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Analyze the figure below. All of the choices are true EXCEPT for one. Identify the exception.* 34.13% -13.59% 2.28% +1 +2 Given its symmetrical nature, the percentages on the right side are exactly the same on the left side. Mean is O and the standard deviation is 1. Looking at the values of the mean and the standard deviation, it is likely that we are dealing with a z-score. Using z-scores implies that the percentages shown cannot be applied to all normal distribution, it woulid depend on the data values.

Analyze the figure below. All of the choices are true EXCEPT for one. Identify the exception.* 34.13% -13.59% 2.28% +1 +2 Given its symmetrical nature, the percentages on the right side are exactly the same on the left side. Mean is O and the standard deviation is 1. Looking at the values of the mean and the standard deviation, it is likely that we are dealing with a z-score. Using z-scores implies that the percentages shown cannot be applied to all normal distribution, it woulid depend on the data values.

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
Problem 1P
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Analyze the figure below. All of the choices are true EXCEPT for one. Identify the exception. * 34.13% -13.59% 2.28% +1 +2 Given its symmetrical nature, the percentages on the right side are exactly the same on the left side. Mean is O and the standard deviation is 1. Looking at the values of the mean and the standard deviation, it is likely that we are dealing with a z-score. Using z-scores implies that the percentages shown cannot be applied to all normal distribution, it woulid depend on the data values.
Transcribed Image Text:Analyze the figure below. All of the choices are true EXCEPT for one. Identify the exception. * 34.13% -13.59% 2.28% +1 +2 Given its symmetrical nature, the percentages on the right side are exactly the same on the left side. Mean is O and the standard deviation is 1. Looking at the values of the mean and the standard deviation, it is likely that we are dealing with a z-score. Using z-scores implies that the percentages shown cannot be applied to all normal distribution, it woulid depend on the data values.
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