Actuarial Outpost Is 50th percentile of data, the mean? or median?
 Register Blogs Wiki FAQ Calendar Search Today's Posts Mark Forums Read
 FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions

 Salary Surveys Property & Casualty, Life, Health & Pension Health Actuary JobsInsurance & Consulting jobs for Students, Associates & Fellows Actuarial Recruitment Visit DW Simpson's website for more info. www.dwsimpson.com/about Casualty JobsProperty & Casualty jobs for Students, Associates & Fellows

#1
02-05-2011, 12:51 PM
 UFActuary Member Join Date: Jul 2005 Posts: 3,431
Is 50th percentile of data, the mean? or median?

What I don't understand is if the median and mean are not the same (there's a skew).....in a sample

Then the amount by which your score is higher than half the sample, would be the median, NOT the mean.

So then if you calculate percentile by taking x - mu/ std deviation

Then use the NDistribution table to get percentile, then is that the wrong thing to do? Or does percentile mean something else here? Since this uses mean, not median, as the 50th percentile, where x = mu and Z-score is 0.
#2
02-05-2011, 12:54 PM
 UFActuary Member Join Date: Jul 2005 Posts: 3,431

Or is there another distribution table for skewed distributions? Isn't this because I am using a normal distribution for a sample that is not normally distributed, and that's the problem?

This is embarassing, but what I'm referring to are Superbowl Square bets. Where you bet on say Packers to have score ending in 7, Steelers to end in 2. And since there are 100 squares, the mean percent chance of winning is 1% assuming one payout at the end of the game.

BUT the median is much closer to .39%, this because the squares like 7,0 and 0,7 are biggies...they are as much as 10 times the percent likelihood to win as other squares.....

so if I take the percentage of my 7,2 which comes out to lets say 0.51%, I am above the median but well below the mean.

If I go 0.0051 - 0.01 / Std Dev I will get a percentile score below the 50% because I'm below the mean... is this logical? Or how SHOULD the percentile score be calculated mathematically?

Last edited by UFActuary; 02-05-2011 at 12:59 PM..
#3
02-05-2011, 01:04 PM
 Monkey D. Luffy Member Join Date: Aug 2006 Location: 魚人島 Studying for HOLY CRAP DID U SEE THAT? College: Not Zoidberg Favorite beer: Anything. Now. Thanks! Posts: 626

Order all of the boxes in order of least likely to hit to most likely to hit, and then wherever your box comes in that order is the percentile it's in.

50th percentile is median.
#4
02-05-2011, 02:08 PM
 abt5 Member CAS Join Date: Dec 2010 Favorite beer: ice ice baby Posts: 5,458

Quote:
 Originally Posted by UFActuary What I don't understand is if the median and mean are not the same (there's a skew).....in a sample Then the amount by which your score is higher than half the sample, would be the median, NOT the mean. So then if you calculate percentile by taking x - mu/ std deviation Then use the NDistribution table to get percentile, then is that the wrong thing to do? Or does percentile mean something else here? Since this uses mean, not median, as the 50th percentile, where x = mu and Z-score is 0.
" x - mu/ std deviation" transforms everything into a standard normal. The mu here has nothing to do with the percentile itself, it's just used to give you the z-score. Once you have the z-score, you get the percentile from the standard table.
50% is always the median
#5
02-05-2011, 02:21 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,817

From the distribution, create a table of the distribution function.

E.g. F(a) = 0
F(b) = 0.01
F(c) = 0.25
F(d) = 0.40
F(e) = 0.50
F(f) = 0.62
F(g) = 0.87
...
F(z) = 1.00

You may have a lot more values whose distribution function you calculate. For example if we took a group of adult males and ordered by weight in pounds, we might have values for F(120), F(121), F(122)...,F(250), and if our scale was sensitive we might have F(160.138) even.

Going back to my original example, since F(e)=0.50, then e is the median.

If I wanted to know what percentile f is, the answer would be .62 since F(f) = .62.

If I wanted to know what is the 40th percentile, I would say d, since F(d) = .40.

All that is true for any distribution, skewed or not. The mean could equal e, or be much bigger than e, or much less than e. Doesn't matter. All the percentiles are based just on the distribution function.

If you don't know enough about the distribution to know what the cumulative distribution function is, you might have to estimate it based on whatever you do know about the distribution. The normal approximation might be something you would use in those estimates. The estimates might or might not be any good.
#6
02-05-2011, 03:15 PM
 PhildeTruth Member CAS Join Date: Dec 2009 Favorite beer: Allagash White Posts: 3,236

lol?
#7
02-05-2011, 03:26 PM
 r. mutt Member Join Date: May 2002 Location: Extreme fever swamps of the left Posts: 10,149

Quote:
 Originally Posted by PhildeTruth lol?
no question mark needed imo
__________________
(\ (\
( ^_^)
(_(")(") It all comes full akchigs eventually. ~ Full-On Devi
#8
02-05-2011, 03:30 PM
 Jasper07734 Member Non-Actuary Join Date: Jun 2008 Posts: 5,992

What's the name for the 42nd percentile?
#9
02-05-2011, 03:35 PM
 whisper Member CAS AAA Join Date: Jan 2002 Location: Chicago Favorite beer: Hefewizen Posts: 35,036

VaR

----

The normal distribution has some nice properties which create easy ways to think of the distribution. The thing to remember is that these properties are unique to the normal distribution. Trying to take extrapolate the normal distribution properties to non-normal distributions is just going to cause problems - if not be flat out wrong.
#10
02-05-2011, 05:25 PM
 r. mutt Member Join Date: May 2002 Location: Extreme fever swamps of the left Posts: 10,149

Quote:
 Originally Posted by whisper VaR
Oof. I laughed at an actuarial joke.
__________________
(\ (\
( ^_^)
(_(")(") It all comes full akchigs eventually. ~ Full-On Devi