A distribution of scores is approximately normal with a mean of 78 and a standard deviation of 8.6. Which of the following equations can be used to find the score \(x\) above which 33 percent of the scores fall?

Summary Submit Skip Question

The z-score that corresponds to the 33rd percentile is 0.44. Using the equation for the standardized test statistic:

$$ z = \frac{x-\mu}{\sigma} $$ $$ 0.44 = \frac{x-78}{8.6} $$