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The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1800 and the standard deviation is $80.

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1800 and the standard deviation is $80.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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Step 1: Determining the given information The population mean is $1800 and standard deviation is $80. It is also given that the graph illustrates a normal distribution. Step 2: Finding the approximate percentage of buyers who paid between $1560 and $1800. The approximate percentage of buyers who paid(X) between $1560 and $1800 is, P(1560X1800)=1 )=P( =P(-3<z < 0) =P(z < 0) - P(z < −3) By using the EXCEL, the required probability is, fx D 0.49865 fx =NORM.S.DIST(0,TRUE)-NORM.S.DIST(-3,TRUE) E D 0.00135 F That is, P(1560 < X < 1800)=P(z < 0) – P(z < −3) -0.4987 x 100% =49.87% 1560-1800 X-μ 80 Thus, the approximate percentage of buyers who paid(X) between $1560 and $1800 is 49.87%. Step 3: Finding the approximate percentage of buyers who paid more than $2040. The approximate percentage of buyers who paid more than $2040 is, P(X > 2040)=1 − P(X < (2040) =1 - P(X-" < =1 - P(Z < 3) By using the EXCEL, the required probability is, =1-NORM.S.DIST(3, TRUE) E F That is, That is, P(X> 2040)=1 - P(Z <3) -0.0014 × 100% =0.14% D 0.02275 G Thus, the approximate percentage of buyers who paid more than $2040 is 0.14%. Step 4: Finding the approximate percentage of buyers who paid less than $1640. The approximate percentage of buyers who paid less than $2040 is, P(X < 1640)=P(X-¹ = P(Z < − 2) By using the EXCEL, the required probability is, fx =NORM.S.DIST(-2, TRUE) E F P(X1640)=P(z < − 2) 2040-1800 80 H 1640-1800 80 1800-1800 80 -0.0228 x 100% = 2.28% Thus, the approximate percentage of buyers who paid less than $1640 is 2.28%. Solution The approximate percentage of buyers who paid(X) between $1560 and $1800 is 49.87%. The approximate percentage of buyers who paid more than $2040 is 0.14%. The approximate percentage of buyers who paid less than $1640 is 2.28%.
Transcribed Image Text:Step 1: Determining the given information The population mean is $1800 and standard deviation is $80. It is also given that the graph illustrates a normal distribution. Step 2: Finding the approximate percentage of buyers who paid between $1560 and $1800. The approximate percentage of buyers who paid(X) between $1560 and $1800 is, P(1560X1800)=1 )=P( =P(-3<z < 0) =P(z < 0) - P(z < −3) By using the EXCEL, the required probability is, fx D 0.49865 fx =NORM.S.DIST(0,TRUE)-NORM.S.DIST(-3,TRUE) E D 0.00135 F That is, P(1560 < X < 1800)=P(z < 0) – P(z < −3) -0.4987 x 100% =49.87% 1560-1800 X-μ 80 Thus, the approximate percentage of buyers who paid(X) between $1560 and $1800 is 49.87%. Step 3: Finding the approximate percentage of buyers who paid more than $2040. The approximate percentage of buyers who paid more than $2040 is, P(X > 2040)=1 − P(X < (2040) =1 - P(X-" < =1 - P(Z < 3) By using the EXCEL, the required probability is, =1-NORM.S.DIST(3, TRUE) E F That is, That is, P(X> 2040)=1 - P(Z <3) -0.0014 × 100% =0.14% D 0.02275 G Thus, the approximate percentage of buyers who paid more than $2040 is 0.14%. Step 4: Finding the approximate percentage of buyers who paid less than $1640. The approximate percentage of buyers who paid less than $2040 is, P(X < 1640)=P(X-¹ = P(Z < − 2) By using the EXCEL, the required probability is, fx =NORM.S.DIST(-2, TRUE) E F P(X1640)=P(z < − 2) 2040-1800 80 H 1640-1800 80 1800-1800 80 -0.0228 x 100% = 2.28% Thus, the approximate percentage of buyers who paid less than $1640 is 2.28%. Solution The approximate percentage of buyers who paid(X) between $1560 and $1800 is 49.87%. The approximate percentage of buyers who paid more than $2040 is 0.14%. The approximate percentage of buyers who paid less than $1640 is 2.28%.
The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1800 and the standard deviation is $80. 1720 1800 1880 1960 2040 Distribution of Prices a What is the approximate percentage of buyers who paid between $1720 and $1880? 1560 1640 196 What is the approximate percentage of buyers who paid between $1640 and $1800? 196 What is the approximate percentage of buyers who paid between $1720 and $1800? 196 What is the approximate percentage of buyers who paid between $1560 and $1800? Question Hels 196 What is the approximate percentage of buyers who paid more than $2040? What is the approximate percentage of buyers who paid less than $1640?
Transcribed Image Text:The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1800 and the standard deviation is $80. 1720 1800 1880 1960 2040 Distribution of Prices a What is the approximate percentage of buyers who paid between $1720 and $1880? 1560 1640 196 What is the approximate percentage of buyers who paid between $1640 and $1800? 196 What is the approximate percentage of buyers who paid between $1720 and $1800? 196 What is the approximate percentage of buyers who paid between $1560 and $1800? Question Hels 196 What is the approximate percentage of buyers who paid more than $2040? What is the approximate percentage of buyers who paid less than $1640?
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