Weekly Challenge 4: Top marks

7032 /www/nrich/html/content/id/7032/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="buttonBar"> <li><a href="http://nrich.maths.org/6149&part=">Try this next</a></li> <li><a href="http://plus.maths.org/issue41/features/parker/index.html">Read all about it</a></li> <li><a href="http://nrich.maths.org/6267&part=">Warm-up problem</a></li> <li><a href="http://nrich.maths.org/7052&part=solution">Last week's solution</a></li> </ul> 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 two cases that there are an even number of examinees and 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. <div class="framework">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 arguaments are of the form: IF the following assumptions hold THEN the following is true.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This challenge uses the mathematics of averaging and expectation, commonly encountered in S1 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. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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. </mdoxml> 2 4 0 0 0 0 1 Weekly Challenge 4: Top marks A weekly challenge concerning statistics and averaging. Collections Weekly Challenge Advanced Probability and Statistics Statistics generally Stage 5 Statistics Mapping Document SM - Statistical Modelling