Question

Julian’s z-core was 1.39 on the mcat. if the mean was 33 and the standard deviation was 16, and the distribution was normal, what percentage of students scored above julian?

Answer 1

To find the percentage of students that scored above Julian on the MCAT, we need to calculate the area under the normal curve to the right of Julian's z-score.

Step-by-step explanation:

To find the percentage of students that scored above Julian, we need to calculate the area under the normal curve to the right of Julian's z-score.

The z-score formula is z = (x - mean) / std_dev.

Substituting the given values into the formula, we get z = (1.39 - 33) / 16 = -1.9625.

Using a standard normal distribution table or calculator, we can find that the area to the left of -1.9625 is approximately 0.0241.

Therefore, the percentage of students that scored above Julian is 100% - (0.0241 * 100) = 97.59%.