Answer A,B,C Given the probability distributions shown to the right, complete the following parts.a. Compute the expected value for each distribution.b. Compute the standard deviation for each distribution.c. What is the probability that x will be at least 3 in Distribution A and Distribution B?d. Compare the results of distributions A and B.a. What is the expected value for distribution A?H=(Type an integer or decimal rounded to three decimal places as needed.)X₁01234Distribution AP(X=x;)0.060.080.130.180.55X₁01234Distribution BP(X=x;)0.550.180.130.080.06

Given data is,Distribution ADistribution…

Answered: a. Compute the expected value for each…

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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Answer A,B,C

Given the probability distributions shown to the right, complete the following parts.
a. Compute the expected value for each distribution.
b. Compute the standard deviation for each distribution.
c. What is the probability that x will be at least 3 in Distribution A and Distribution B?
d. Compare the results of distributions A and B.
a. What is the expected value for distribution A?
H=
(Type an integer or decimal rounded to three decimal places as needed.)
X₁
0
1
2
3
4
Distribution A
P(X=x;)
0.06
0.08
0.13
0.18
0.55
X₁
0
1
2
3
4
Distribution B
P(X=x;)
0.55
0.18
0.13
0.08
0.06

Expert Solution

Step 1: we first take the given data then we find the each part

Given data is,

Distribution A		Distribution B
x_i	P(x_i=x)	x_i	P(x_i=x)
0	0.06	0	0.55
1	0.08	1	0.18
2	0.13	2	0.13
3	0.18	3	0.08
4	0.55	4	0.06

(a)

The expected value for distribution A is,

$mu equals E left parenthesis x subscript i right parenthesis equals sum x subscript i p subscript i space space space space space space space space space space space space space space space equals 0 asterisk times 0.06 plus 1 asterisk times 0.08 plus 2 asterisk times 0.13 plus 3 asterisk times 0.18 plus 4 asterisk times 0.55 space space space space space space space space space space space space space space space equals 0 plus 0.08 plus 0.26 plus 0.54 plus 2.2 mu equals E left parenthesis x subscript i right parenthesis equals 3.08$

The expected value for distribution A is $mu equals E left parenthesis x subscript i right parenthesis equals 3.08$

The expected value for distribution B is,

$mu equals E left parenthesis x subscript i right parenthesis equals sum x subscript i p subscript i space space space space space space space space space space space space space space space equals 0 asterisk times 0.55 plus 1 asterisk times 0.18 plus 2 asterisk times 0.13 plus 3 asterisk times 0.08 plus 4 asterisk times 0.06 space space space space space space space space space space space space space space space equals 0 plus 0.18 plus 0.26 plus 0.24 plus 0.24 mu equals E left parenthesis x subscript i right parenthesis equals 0.92$

The expected value for distribution B is $mu equals E left parenthesis x subscript i right parenthesis equals 0.92$

(b)

The standard deviation for distribution A is,

but first we need to find the $E left parenthesis x subscript i superscript 2 right parenthesis$ for distribution A

$E left parenthesis x subscript i superscript 2 right parenthesis equals sum x subscript i superscript 2 p subscript i space space space space space space space space space equals 0 squared asterisk times 0.06 plus 1 squared asterisk times 0.08 plus 2 squared asterisk times 0.13 plus 3 squared asterisk times 0.18 plus 4 squared asterisk times 0.55 space space space space space space space space space equals 0 asterisk times 0.06 plus 1 asterisk times 0.08 plus 4 asterisk times 0.13 plus 9 asterisk times 0.18 plus 16 asterisk times 0.55 space space space space space space space space space equals 0 plus 0.08 plus 0.52 plus 1.62 plus 8.8 E left parenthesis x subscript i superscript 2 right parenthesis equals 11.02 space space space space space space space space space space space space space$

the $E left parenthesis x subscript i superscript 2 right parenthesis$ for distribution A is $E left parenthesis x subscript i superscript 2 right parenthesis equals 11.02 space space space$

Hence,

The standard deviation for distribution A is,

$S D space o f space A equals square root of E left parenthesis x subscript i superscript 2 right parenthesis minus left square bracket E left parenthesis x right parenthesis right square bracket squared end root space space space space space space space space space space space space space space space equals square root of 11.02 minus left square bracket 3.08 right square bracket squared end root S D space o f space A equals 1.238386$

The standard deviation for distribution A is $S D space o f space A equals 1.238386$

similarly for distribution B is,

but first we need to find the $E left parenthesis x squared right parenthesis$ for distribution B

$E left parenthesis x subscript i superscript 2 right parenthesis equals sum x subscript i superscript 2 p subscript i space space space space space space space space space space equals 0 squared asterisk times 0.55 plus 1 squared asterisk times 0.18 plus 2 squared asterisk times 0.13 plus 3 squared asterisk times 0.08 plus 4 squared asterisk times 0.06 space space space space space space space space space space equals 0 asterisk times 0.55 plus 1 asterisk times 0.18 plus 4 asterisk times 0.26 plus 9 asterisk times 0.24 plus 16 asterisk times 0.24 space space space space space space space space space space equals 0 plus 0.18 plus 0.52 plus 0.72 plus 0.96 E left parenthesis x subscript i superscript 2 right parenthesis equals 2.38$

The $E left parenthesis x subscript i superscript 2 right parenthesis$ for distribution B is $E left parenthesis x subscript i superscript 2 right parenthesis equals 2.38$

The standard deviation for distribution B is,

$S D space o f space B equals square root of E left parenthesis x subscript i superscript 2 right parenthesis minus left square bracket E left parenthesis x subscript i right parenthesis right square bracket squared end root space space space space space space space space space space space space space space equals square root of 2.38 minus left square bracket 0.92 right square bracket squared end root S D space o f space B equals 1.238386$

The standard deviation for distribution B is $S D space o f space B equals 1.238386$

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a. Compute the expected value for each distribution. b. Compute the standard deviation for each distribution. c. What is the probability that x will be at least 3 in Distribution A and Distribution B? . Compare the results of distributions A and B.