Let’s say you have a bell curve distribution. You cut off the bottom 25%. What happens to the mean and median? How do you calculate that?
What if you only take the top 10%?
Is there a website where I can put I number to see how it changes the distribution?
I don’t have an easy answer - really because at first glance this is a rather odd thing to want to do. I’m rather interested in the actual thing you’re looking at – what kind of random variable has the property of never falling below a certain value, but follows a “chopped off” Gaussian distribution above that value? I can almost imagine a few things, but I’m not really sure how plausible they are.
Some things that don’t answer your question that may be relevant are the lognormal distribution, or deviations such as kurtosis and heteroskedasticity.
Look up conditional value at risk. Also called expected shortfall. Basically it is the expected value of a random variable, on the condition that it is less than some value. It is a popular measure of risk in finance.
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By any chance, are you asking about the following type of situation:
Assume there is a college class where the students’ ‘intelligence’ and skill at taking exams are both normally distributed (with the distributions bell curve shaped).
If you remove, say, the bottom 25 percent, is there way of making the remaining 75 percent of students normally distributed with respect to intelligence and exam taking skill? I think the answer must be ‘no’. (As a result, I was frequently upset when profs tried to effect such miracles, e.g. when they tried to make a bell curve out of the remaining students (for the evil purpose, of course, of ‘belling’ our marks).
One practical but brutish way to do this is by using standard tools to just generate some random data and then calculate their statistics. For example
rnvars 400000 100 15 | statmoment
generates 400,000 normal variates of mean 100 and s.d. 15, and passes them to a statistics program. This prints out, about as expected,
400000 items: mean 99.992430 sd 15.011881 skew 0.008847 kurt 3.002609 min 26.098508 max 167.497638
Now do the same thing, but sort the random data and discard the bottom 25% before showing the statistics:
rnvars 400000 100 15 | sort -nr | head -300000 | tee tmp | statmoment
printing
300000 items: mean 106.343541 sd 11.004399 skew 0.732533 kurt 3.229362 min 89.838353 max 167.497638
statmoment doesn’t show the median, but you can get that with standard Unix tools:
echo -n Median is" " ; head -150000 tmp | tail -1
prints
Median is 104.752155
If the 25% truncation case is all you wanted, you should be able to just scale the above numbers and use them.
You can also similarly brute-force it in excel, if that’s more the OP’s game.
Not directly responsive, but reminded me of Gould’s book Full House where he discussed changes in mean/median/mode. He was generally speaking of distributions with a hard minimum or maximum, which would be skewed one way or the other. Discussed the effect on the mean/median/mode as the tail extended. As a general rule, I believe the median and mode are less responsive than the mean to changes at the extreme. So if you chopped off one of the tails, I would imagine the mean would change the most, the median less so, and the mode likely not at all.
That’s all this math-impaired lawyer can add (other than that I highly recommend that book)! 
Wikipedia to the rescue
I just noticed that the page I gave didn’t have the median calculated, but it’s pretty easy.
Using EXCEL’s terminology, let NORM.INV be the inverse quantile of the normal distribution NORM.INV(0.5) = 0, NORM.INV(0.025)=-1.96
then then if you cut off the lower q% of the tail, and the upper r% of the tail the pth quantile will be
NORM.INV(q+p*(1-r-q))
The median will be NORM.INV(q+(1-r-q)/2)
Wow, two great words at once. (I always thought kurtosis had something to do with scoliosis/orthopedics. Maybe it does.) And now I see it can be an adjective with lepto and platy! (An easy descriptive pdf: http://www.uky.edu/Centers/HIV/cjt765/9.Skewness%20and%20Kurtosis.pdf
Heteroskeda…I have to work on.
Yeah, something like that.
I know the new distribution will not be a bell curve. But what happens to the average and median values if you cut off the bottom 20 or 30%? How much does it improve?
How much does it go down if you cut off the top 5%?
etc.
Thanks. Math was never something I was good at but I found this site.
http://www.ntrand.com/truncated-normal-distribution/
I believe you need the bottom end of the distribution, top end, mode and standard deviation to calculate the new mean of a truncated distribution.
However where it says to calculate the mean, I’m not sure how to get the probability density function and cumulative distribution function.
A can be obtained from a, m & sigma. B from b, m & sigma (I assume that is a typo where the calculation for A and B are identical. (a-m)/sigma for both. If it isn’t a typo wouldn’t the answer for all the equations end up as 0?) But getting delta or the mean requires the two functions above, I’m not sure how they are obtained.
The probability density function it is referring to is the probablility density furnction of the standard normal distribution.
phi(x)=1/sqrt(2pi)*exp(-x^2)
with capital PHI being the integral of that, which doesn’t have a nice closed form, but for which there are look up tables, or function calls in most programming languages. In Excel, but phi and PHI can be calculated with the function NORM.DIST, with mean 0 standard deviation 1 and the flag for cumulative set to 0 and 1 respectively.
So in EXCEL notation.
DELTA would be PHI(B)-PHI(A) = NORM.DIST(B,0,1,1)-NORM.DIST(A,0,1,1)
and
mean=m+(NORM.DIST(B,0,1,0)-NORM.DIST(A,0,1,0))*sigma/DELTA
The issue with A=B is not a typo. According the the notation, in the site you included the function is cut below A and above B, so if A=B, you have cut the entire distribution out, and so there is no distribution left, giving you a mean of (0/0)=Undefined.
Based on your OP it looked like you were more interested in cutting off at perticular percentiles (cutting off the top and bottom 25%) rather than particular values (cutting off values below 2 and above 3). If you want to cut off the lower q% and the upper p% then you can simplify things by replacing the equations for A and B and DELTA with the following EXCEL equivalents
A=NORM.INV(q,0,1) and B=NORM.INV(1-p,0,1) DELTA=1-p-q
Hope this helps.
ETA: type and formatting issues in last post
should be
mean=m+(NORM.DIST(A,0,1,0)-NORM.DIST(B,0,1,0))*sigma/DELTA
and the last line should read
A=NORM.INV(q,0,1)
B=NORM.INV(1-p,0,1)
DELTA=1-p-q
nm double post