Probabilities of test result scores A Big 5 personality test has five orthogonal normally distributed scales. 
What I would like to know is the likelihood of exactly zero, one, two, three, four or five scores outside of the median. E.g how likely is it that a test taker has one score higher than one standard deviation above the mean, and one score less than one standard deviation above the mean?
This requires a firmer grasp of probability calculus than I have, so I simulated using R:
mydata <- matrix(rnorm(50000,0,1),ncol=5)

res <- apply(abs(mydata) > 1,1,which)

res <- sapply(0:5,function(X){length(which(sapply(res,length)==X))})/100

names(res) <- c(0:5)

pie(res)


I don't feel that this makes any sense. The likelihood of exactly one and exactly two scores outside the mean is the same? 
Any and all suggestions or comments appreciated. Thank you.
 A: If the scores are approximately normally distributed, that implies a probability of 0.68 that a given score falls within one standard deviation of the median and, so, a probability of 0.32 that a given score lies more than one standard deviation away from the median (above or below).
Assuming the scores are independent of one another, we can use those values to derive the quantities you're after. The probability that all five scores are within 1 s.d. is the product of the probabilities of the five independent events, so 0.68^5, or 0.145.
The probability of exactly one score lying more than 1 s.d. away from the median is the product of the probability that four lie within 1 s.d. (0.68^4) and the probability of a single score landing more than 1 s.d. away (0.32), which is 0.68^4 * 0.32 = 0.068. But there are five ways that can happen (once for each question), so the total probability of any single answer being more than 1 s.d. away from the median is 0.34.
Now you start to see the pattern.


*

*The probability that any pair of scores falls more than 1 s.d. away while the others land within that range is 0.68^3 * 0.32^2, or 0.032, times the 10 ways that can happen (AB vs. CDE, AC vs. BDE, etc.), which equals 0.322.

*The probability that any trio of scores fall more than 1 s.d. away is 0.68^2 * 0.32^3, or 0.015, times the 10 ways that can happen, which is 0.152.

*The probability for a quartet of scores is 0.68 * 0.32^4, and there are five ways that can happen, which produces 0.036.

*The probability that all five scores are at least 1 s.d. away is 0.32^5, which is 0.003.


So I can't tell exactly what values lay behind your pie chart, but there are two slices that are approximately equal: exactly one and exactly two. That may seem counterintuitive, but then remember that there are twice as many ways to get pairs of outlier answers as there are ways to get single outliers, and then I think it makes a little more sense.
