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$24$ students took a test and they each scored a mark from $0$ to $100$.

Gillian scored $71$.

$18$ students scored lower than her and $5$ students scored higher than her. Gillian is the $19$th mark.

At what percentile $P_n$ was her mark?

$$\frac{P_n}{100}24=N$$

For $N$ to round up to $19$, $\frac{P_n}{100}24$ must be a value between $18$ and $19$, but not including $18$ or $19$.

For $N=18$, $P_n=75$.

For $N=19$, $P_n=79.1\dot6$.

The answer must be between but not include $75$ and $79.1\dot6$.

In other words Gillian's score is at a range of percentiles.

My textbook gives an answer of $75$.

But $76$, $77$, $78$, $79$ and everything in between will all return a value of $N=19$.

How does one answer this kind of question? At what percentile was her mark?

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One double answer is: What rule does your (unstated) textbook specify? And -- if different -- what do your teachers expect? We can't answer either from your post.

A longer answer is that several rules can be found, for example under the heading of percentile rank. I suggest as starting points two linked principles for good rules:

  1. Low values and high values should be treated symmetrically and consistently. So the person with rank 1 out of 24 and the person with rank 24 of 24 would have percentile ranks that add to 100. A rank of 100/24 $\approx$ 4% for the lowest and of 2400/24 = 100% for the highest is not a procedure that satisfies this principle.

  2. Specifically, and consistent with #1, if the number of marks is odd then the middlemost mark -- say for the person with rank 12 out of 23, with 11 marks higher and 11 marks lower -- is assigned percentile rank 50% as being the median mark.

Several rules satisfy these principles, including 100 $\times$ (rank $-$ 0.5) / sample size and 100 $\times$ rank / (sample size $+$ 1). In statistics, these details are sometimes discussed under the heading of plotting positions, jargon for the precise cumulative probabilities to be used on quantile plots (historically often called probability plots).

Those two rules return 77% and 76% respectively for rank 19 out of 24, rounding to the nearest integer. For large sample sizes, results will be close.

Nothing said yet about tied ranks, but it is hard to defend any procedure except the same mark being given the same rank, conventionally by averaging to preserve the sum of ranks. Hence marks 42, 56, 56, 56, 98 would have been ranked 1, 2, 3, 4, 5 had the 56s been slightly different, so are assigned ranks 1, 3, 3, 3, 5 given that (2 + 3 + 4) / 3 $=$ 3.

In reporting marks to students or anyone else, there may well be a polite convention of rounding up, but that's politics and psychology, not statistics.

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