There are $450$ students in our class. I have to choose the students who are ranked in the $80$th percentile or above.

Question

There are $450$ students in our class. I have to choose the students who are ranked in the $80$th percentile or above.

Which measure of average would be the most meaningful – mean, median, mid-range, or mode? Do we at all need any measure of average?

can anyone please help me ?

Answer 1

First, you need to decide what kind of score you will use to represent a student's accomplishments in a class. That score could be based on the mean of three exams, the median of five exams, the highest score among four exams, or whatever the instructor thinks is the best measure of student performance.

Then you will have $n = 450$ scores. Sort then from smallest to largest. Roughly speaking, the 80th percentile will lie at the the $360$th of the sorted scores. The top $90$ scores will be at or above the 80th percentile.

If scores are integers, then it is possible to have several sorted scores tied at the 89th percentile. If that happened, you would have to decide whether to include all or none of the students with these tied scores. So if you know in advance you will be interested in percentiles, it is best not to round scores to be integers.

Example in R without rounding:

set.seed(223)
x = 100*rbeta(450, 7, 3)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  17.45   59.79   70.72   68.75   79.44   96.30 
quantile(x, .8)
     80% 
82.22081 

Here is a stripchart of all 458 scores with a vertical red line at the 80th percentile:

stripchart(x, pch="|")
abline(v = 82.22, col="red")

The plot is somewhat congested around the 80th percentile, so let's take a detailed look at the actual values in the vicinity of the 90th percentile. We begin by sorting the 450 scores from smallest to largest. Then we print scores in positions 355 to 365.

x.s = sort(x)
x.s[355:365]
[1] 81.84596 81.98908 82.17073 82.17978 82.21221 82.22057
[7] 82.22177 82.23840 82.32506 82.38705 82.59776

The 80th percentile $82.22081$ is between scores $82.22057$ and $82.221777.$ There are no ties, so there is no ambiguity about drawing the line between the lower 80% and the upper 20% of the sample. [Caution: Various statistical computer programs and textbooks have slightly different ways of giving an exact value at the 'borderline' between scores of rank 360 and 361, but that usually makes no practical difference in interpretation.]

With rounding to integers:

Call the rounded scores y and the sorted rounded scores y,s:

y.r = round(x.s)
summary(y.s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  17.00   60.00   71.00   68.77   79.00   96.00 
quantile(y.s, .80)
80% 
 82

stripchart(y.s, ylim=c(-.2,20), meth="stack", offset=1/6, pch="-")
 abline(v=82, col="red")

Now, the 80th percentile is at rounded score $82.$ However, several students have rounded score $82.$ To be precise, there are 14 scores tied at $82.$

sum(y.s == 82)
[1] 14

y.s[350:365]
[1] 81 82 82 82 82 82 82 82 82 82 82 82 82 82 82 83

So we have a choice to include or exclude score $82$ among the "top 20%". Inclusion leads to 80.89% below and exclusion leads to 77.78% below.

mean(y.s < 82);  mean(y.s <= 82)
[1] 0.7777778
[1] 0.8088889

In the grand scheme of things the difference may not matter. But if the "top 20%" get A's and those just below get B's, it will matter to the 14 students tied at the borderline. So it is a good idea to realize there there may be ties at the borderline and to have a clear policy how to handle them.

Answer 2

If you're actually running a class

The median would be most helpful to understand. Let's suppose you have homework assignments (30%), quizzes (5%), midterms (25%), and finals (40%) for students and you weighted those to come up with a score for each student $S$ between 0 and 100. Now you sort the scores and find the scores that are at the 225th and 226th. These are the medians. Or you could take the average of them which is common. That is the median score. It is the 50th percentile score. Anyone who scored above the median is in the top 50th percentile. Now do the same thing for the 80th percentile. You have 450 students: $450*.8=360$ so take the average of the 360th and 361th score to find the 80th percentile score. Anyone who scored above that score is in the top 80th percentile of the class.