Science:Math Exam Resources/Courses/MATH103/April 2009/Question 01 (d)

MATH103 April 2009

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Question 01 (d)
Multiple Choice question: Select ONE correct answer. You will not be graded for any work. A Math 103 class of 100 students took a test. The following graph shows how many students' marks were in each of the intervals 1-10, 11-20, ..., 91-100. What was (approximately) the median? (a) 50 (b) 60 (c) 70 (d) 80 (e) 90

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

For a random variable $\ X$ the median $\ m$ is defined as the value in the middle when all of the numbers $a_{1},a_{2},\dots ,a_{100}$ lay sorted in front of you: $a_{1}\leq a_{2}\leq a_{3}\leq \dots \leq a_{49}\leq \underbrace {a_{50}} _{=m}\leq a_{51}\leq \dots \leq a_{99}\leq a_{100}.$ This means for us: Find a number such that half of the students have less points and half of the students have more points than this number. E.g. the median of 1, 1, 2, 3, 17 would be 2.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. In total we have 100 students. So we need to find the number of points such that 50 students have more and 50 students have less points. We line up the marks of the students: $\ \underbrace {\underbrace {20,20,20,20,20} _{5{\text{times}}},\underbrace {30,30,\dots ,30} _{15{\text{ times}}},\underbrace {40,\dots ,40} _{10{\text{ times}}},\underbrace {50,\dots ,50} _{15{\text{ times}}},\underbrace {60,\dots ,60} _{5{\text{ times}}}} _{50{\text{ students}}},\underbrace {70,\dots ,70} _{15{\text{ times}}},\dots$ Unfortunately we end up between 60 and 70. So, we need to think closer. The column at the figure between 20 and 30 marks means, that the student got an amount of marks in the interval $\ [20,30]$. So, the students in the column between 60 and 70 got an amount of marks between 60 and 70. The students in the column between 70 and 80 got an amount of marks between 70 and 80. So, the 70 lies exaclty in the middle. Hence the mean is about 70 and answer c) is correct.

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