Top marks
Can you make all of these statements about averages true at the same time?
Problem
Some students sit an examination with $50$ compulsory multiple choice questions, scoring $+2$ for a correct answer and $-1$ for an incorrect answer, with a minimum score of zero for the overall test.
Three of the students are discussing the possible marks:
Tyler says "Nobody will score the average mark".
Sadia says "Nobody will score higher than the average mark".
Joseph says "I will be the only person to score the average mark."
Each student chooses their own definition of average from arithmetic mean, median and mode.
Can you create a set of scores, and choices of average for which they are all simultaneously correct, in the following two cases:
there are an even number of examinees?
there are an odd number of examinees?
Prove, in the two cases even/odd, that the 'choices' of averages made by Tyler, Sadia and Joseph are fixed, if it is possible that they are simultaneously correct.
Extension: Consider whether it is always possible simultaneously to meet these conditions for any number of students.
Did you know ... ?
There are always many underlying assumptions in statistical modelling. A good statistician is very aware of the need for clarity in making statistical statements and good statistical arguments are of the form: IF the following assumptions hold THEN the following is true.
Getting Started
This challenge uses the mathematics of averaging and expectation, commonly encountered in post-16 courses. There is an expectation of greater rigour in thinking than will be encountered at GCSE.
This mathematics follows on from mean, mode and median calculations.
Student Solutions
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.
Teachers' Resources
Why do this problem?
This problem encourages students to think very carefully about the definitions of mean, median and mode and how they relate to one another. It follows on from some of the ideas raised in Statistical Shorts and Stats Statements.
Possible approach
Share the problem in stages, starting with the format of the test. Have students work out possible scores and deal with any misconceptions about scoring.
Then examine each of the students' statements in turn and discuss how it might be possible for the statement to be true (or why it might not be possible) for each of the possible averages. The issue of even numbers of scores and odd numbers of scores in relation to the median should arise naturally. It may be preferable for these discussions to take place in pairs or small groups to ensure all students reach conclusions independently with fully formed reasoning rather than just accepting a statement made by someone else.
After some initial thinking time about the requirements, it should be possible for students to construct sets of scores which meet the criteria individually or in pairs. They might then swap these sets of scores with another pair who can check whether the conditions are met and by which choice of average for each student named in the problem.
Key questions
Is it possible for "Nobody to have scored the average mark" if the average is the
- mean
- median
- mode?
If it is possible, do any other conditions need to be met?
Is it possible for "Nobody to have scored higher than the average mark" if the average is the
- mean
- median
- mode?
If it is possible, do any other conditions need to be met?
Is it possible for "Only one person to score the average mark" if the average is the
- mean
- median
- mode?
If it is possible, do any other conditions need to be met?
Possible support
If students are struggling with constructing actual sets of scores, they could be encouraged to use as few scores as possible. Three students are named in the problem, so there must be at least three scores. But having only three scores will imply that the mode and the median are the same, which will limit their options if not make it impossible to fulfil the requirements. (They don't need to prove that at this stage, but could be directed towards more than three scores, so ideally four and five as even and odd cases.)
They can also be encourgaed to start with the top scores and work down, which helps when attempting to meet Sadia's requirement.