Percentages

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A data set is normally distributed with the *mean* value $\mu = 68.00$ and *standard deviation* $\sigma = 8.00$. What percentage of data values are less than $x = 83.00$

**Answer**

*z-score* of $x=83.00 = z = \frac{x-\mu}{\sigma} = \frac{83.00 - 68.00}{8.00}=1.88$

The area under the *standard normal distribution curve* to the left of $z=1.88$ is $0.9699$

The percentage of data values that are less than $x=83.00$ is $96.99$