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August 3rd, 2017, 09:06 AM  #1 
Member Joined: Mar 2017 From: Israel Posts: 30 Thanks: 2  Interquartile range
Hello Can you help me please about the following exercise: In a certain distribution, the interquartile range is zero. Therefrom, necessarily: a. At least half of the observations are equal. b. All observations are equal. c. Interquartile range can not be zero. d. The standard deviation is also zero. e. None of the answers is correct. I think the answer is b, because if Q1 (first quartile) = Q3 (third quartile), then the interquartile range is 0, but I'm not sure if I'm right. Thanks a lot! 
August 3rd, 2017, 09:31 AM  #2 
Senior Member Joined: Oct 2009 Posts: 142 Thanks: 60 
What is the IQR of $$1,2,2,2,2,2,2,2,2,2,2,2,2,2,10$$ 
August 3rd, 2017, 09:43 AM  #3 
Member Joined: Mar 2017 From: Israel Posts: 30 Thanks: 2  
August 3rd, 2017, 10:26 AM  #4 
Member Joined: Mar 2017 From: Israel Posts: 30 Thanks: 2  Oh sorry, I was confused. So if i understood, the answer is e, because a and b aren't true because there are different cases which IQR can be zero, c is not true because IQR can be zero, d is not true because there is no connection between IQR to standard deviation, so the answer is e. Right? Last edited by IlanSherer; August 3rd, 2017 at 10:31 AM. 
August 3rd, 2017, 10:30 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 803 Thanks: 320 
Let the number of observations = 4n + k, where k = 0, 1, 2, or 3. Let a, b, c, and d be the least elements in the first, second, third, and fourth quartiles respectively. Let w, x, y, and z be the greatest elements in the first, second, third, and fourth quartiles respectively. $\therefore a \le w \le Q_1 \le b \le x \le Q_2 \le c \le y \le Q_3 \le d \le z.$ $\therefore a \le w \le Q_1 =b = x = Q_2 = c = y = Q_3 \le d \le z.$ If k = 0, then there are n elements in each quartile. So at least half the observations are equal. If k = 1, then we must assign one quartile n + 1 elements. If that assignment is made to the second or third quartiles, more than half the elements are equal. If that assignment is to the first quartile, then $Q_1 = \dfrac{w + b}{2} = b \implies w + b = 2b \implies w = b.$ In that case the number that are equal is at least 2n + 1, which is greater than 4n / 2. But the same thing happens if the extra element is assigned to fourth quartile. So you can work out what happens if k = 2 or 3. Lots of cases, but we get to answer a in the end (I presume because I have not worked it out all the way). 
August 3rd, 2017, 10:37 AM  #6  
Member Joined: Mar 2017 From: Israel Posts: 30 Thanks: 2  Quote:
But the situation can be 1,1,1,1,1,1,2,2,2,2  this case belongs to at least half of the observations are equal, doesn't it? and IQR is 1. Last edited by IlanSherer; August 3rd, 2017 at 10:40 AM.  
August 3rd, 2017, 10:40 AM  #7 
Senior Member Joined: May 2016 From: USA Posts: 803 Thanks: 320 
Perhaps my previous post was not elegant enough. If the number of observations is not evenly divisible by 4, there may be a way to show that either $w = b$ or $d = y$ or both without going through a whole bunch of cases.

August 3rd, 2017, 10:45 AM  #8  
Senior Member Joined: May 2016 From: USA Posts: 803 Thanks: 320  Quote:
If half the people in the group are mothers, at least half are necessarily women. If half the people in the group are women, perhaps none are mothers. Order is important.  
August 3rd, 2017, 10:49 AM  #9 
Senior Member Joined: May 2016 From: USA Posts: 803 Thanks: 320 
I do not like this problem because there are different ways to calculate medians when the number of elements is odd. Without some specification on which method is to be used, the problem becomes very complex.

August 3rd, 2017, 12:18 PM  #10  
Member Joined: Mar 2017 From: Israel Posts: 30 Thanks: 2  Quote:
 

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