Scores on the Stanford-Binet Intelligence scale have a mean of 100 and a standard deviation of 16, and are presumed to be normally distributed. A person who scores 68 on this scale has what percentile rank within the population?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score. Convert to percentage.

z = (68-100)/16

z = -32/16 = -2
So would it be 95%?

Percentile rank is percent ≤ a score. Is that what you have?

It says to calculate using this:

Nk/100

So im not sure if i did it correctly....

To find the percentile rank of a score on the Stanford-Binet Intelligence scale, we need to calculate the z-score first. The z-score tells us how many standard deviations a particular score is away from the mean.

The formula to calculate the z-score is given by:
z = (x - μ) / σ

Where:
x is the individual score (68 in this case)
μ is the mean (100)
σ is the standard deviation (16)

Substituting the given values:
z = (68 - 100) / 16
z = -32 / 16
z = -2

Once we have the z-score, we can use a standard normal distribution table (also known as a Z-table) to determine the percentile rank.

By looking up the z-score of -2 in the Z-table, we find that the cumulative probability associated with it is approximately 0.0228. This means that the score of 68 corresponds to the 2.28th percentile.

In other words, the person who scores 68 on the Stanford-Binet Intelligence scale has a percentile rank of approximately 2.28% within the population.