3rdyearbiostudent Posted June 12, 2014 Report Share Posted June 12, 2014 hey everyone, any idea how to interpret this score? I know it's above the passing point (390); but how does it translate in terms of percentile? Link to comment Share on other sites More sharing options...
supafield Posted June 13, 2014 Report Share Posted June 13, 2014 Generally the mean is set close to 500... and 100 pts in either direction is a standard deviation away from the mean. At least, that is how I understood it. Link to comment Share on other sites More sharing options...
burrb Posted June 13, 2014 Report Share Posted June 13, 2014 ^ Assuming a mean of 500 and sd of 100, your score of 483 would give you ~43%ile using z score calculations. Edit: In a week or so when they release the official data for this year's LMCC you can calculate your actual %ile using the mean and sd. Just goggle z score calculation and z score chart. Link to comment Share on other sites More sharing options...
justletmein Posted June 13, 2014 Report Share Posted June 13, 2014 It means you passed. Congrats, go have a drink and move on with your life. Who cares what percentile you are. Link to comment Share on other sites More sharing options...
leviathan Posted June 13, 2014 Report Share Posted June 13, 2014 ^ Assuming a mean of 500 and sd of 100, your score of 483 would give you ~57%ile using z score calculations. Edit: In a week or so when they release the official data for this year's LMCC you can calculate your actual %ile using the mean and sd. Just goggle z score calculation and z score chart. You can't be above 50th percentile if you're below the mean. Link to comment Share on other sites More sharing options...
A-Stark Posted June 13, 2014 Report Share Posted June 13, 2014 That's only true of a symmetric distribution. You're thinking of the median. Link to comment Share on other sites More sharing options...
leviathan Posted June 13, 2014 Report Share Posted June 13, 2014 True, but if it's not normally distributed then we can't estimate percentiles. If we are, then the percentile is 43rd, not 57th. Link to comment Share on other sites More sharing options...
A-Stark Posted June 13, 2014 Report Share Posted June 13, 2014 You can always calculate percentiles for any continuous distribution. The median is simply the 50th percentile. Link to comment Share on other sites More sharing options...
burrb Posted June 13, 2014 Report Share Posted June 13, 2014 You can't be above 50th percentile if you're below the mean. Oops I was on the wrong side of the curve. So it'd be 43%ile. Link to comment Share on other sites More sharing options...
leviathan Posted June 13, 2014 Report Share Posted June 13, 2014 You can always calculate percentiles for any continuous distribution. The median is simply the 50th percentile. How can you calculate a percentile with just a mean and SD if the distribution is not normal? My stat knowledge is kinda rusty but I thought that was not possible. Link to comment Share on other sites More sharing options...
supafield Posted June 13, 2014 Report Share Posted June 13, 2014 How can you calculate a percentile with just a mean and SD if the distribution is not normal? My stat knowledge is kinda rusty but I thought that was not possible. Take a look at this. http://allpsych.com/researchmethods/images/normalcurve.gif Link to comment Share on other sites More sharing options...
leviathan Posted June 13, 2014 Report Share Posted June 13, 2014 Take a look at this. http://allpsych.com/researchmethods/images/normalcurve.gif That's a normal distribution. Link to comment Share on other sites More sharing options...
A-Stark Posted June 13, 2014 Report Share Posted June 13, 2014 How can you calculate a percentile with just a mean and SD if the distribution is not normal? My stat knowledge is kinda rusty but I thought that was not possible. The definition of the p-th percentile is simply whatever value that is greater than or equal to p-percent of the observations. So if 80% of LMCC Part I scores are at or below 600 than we find the 80th percentile at 600. Likewise if the median score is 495 than this is the 50th percentile. The helpful thing about a normal distribution (or observations that can be readily approximated with one) is that we can "standardize" each observation with the observed mean and SD to a "z-score", which is a simply a standard normal (mean 0, SD 1) value which has a standard, known associated percentile. Link to comment Share on other sites More sharing options...
supafield Posted June 14, 2014 Report Share Posted June 14, 2014 That's a normal distribution. oops apologies Link to comment Share on other sites More sharing options...
leviathan Posted June 14, 2014 Report Share Posted June 14, 2014 The definition of the p-th percentile is simply whatever value that is greater than or equal to p-percent of the observations. So if 80% of LMCC Part I scores are at or below 600 than we find the 80th percentile at 600. Likewise if the median score is 495 than this is the 50th percentile. The helpful thing about a normal distribution (or observations that can be readily approximated with one) is that we can "standardize" each observation with the observed mean and SD to a "z-score", which is a simply a standard normal (mean 0, SD 1) value which has a standard, known associated percentile. None of that is helpful if the scores aren't normally distributed and if we don't know what that distribution looks like (with the raw data). So again, I don't see how you can calculate percentile based on mean and SD only if you don't have a normal distribution of data. Link to comment Share on other sites More sharing options...
