Dtjytyk

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Sample-quartile

sample variance, co-variance and correlation coefficientAsymptotic correlation between sample mean and sample medianKnowing that the weight corresponding to the third quartile […] calculate the value of μ and the value of σ.Analysis of IQ scores given mean, median, sd, quartilesWhat to call & how to compute errors in a very asymmetric sampleWhat is the third quartile when there are no data above the median?box and whiskers plotBasic Quantile CalculationLower/Upper quartile problems(Histogram) Why is the lower quartile in interval 36-43?

0

$begingroup$

I don't know : Is there a sample such that the mean does not lie between the lower and upper quartile?
Is there a sample such that the median does not lie between the lower and upper quartile?

statistics median

share|cite|improve this question

asked 2 days ago

Alicja CAlicja C

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hello and welcome to math.stackexchange. What have you tried so far? What are your thoughts?
  $endgroup$
  – Hans Engler
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that second is right but i don't know how to proof this. And I think the first is not good but i didn't find an example. :/
  $endgroup$
  – Alicja C
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: Consider the sequences $1,2,3,4,5,10^{10^{10}}$ and $1,1,1,1,1,1$
  $endgroup$
  – Brian
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the second question: Remember that the quartiles and the median are all examples of quantiles of a sample. Therefore they are ordered in a very specific way.
  $endgroup$
  – Hans Engler
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey @BrianS , this is a dumb question but... If I said "Pick a number between 1 and 1", does that mean there are no numbers to pick or that I could pick 1?
  $endgroup$
  – randomgirl
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

I don't know : Is there a sample such that the mean does not lie between the lower and upper quartile?
Is there a sample such that the median does not lie between the lower and upper quartile?

statistics median

share|cite|improve this question

asked 2 days ago

Alicja CAlicja C

62

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hello and welcome to math.stackexchange. What have you tried so far? What are your thoughts?
  $endgroup$
  – Hans Engler
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think that second is right but i don't know how to proof this. And I think the first is not good but i didn't find an example. :/
  $endgroup$
  – Alicja C
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: Consider the sequences $1,2,3,4,5,10^{10^{10}}$ and $1,1,1,1,1,1$
  $endgroup$
  – Brian
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the second question: Remember that the quartiles and the median are all examples of quantiles of a sample. Therefore they are ordered in a very specific way.
  $endgroup$
  – Hans Engler
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey @BrianS , this is a dumb question but... If I said "Pick a number between 1 and 1", does that mean there are no numbers to pick or that I could pick 1?
  $endgroup$
  – randomgirl
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I don't know : Is there a sample such that the mean does not lie between the lower and upper quartile?
Is there a sample such that the median does not lie between the lower and upper quartile?

statistics median

share|cite|improve this question

asked 2 days ago

Alicja CAlicja C

62

$endgroup$

share|cite|improve this question

asked 2 days ago

Alicja CAlicja C

62

share|cite|improve this question

asked 2 days ago

Alicja CAlicja C

62

asked 2 days ago

Alicja CAlicja C

62

asked 2 days ago

Alicja CAlicja C

62

1 Answer
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0

$begingroup$

Is there a sample such that the mean does not lie between the lower and upper quartile?

One such sample is ${1,2,3,4,5,120}$, which has quartiles $Q_1=2$ and $Q_3=5$, but a mean of $22.5$. Note that the mean of a data set is heavily influenced by outliers while the quartiles are generally resistant to extreme values.

Is there a sample such that the median does not lie between the lower and upper quartile?

Consider the definitions of the median and quartiles. The first quartile of a data set separates the bottom $25%$ of observations from the top $75%$. Similarly, the third quartiles separates the bottom $75%$ of observations from the top $25%$. As noted in the comments, the median is simply the "second quartile" of a data set and separates the bottom $50%$ from the top $50%$. If there was a data set that had a median less than the first quartile, then the observation at the $50$th percentile would be less than an observation at the $25$th percentiles, which makes no sense. Similar logic shows that the median cannot be greater than the third quartile.

This can be shown more rigorously by providing a formal definition for the median of a data set. We can define median $tilde x$ of a set of $n$ observations by
$$
tilde x = frac{1}{2} left(a_{lceil n/2 rceil} + a_{lceil (n/2)+1 rceil} right)
$$
where $a_i$ is the $i$th element of the order sequence of observations and $lceil cdot rceil:mathbb R to mathbb Z$ is the ceiling function. The first quartiles is median of the sequence $a_1,...,a_{lceil n/2 rceil}$ and the third quartiles is the median of the sequence $a_{lceil (n/2)+1 rceil},...,a_n$. Hence, we have the ordering $Q_1 leq tilde x leq Q_3$, meaning that the median can equal the first or third quartiles, but cannot be less than the first nor more than the third.

share|cite|improve this answer

answered 2 days ago

BrianBrian

35812

New contributor

Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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add a comment | 

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0

$begingroup$

Is there a sample such that the mean does not lie between the lower and upper quartile?

One such sample is ${1,2,3,4,5,120}$, which has quartiles $Q_1=2$ and $Q_3=5$, but a mean of $22.5$. Note that the mean of a data set is heavily influenced by outliers while the quartiles are generally resistant to extreme values.

