The probability that a normal random variable has an outcome within 2.5 standard deviations of the mean is 98.76%
Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \sigma=standard\ deviation[/tex]
Given an outcome within 2.5 standard deviations of the mean. Hence:
P(-2.5 < z < 2.5) = P(z < 2.5) - P(z < -2.5) = 0.9938 - 0.0062 = 98.76%
The probability that a normal random variable has an outcome within 2.5 standard deviations of the mean is 98.76%
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