An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, XX, is as follows: xixi 1 2 3 4 5 6 Total P(X=xi)P(X=xi) 0.15 0.16 0.33 0.2 0.11 0.05 1 The expected value is computed, E[X]=3.11E[X]=3.11. To find the variance and standard deviation the following table was set up: xixi P(X=xi)P(X=xi) (xi−E[X])2(xi-E[X])2 (xi−E[X])2P(X=xi)(xi-E[X])2P(X=xi) 1 0.15 (1−3.11)2=(−2.11)2=4.4521(1-3.11)2=(-2.11)2=4.4521 4.4521⋅0.15=0.66784.4521⋅0.15=0.6678 2 (2−3.11)2=(−1.11)2=1.2321(2-3.11)2=(-1.11)2=1.2321 1.2321⋅0.16=0.19711.2321⋅0.16=0.1971 3 0.33 (3−3.11)2=(−0.11)2=(3-3.11)2=(-0.11)2= 0.0121⋅0.33=0.0040.0121⋅0.33=0.004 4 0.2 (4−3.11)2=(0.89)2=0.7921(4-3.11)2=(0.89)2=0.7921 0.7921⋅0.2=0.7921⋅0.2= 5 (5−3.11)2=(1.89)2=3.5721(5-3.11)2=(1.89)2=3.5721 3.5721⋅0.11=0.39293.5721⋅0.11=0.3929 6 0.05 8.3521⋅0.05=0.41768.3521⋅0.05=0.4176 E[X]=3.11E[X]=3.11 Total: Fill out the blanks (round to 4 decimals) in the table above and compute the VAR[X] and SD[X]: VAR[X]=VAR[X]= (Round the answer to 4 decimals) SD[X]=SD[X]= (Round the answer to 2 decimals)

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

An unfair die looks like an ordinary six-sided die but the outcomes are not equally likely. The probability distribution of the face value, XX, is as follows:

xixi	1	2	3	4	5	6	Total
P(X=xi)P(X=xi)	0.15	0.16	0.33	0.2	0.11	0.05	1

The expected value is computed, E[X]=3.11E[X]=3.11. To find the variance and standard deviation the following table was set up:

xixi	P(X=xi)P(X=xi)	(xi−E[X])2(xi-E[X])2	(xi−E[X])2P(X=xi)(xi-E[X])2P(X=xi)
1	0.15	(1−3.11)2=(−2.11)2=4.4521(1-3.11)2=(-2.11)2=4.4521	4.4521⋅0.15=0.66784.4521⋅0.15=0.6678
2		(2−3.11)2=(−1.11)2=1.2321(2-3.11)2=(-1.11)2=1.2321	1.2321⋅0.16=0.19711.2321⋅0.16=0.1971
3	0.33	(3−3.11)2=(−0.11)2=(3-3.11)2=(-0.11)2=	0.0121⋅0.33=0.0040.0121⋅0.33=0.004
4	0.2	(4−3.11)2=(0.89)2=0.7921(4-3.11)2=(0.89)2=0.7921	0.7921⋅0.2=0.7921⋅0.2=
5		(5−3.11)2=(1.89)2=3.5721(5-3.11)2=(1.89)2=3.5721	3.5721⋅0.11=0.39293.5721⋅0.11=0.3929
6	0.05		8.3521⋅0.05=0.41768.3521⋅0.05=0.4176
E[X]=3.11E[X]=3.11		Total: