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QuestionA data set is normally distributed with the
mean value $\mu = 69.00$ and
standard deviation $\sigma = 14.00$. What percentage of data values are less than $x = 74.00$
Answerz-score of $x=74.00 = z = \frac{x-\mu}{\sigma} = \frac{74.00 - 69.00}{14.00}=0.36$
The area under the
standard normal distribution curve to the left of $z=0.36$ is $0.6406$
The percentage of data values that are less than $x=74.00$ is $64.06$