0.11
2.0.18 0.36
4.0.2
5
0.12
=
=
1.02
5.0.12 0.6
=
(Round the answer to 2 decimals)
6
0.05
Total
1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd372abab-652d-4bea-b0d9-4c3265961c0a%2F3f52e38f-5b0b-49bc-9d7c-304df07a5903%2Fo6eezls_processed.png&w=3840&q=75)
Transcribed Image Text:Question 1
An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability
distribution of the face value, X, is as follows:
1
2
3
0.34
Xi
P(X = x₁)
0.11
0.18
Xi
1
To find the expected value the following table was set up:
P(X = x₂)
0.11
2
3
4
5
6
E[X]
0.34
=
0.2
4
0.05
0.2
3.
Total:
Fill out the blanks in the table above and compute the E[X]:
x₁ · P(X = x₁)
1.0.11 0.11
2.0.18 0.36
4.0.2
5
0.12
=
=
1.02
5.0.12 0.6
=
(Round the answer to 2 decimals)
6
0.05
Total
1
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