My Math Forum Question About Median

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 July 4th, 2017, 08:26 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,652 Thanks: 120 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 dead-end after that. Please someone help me.
 July 4th, 2017, 10:11 PM #2 Senior Member   Joined: May 2016 From: USA Posts: 881 Thanks: 353 $b > d \implies b = d + x,\ x > 0.$ $a + d > b + c \implies a + d = b + c + y,\ y > 0.$ $d > 2b - a \implies d = 2b - a + z,\ z > 0.$ $a + d = b + c + y = d + x + c + y \implies$ $a = c + x + y \implies a > c.$ $d = 2b - a + z = 2d + x + x - a + z \implies$ $a = d + x + x + z = b + x + z \implies a > b > d.$ $a + d = b + c + y \implies b + x + z + d = b + c + y \implies$ $d = c + y - x - z.$ $\therefore y - x - z \ge 0 \implies d \ge c \implies a > b > d \ge c \implies$ $median = \dfrac{b + d}{2} = \dfrac{d + x + d}{2} = d + \dfrac{x}{2}.$ $\text {But } y - x - z < 0 \implies d < c \implies a > b > d \text { and } a > c > d \implies$ $median = \dfrac{b + c}{2} = \dfrac{d + x + d - y + x + z}{2} = d + x + \dfrac{z - y}{2}.$ $median  July 5th, 2017, 12:06 AM #3 Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,652 Thanks: 120 Math Focus: Trigonometry and Logarithm Can we determine the median's exact number? July 5th, 2017, 05:52 AM #4 Senior Member Joined: May 2016 From: USA Posts: 881 Thanks: 353 Quote:  Originally Posted by Monox D. I-Fly Can we determine the median's exact number? Not that I can see. We only have three inequations and four unknowns. If there were sufficient additional independent pieces of information, we could do so. It is impossible to say how many additional pieces are sufficient because inequations give less information than do equations. In this case, we do not even know whether the scores are all non-negative or whether they are all integers. Let's alter the problem slightly by substituting equations for inequations.$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 \impliesa = 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, 06:15 PM #5 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,652 Thanks: 120 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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