Percentages

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A data set is normally distributed with the *mean* value $\mu = 79.00$ and *standard deviation* $\sigma = 18.00$. What percentage of data values are less than $x = 79.00$

**Answer**

*z-score* of $x=79.00 = z = \frac{x-\mu}{\sigma} = \frac{79.00 - 79.00}{18.00}=0.00$

The area under the *standard normal distribution curve* to the left of $z=0.00$ is $0.5000$

The percentage of data values that are less than $x=79.00$ is $50.00$