Question 613683
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Hi, Previously Posted
What percentage of the area of the standard normal distribution is between z = -2.00 and z = +2.00? 
two standard deviations from the mean account for about 95.45%
 P(-2< z <2 ) = NORMSDIST(2) - NORMSDIST(-2) = .97725 - .02275 = .9545 OR 95.45%

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