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Question

A data set is normally distributed with the mean value $\mu = 62.00$ and standard deviation $\sigma = 12.00$. What percentage of data values are less than $x = 71.00$

Answer

z-score of $x=71.00 = z = \frac{x-\mu}{\sigma} = \frac{71.00 - 62.00}{12.00}=0.75$

The area under the standard normal distribution curve to the left of $z=0.75$ is $0.7734$

The percentage of data values that are less than $x=71.00$ is $77.34$