A-Stark Posted June 14, 2014 Report Share Posted June 14, 2014 You don't calculate any percentiles based on mean and SD directly. Computing z-scores ("normalization") allows for conversion to standard normal values which are then compared to known percentiles for the standard normal distribution. A p-th percentile is defined as p/100 = int (- inf to x) F(a) da where x is the p-th percentile of the cumulative distribution function F. It may also be expressed by F^(-1)(p/100) = x, where F^(-1) represents the inverse cdf. Since for empirical (and, indeed, many theoretical*) distributions there is no closed form of F^(-1), numerical methods are employed to approximate x. The bottom line is that you can always compute the empirical distributions based on random samples and derive descriptive statistics (mean, median, quartiles, percentiles, variance, standard deviation) irrespective of the theoretical "underlying" distribution. *The normal distribution has a closed-form probability distribution function but doesn't have a closed form cdf (the integral of the pdf does not exist). Anyway that's enough mathematical statistics for a Friday night. Link to comment Share on other sites More sharing options...
Sarvish Posted June 14, 2014 Report Share Posted June 14, 2014 so basically this is the ~40th penticle and yet it's still a passing score..? You wouldn't want 40% of test takers failing, would you? Link to comment Share on other sites More sharing options...
mavrik13 Posted June 14, 2014 Report Share Posted June 14, 2014 None of that is helpful if the scores aren't normally distributed and if we don't know what that distribution looks like (with the raw data). So again, I don't see how you can calculate percentile based on mean and SD only if you don't have a normal distribution of data. +1 - I'm not sure why people are arguing... the mean and SD are used to calculate a Z score, assuming a normal distribution, and subsequently this can be used to estimate percentiles. There is no easy way that I know of to estimate percentiles of a non-normally distributed sample. That being said, it is probably a relatively safe assumption that the LMCC scores are close enough to being normally distributed that we can use z scores to calculate. Link to comment Share on other sites More sharing options...
burrb Posted June 14, 2014 Report Share Posted June 14, 2014 ^ PRECISELY. If you are not given percentiles, you can assume normal distribution and use median/sd. For a large exam such as the MCCQE1, it's almost a given that the results will be close to a normal distribution. so basically this is the ~40th penticle and yet it's still a passing score..? A pass is set such that something like 2% of CMGs fail, I believe. So 40th percentile is below the mean, but nothing wrong with that. Link to comment Share on other sites More sharing options...
leviathan Posted June 14, 2014 Report Share Posted June 14, 2014 I'm not arguing that it isn't normally distributed. This huge tangent all came when I used the word mean when I meant to say median, though the two should equal each other in a normal distribution so it's a moot point. Link to comment Share on other sites More sharing options...
A-Stark Posted June 14, 2014 Report Share Posted June 14, 2014 There is no easy way that I know of to estimate percentiles of a non-normally distributed sample. It's quite easy to compute descriptive statistics for a sample with an unknown distribution. It is, however, time-consuming without a computer and to do so by hand. We can readily come up with a robust formula to give empirical percentiles and write a little script for a program like R to compute it. I'm not arguing that it isn't normally distributed. This huge tangent all came when I used the word mean when I meant to say median, though the two should equal each other in a normal distribution so it's a moot point. Yes. But I reserve the right to be pedantic about probability. Link to comment Share on other sites More sharing options...
burrb Posted June 14, 2014 Report Share Posted June 14, 2014 ^ how the fudge do you know so much about math? me being bad at math is one of the prime reasons I went into medicine haha Link to comment Share on other sites More sharing options...
A-Stark Posted June 14, 2014 Report Share Posted June 14, 2014 I have some degrees in it or something. It's certainly been my observation that most physicians aren't very comfortable with this stuff. Link to comment Share on other sites More sharing options...
ChemPetE Posted June 18, 2014 Report Share Posted June 18, 2014 Detailed MCCQE Part 1 results are now released. Mean ~= 480 Standard deviation = 100 Pass = 390 Assuming a normal distribution means that a passing score is above the 16th percentile. So for the OP, a score of 483 is about the 51st percentile given the assumptions above. Link to comment Share on other sites More sharing options...
newgrad2014 Posted June 18, 2014 Report Share Posted June 18, 2014 Do we need to send the result report to any organizations, such as the College? I heard that if you don't pass Part I, you cannot sign orders as a R1. Does this mean the College automatically knows our exam results or do we have to send the report? Link to comment Share on other sites More sharing options...
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