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?

Answer 1

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.

Answer 2

z = (68-100)/16
z = -32/16 = -2
So would it be 95%?

Answer 3

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

Answer 4

It says to calculate using this:

Nk/100

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