[Math] Is it possible to find the mean and standard deviation when given the normal z-scores for the bottom 25% and top 25%

statistical-inferencestatistics

I want to apply to a university where the 25th /75th percentiles for the SAT Math are 490 / 620 respectively, but I am curious how would I find the mean and the standard deviation assuming that the data is normally distributed? I know that we are talking about the middle 50th-percentile and $z_{0.25} = 620$ and $z_{0.75} = 490$

Best Answer

Using z-score tables for standard normal variable, $$ z_{0.25} = \mu - \sigma \cdot 0.675 \qquad z_{0.75} = \mu + \sigma \cdot 0.675 $$ You now have two equations for two unknowns.

In[213]:= Solve[{Quantile[NormalDistribution[mu, si], 0.25] == 490, 
  Quantile[NormalDistribution[mu, si], 0.75] == 620}, {mu, si}]

Out[213]= {{mu -> 555., si -> 96.3691}}

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] Finding the percentile of a value given other percentiles

Your final observation about the $25$th, $50$th, and $75$th percentiles just says the distribution is not symmetric. Do you have a reason to believe that it should be? Using the standard deviation assumes the distribution is Gaussian, which the evidence contradicts. All you really know is that $350.2$ is somewhere between the $75$th and $90$th percentile. If the events between $260.4$ and $736.0$ are equally distributed, you would have $350.2$ at the $\frac {350.2-260.4}{736.0-260.4}\cdot 15+75\approx 78$th percentile. It is probably rather higher than that-if the data is mostly near the middle it will be. It is hard to give a nice justifiable answer.

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Related Solutions

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

[Math] Finding the percentile of a value given other percentiles

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