The average (arithmetic mean) of a normal distribution of a school's test scores is 65, and standard deviation of the distribution is 6.5. A student scoring a 78 on the exam is in what percentile of the school?

Her score is 2 standard deviations above the mean. Only 2.3% are higher.

Call it the 97th percentile

For a useful online tool for this type of question, see

http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html

To determine the percentile of a specific score in a normal distribution, we can use the Z-score formula. The Z-score represents how many standard deviations a particular score is away from the mean.

The formula for calculating the Z-score is:
Z = (X - μ) / σ

Where:
Z is the Z-score
X is the score you want to find the percentile for
μ is the mean of the distribution
σ is the standard deviation of the distribution

In this case, the mean (μ) of the distribution is 65, the standard deviation (σ) is 6.5, and we want to find the percentile for a score of 78.

First, calculate the Z-score:
Z = (78 - 65) / 6.5
Z = 13 / 6.5
Z = 2

Next, we can look up the percentile associated with a Z-score of 2 in a standard normal distribution table. A standard normal distribution table provides the area under the curve to the left of a given Z-score.

A Z-score of 2 corresponds to a percentile of approximately 97.72. Therefore, a student scoring 78 on the exam is in the 97.72th percentile of the school.