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July 4th, 2017, 07:26 PM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,799 Thanks: 122 Math Focus: Trigonometry and Logarithm  Question About Median
In a class, Budi's score is greater than Doni's. The sum of Adi's and Doni's scores is greater than the sum of Budi's and Coki's scores. Meanwhile, Doni's score is greater than two times Budi's score substracted by Adi's score. Determine the median of those four students' scores. All I know, was, by using their initials that: B > D A + D > B + C D > 2B  A And by using the second and third info I got that their score from lowest to highest is either C, D, B, A or D, B, C, A. However, I met a deadend after that. Please someone help me. 
July 4th, 2017, 09:11 PM  #2 
Senior Member Joined: May 2016 From: USA Posts: 916 Thanks: 366 
$b > d \implies b = d + x,\ x > 0.$ $a + d > b + c \implies a + d = b + c + y,\ y > 0.$ $d > 2b  a \implies d = 2b  a + z,\ z > 0.$ $a + d = b + c + y = d + x + c + y \implies$ $a = c + x + y \implies a > c.$ $d = 2b  a + z = 2d + x + x  a + z \implies$ $a = d + x + x + z = b + x + z \implies a > b > d.$ $a + d = b + c + y \implies b + x + z + d = b + c + y \implies$ $d = c + y  x  z.$ $\therefore y  x  z \ge 0 \implies d \ge c \implies a > b > d \ge c \implies$ $median = \dfrac{b + d}{2} = \dfrac{d + x + d}{2} = d + \dfrac{x}{2}.$ $\text {But } y  x  z < 0 \implies d < c \implies a > b > d \text { and } a > c > d \implies$ $median = \dfrac{b + c}{2} = \dfrac{d + x + d  y + x + z}{2} = d + x + \dfrac{z  y}{2}.$ $median 
July 4th, 2017, 11:06 PM  #3 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,799 Thanks: 122 Math Focus: Trigonometry and Logarithm 
Can we determine the median's exact number?

July 5th, 2017, 04:52 AM  #4 
Senior Member Joined: May 2016 From: USA Posts: 916 Thanks: 366  Not that I can see. We only have three inequations and four unknowns. If there were sufficient additional independent pieces of information, we could do so. It is impossible to say how many additional pieces are sufficient because inequations give less information than do equations. In this case, we do not even know whether the scores are all nonnegative or whether they are all integers. Let's alter the problem slightly by substituting equations for inequations. $b = d + 25,\ a + d = b + c,\ d = 2b  a + 5.$ $\therefore a + d = d + 25 + c \implies a = c + 25.$ $d = 2b  a + 5 \implies d = 2d + 50  a + 5 \implies$ $a = d + 45 = b + 20 \implies a > b > c > d.$ In other words, just by substituting equations for inequations we can determine the order of the four unknowns, but not their respective values. We gain information from moving to equations from inequations, but not enough to compute a numeric value for the median, which equals $a  22.5.$ 
July 5th, 2017, 05:15 PM  #5 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,799 Thanks: 122 Math Focus: Trigonometry and Logarithm 
Thank you everyone who has helped me with this question. I have asked this question in 4 different forums including this one and all confirm that it is impossible to find the exact number of the median's value.


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