You may also like

problem icon

Very Old Man

Is the age of this very old man statistically believable?

problem icon

Into the Exponential Distribution

Get into the exponential distribution through an exploration of its pdf.

problem icon

Into the Normal Distribution

Investigate the normal distribution

Weekly Challenge 4: Top Marks

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

If the scores are $2$ points for a correct answer and $-1$ point for a wrong answer then the score is $100$ for $50$ correct, $97$ for $49$ correct and so on, giving possible scores of
$$0, 1, 4, 7, \dots, 97, 100$$

Suppose that 4 students take the test and score
 
$$ 91, 97, 100, 100$$
Sadia chooses mode, Tyler chooses median and Joseph chooses arithmetic mean.
 
Suppose that 5 students take the test and score
$$91, 94, 97, 100, 100$$
Sadia chooses mode, Joseph chooses median and Tyler chooses arithmetic mean.
 
In these two cases each student is correct.
 
 
Proof that there is no choice in the averages.
 
All students cannot score the same mark as then all three averages are the same, which is inconsistent with Tyler's position.
 
Sadia cannot then choose arithmetic mean, as this must be less than the maximum if all scores are not the same.
 
If Sadia chose the median then the median must be the maximum. If this is the case then at least half of the values equal the median, in which case Joseph must choose the arithmetic mean. However, this means that someone scores the mean, median and mode, which is inconsistent with Tyler's position. Therefore Sadia cannot choose the median.
 
Thus, if all three are to be simultaneously correct Sadia must always choose the mode, which must equal the highest score.
 
 
Thus, neither Joseph nor Tyler can then choose the mode, and they clearly cannot choose the same average.
 
Thus we have two cases:
 
1) Joseph chooses arithmetic mean and Tyler the median. 
2) Joseph chooses median and Tyler chooses arithmetic mean.
 
The median is always an achieved score for an odd number of students.

Thus, if we have an odd number of students we must select case 2) to have a chance of satisfying Tyler. In this case the median must be unique to satisfy Joseph.
 
If we have an even number of students Joseph cannot choose the median: the median is only an achieved score for an even number of students if more than one student scores the median mark. 
 
Thus, if we have an even number of students we must select case 1) to have a chance of satisfying Joseph. In this case the median must not be an achieved score to satisfy Tyler.



Note: This does not prove that the conditions can in fact be simultaneously met for various numbers of students; merely that there is no choice IF it is met.