Quartiles of box plot

quantilestatistics

I'm struggling with the meaning of quartiles , in this box plot how to compare these series like should I compare the upper $50$% of each series or the lower or the $50$% between $1$st quartile and $3$rd quartile ?

a: $50$% of the values in series $A$ are greater than or equal to $50$% of the values in series $B$.

b: $75$% of the values in series $A$ are greater than or equal to $25$% of the values in series $B$.

c: $50$% of the values in series $B$ are less than or equal to $50$% of the values in series $A$.

d: $75$% of the values of series $A$ are less than or equal to $50$% of the values of series $B$

Best Answer

Let's examine the statements in turn:

a: 50% of the values in series A are greater than or equal to 50% of the values in series B.

So look at the middle mark in Series A - the median - compare with the middle mark in Series B. It's False: 4 < 6.

b: 75% of the values in series A are greater than or equal to 25% of the values in series B.

Look at the Upper Quartile ($UQ_A$) for Series A, against the Lower Quartile ($LQ_B$) for Series B. Technically, it's true, as $6 \geq 4$, but as statement's go it is a weak one, as it would make more sense to compare $UQ_A$ with the median of Series B. I suggest you keep looking.

I will leave you to check c and d for yourself in the same way. One is True, the other False.

Related Solutions

[Math] Is it possible to calculate the mean and standard deviation from a median and quartiles

It's mathematically impossible to deduce mean or standard deviation from median/quartiles, because medians and quartiles discard most of the data on which the mean and standard deviation are based.

Example:

data   frequency  
   0       50      
 1.4        4     
   2       50

That has a mean of 1.0 and standard deviation of 0.9. (I'm using 2 significant figures so I don't have to go into population versus sample standard deviation.)

data     frequency    
   0       30        
 1.4       44        
   2       30

That data also has the median and quartiles the same as in your example, but now the mean is 1.2 and the standard deviation is 0.8.

data     frequency        
   0       30        
 1.4        3        
   2       70        
10000000    1

Now I've changed my maximum without changing the median or quartiles, you can see even more clearly how the median and quartiles exclude extreme data, because the mean is now 96000 and the standard deviation is 98000 (still 2 sig.fig.).

[Math] Getting the upper and lower quartiles in data with an even number of observations, or where the quartile lands on a decimal number

If taking the $25$th percentile of the data gives you the value $3.5$, then the value of the first quartile is $3.5$. We would usually expect that the number of observations in each of your frequency classes is a whole number, but there is nothing wrong with having a quartile or median value that is not a whole number, so further rounding is neither needed nor desired.

Best Answer

Related Solutions

[Math] Is it possible to calculate the mean and standard deviation from a median and quartiles

[Math] Getting the upper and lower quartiles in data with an even number of observations, or where the quartile lands on a decimal number

Related Question