Dorian's score of 558 on the writing portion of the 2009 SAT exam placed him in the 72nd percentile. If the mean on this exam was 493, what was the standard deviation?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.72) and its Z score. Insert the Z score and other data in the equation above and solve for SD.
To find the standard deviation, we first need to convert Dorian's score from the percentile to a z-score. The z-score represents the number of standard deviations that a value is away from the mean.
The percentile tells us that Dorian scored higher than 72% of the students who took the exam. This means that his z-score is the value that corresponds to the 72nd percentile on a standard normal distribution.
To find the z-score, we can use a z-score table or a calculator. The z-score corresponding to the 72nd percentile is approximately 0.55.
The z-score formula is: z = (x - μ) / σ
Where:
- z is the z-score
- x is the value of interest (Dorian's score)
- μ is the population mean
- σ is the population standard deviation (what we need to find)
Rearranging the formula, we get: σ = (x - μ) / z
Plugging in the values, we have:
x = 558 (Dorian's score)
μ = 493 (mean)
z = 0.55 (z-score)
Substituting these values, we get:
σ = (558 - 493) / 0.55
Calculating this, we find that the standard deviation (σ) is approximately 119. Therefore, the standard deviation of the writing portion of the 2009 SAT exam is 119.