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 = (68-100)/16
z = -32/16 = -2
So would it be 95% or 62%??
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score. Remember that it is below the mean, and percentile rank is % ≤ that score.
To find the percentile rank of a person who scores 68 on the Stanford-Binet Intelligence scale, you need to calculate the z-score and then use a standard normal distribution table to find the corresponding percentile.
The z-score formula is:
z = (x - mean) / standard deviation
Let's substitute the values to calculate the z-score:
z = (68 - 100) / 16
z = -32 / 16
z = -2
Now, to find the percentile rank, you need to find the area to the left of the z-score in the standard normal distribution table. Since the z-score is negative, you will be looking for the area in the left tail.
Looking up the absolute value of the z-score (-2), in a standard normal distribution table, you would find that the area to the left of -2 is approximately 0.02275.
To convert this area to a percentile, subtract the area from 1, and then multiply by 100.
Percentile rank = (1 - 0.02275) * 100
Percentile rank = 0.97725 * 100
Percentile rank = 97.725%
Therefore, a person who scores 68 on the Stanford-Binet Intelligence scale has a percentile rank of approximately 97.725%, or you can round it to 98% for simplicity.