SD is the sum of square of difference btw each score and average divided by (#experiments - 1). If you want to represent a range, then confidence interval estimation is what you need.
Cheers
Entrants to a school take an exam and their points are given the following importance for each respective area, with their average scores and standard deviations included. The averages included represent the scores of applicants admitted.
Biologia - 20% Avg. 68.68 SD 6.59
Filosofia - 5% Avg. 61.44 SD 7.63
Fisica - 10% Avg. 63.47 SD 10.0
Linea Interd. - 10% Avg. 59.12 SD 8.30
Lenguage - 15% Avg. 64.53 SD 8.30
Matematica - 20% Avg. 78.83 SD 11.30
Quimica - 10% Avg. 65.19 SD 6.49
Ciencias Sociales - 10% Avg. 60.12 SD 6.41
It's easy to find the average total score by multiplying each average by its percentage worth (i.e. biologia x .2) and adding the total. What I don't know how to do is find out the standard deviation of the average total score. In other words, what range of total scores represents the applicants who were admitted? Or with the given data is this not possible? Thanks so much
Hi Mike,
I've been reading on the central limit theorem, but am finding myself more confused trying to estimate the range of admitted scores. It makes sense intelectually but I don't know how to do the math.
I know that the average score admitted is 67.0435, based on doing the basic math of (.2)(68.68) + (.05)(61.44) + (.1)(63.47) + (.1)(59.12) + (.15)(64.53) + (.2)(78.83) + (.1)(65.19) + (.1)(60.12) from the percentage weights and scores provided. I also know the standard deviations of each of those individual scores that compose the total, and that 85 applicants were accepted. I just don't know how to put it all together to estimate the range of scores in admitted applicants.
From the formula on wikipedia, it seems like I take the sums of the totals, which would have to equal (85)(67.0435)=5698.6975 and subtract the mean by the sample size, so (85)(67.0435)=5698.6975 again, which makes zero. Then I apparently divide by sigma multiplied by the square root of n, 9.22. I have no idea what sigma stands for, and the top part of the equation seems to be zero. But if I did know what sigma was and the top wasn't zero, I'm not sure what exactly the solved equation gives me.
I'm so confused. I was thinking that maybe I could calculate the standard deviation of the entire score by multiplying them by the percentage weights like I did to get the average total score, which would give me a rough idea of at least where the vast majority of total scores fit. Would that work?