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A data set is normally distributed with the mean value $\mu = 64.00$ and standard deviation $\sigma = 9.00$. What percentage of data values are less than $x = 47.00$


z-score of $x=47.00 = z = \frac{x-\mu}{\sigma} = \frac{47.00 - 64.00}{9.00}=-1.89$

The area under the standard normal distribution curve to the left of $z=-1.89$ is $0.0294$

The percentage of data values that are less than $x=47.00$ is $2.94$