Is there a sample such that the median does not lie between the lower and upper quartile?

Consider the definitions of the median and quartiles. The first quartile of a data set separates the bottom $25%$ of observations from the top $75%$. Similarly, the third quartiles separates the bottom $75%$ of observations from the top $25%$. As noted in the comments, the median is simply the "second quartile" of a data set and separates the bottom $50%$ from the top $50%$. If there was a data set that had a median less than the first quartile, then the observation at the $50$th percentile would be less than an observation at the $25$th percentiles, which makes no sense. Similar logic shows that the median cannot be greater than the third quartile.

This can be shown more rigorously by providing a formal definition for the median of a data set. We can define median $tilde x$ of a set of $n$ observations by
$$
tilde x = frac{1}{2} left(a_{lceil n/2 rceil} + a_{lceil (n/2)+1 rceil} right)
$$
where $a_i$ is the $i$th element of the order sequence of observations and $lceil cdot rceil:mathbb R to mathbb Z$ is the ceiling function. The first quartiles is median of the sequence $a_1,...,a_{lceil n/2 rceil}$ and the third quartiles is the median of the sequence $a_{lceil (n/2)+1 rceil},...,a_n$. Hence, we have the ordering $Q_1 leq tilde x leq Q_3$, meaning that the median can equal the first or third quartiles, but cannot be less than the first nor more than the third.

share|cite|improve this answer

answered 2 days ago

BrianBrian

35812

New contributor

Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

Is there a sample such that the mean does not lie between the lower and upper quartile?

One such sample is ${1,2,3,4,5,120}$, which has quartiles $Q_1=2$ and $Q_3=5$, but a mean of $22.5$. Note that the mean of a data set is heavily influenced by outliers while the quartiles are generally resistant to extreme values.

Is there a sample such that the median does not lie between the lower and upper quartile?

Consider the definitions of the median and quartiles. The first quartile of a data set separates the bottom $25%$ of observations from the top $75%$. Similarly, the third quartiles separates the bottom $75%$ of observations from the top $25%$. As noted in the comments, the median is simply the "second quartile" of a data set and separates the bottom $50%$ from the top $50%$. If there was a data set that had a median less than the first quartile, then the observation at the $50$th percentile would be less than an observation at the $25$th percentiles, which makes no sense. Similar logic shows that the median cannot be greater than the third quartile.

This can be shown more rigorously by providing a formal definition for the median of a data set. We can define median $tilde x$ of a set of $n$ observations by
$$
tilde x = frac{1}{2} left(a_{lceil n/2 rceil} + a_{lceil (n/2)+1 rceil} right)
$$
where $a_i$ is the $i$th element of the order sequence of observations and $lceil cdot rceil:mathbb R to mathbb Z$ is the ceiling function. The first quartiles is median of the sequence $a_1,...,a_{lceil n/2 rceil}$ and the third quartiles is the median of the sequence $a_{lceil (n/2)+1 rceil},...,a_n$. Hence, we have the ordering $Q_1 leq tilde x leq Q_3$, meaning that the median can equal the first or third quartiles, but cannot be less than the first nor more than the third.

share|cite|improve this answer

answered 2 days ago

BrianBrian

35812

New contributor

Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Is there a sample such that the mean does not lie between the lower and upper quartile?

One such sample is ${1,2,3,4,5,120}$, which has quartiles $Q_1=2$ and $Q_3=5$, but a mean of $22.5$. Note that the mean of a data set is heavily influenced by outliers while the quartiles are generally resistant to extreme values.

Is there a sample such that the median does not lie between the lower and upper quartile?

Consider the definitions of the median and quartiles. The first quartile of a data set separates the bottom $25%$ of observations from the top $75%$. Similarly, the third quartiles separates the bottom $75%$ of observations from the top $25%$. As noted in the comments, the median is simply the "second quartile" of a data set and separates the bottom $50%$ from the top $50%$. If there was a data set that had a median less than the first quartile, then the observation at the $50$th percentile would be less than an observation at the $25$th percentiles, which makes no sense. Similar logic shows that the median cannot be greater than the third quartile.

This can be shown more rigorously by providing a formal definition for the median of a data set. We can define median $tilde x$ of a set of $n$ observations by
$$
tilde x = frac{1}{2} left(a_{lceil n/2 rceil} + a_{lceil (n/2)+1 rceil} right)
$$
where $a_i$ is the $i$th element of the order sequence of observations and $lceil cdot rceil:mathbb R to mathbb Z$ is the ceiling function. The first quartiles is median of the sequence $a_1,...,a_{lceil n/2 rceil}$ and the third quartiles is the median of the sequence $a_{lceil (n/2)+1 rceil},...,a_n$. Hence, we have the ordering $Q_1 leq tilde x leq Q_3$, meaning that the median can equal the first or third quartiles, but cannot be less than the first nor more than the third.

share|cite|improve this answer

answered 2 days ago

BrianBrian

35812

New contributor

Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 2 days ago

BrianBrian

35812

New contributor

Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

BrianBrian

35812

New contributor

Brian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 2 days ago

BrianBrian

35812