NBME 3 biostatistics mean median mode question

[zeevee]

Can you tell me how you solved it .I dont think its a regular negative skew .

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[Kobebucsfan]

mode is 1, median is 1. so we r left with Choices A and B. i dont think this is a SKEW type question, someone correct me


using rough estimates, i got mean to be 1.6. so Choice A

 

[iceman55]

It's a positive skew. The mean is greater the median. No need for calculations. Just look at graph, three (8,9,10) patients have huge creatinine levels that skews the mean positively, but most of the number are clustered at the lower end.


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[zhopv10]

Be careful looking at the graph straight up. Skew is generally best seen with a frequency histogram or box plot, so you would say (if the graph looked this way but was in fact a frequency histogram and say the x axis was creatinine level and the y axis was number of people) now then you would say there is left skew (just the shape not the literal translation, this data would be right skew), the median would stay stationary and the mean would be shifted down by few smaller outliers with exceptionally small values (I like to think of how the box plot would look). Anyway, since this is just patients 1-10, and their respective creatinine levels I would do as kobebucsfan did and do a couple quick rough calculations to get a. As a side note if this was a frequency histogram it would be right skewed and this would mean that the mean would be greater than the median (pulled by the low values). Basically the mean follows the tail. Here is a nice simple explanation http://m.dummies.com/how-to/content/how-to-identify-skew-and-symmetry-in-a-statistical.html and here http://pirate.shu.edu/~wachsmut/Teaching/MATH1101/Descriptives/box.html


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[zeevee]

@zhopv10 The mean is larger than the median I chose A .What do you think ?

 

[zhopv10]

@zeevee Ya I think A is correct, in this case you just need to do the calculations the median looks to be about 1 and the mean looks to be about 1.6ish maybe, thus the mean is bigger (just using rough calculation for the mean). I was just pointing out that skew isn't involved here since it is not a frequency histogram (it is just a bar graph of creatinine levels). I also wanted to point out that in frequency graphs the tail points towards the skew (left or right) and the mean follows that skew too (so left skew the mean is lower than the median due to the far out low values in a few individuals). Hopefully that was clear, sorry if it wasn't above I wrote that last post a little late last night :/ haha


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[zhopv10]

Now also you could convert the graph in your mind like iceman55 did and note that most of the patients are in the lower number range and there are 3 patients with really high numbers. Thus, they are the outliers and the frequency graph would be right skewed and then by definition the mean would be larger than the median. That also works as well. Whichever is fast for you, I think iceman' method he pointed out is probably faster and easier in a test setting if you are comfortable thinking in those terms.


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[worldbeater]

IF they wanted this to be a skew, it would be a negative skew (towards the left). In a negative skew, the mean is at the left (think of it as a "mean" eye), and it would be lower than the median and mode. Mean < median < mode. That's how the concept works and B would be the correct answer (median is larger than mean).

But if this isn't a skew, you would have to go with the posters explanation above.

 

[dempty]

This question absolutely involves skew (Iceman55 has it right). If the mean and the median are not equal, the distribution is skewed (irrespective of how the data are presented). The image is very clear in showing several values that are much larger than the majority of the other values-- this is causing a positive skew on the distribution (there's no need for a calculation in this problem). Positive skew (generally) implies that the mean is greater than the median.

For those that are saying it is negatively skewed: don't be fooled by the way the data are presented-- each observation here only carries a relative frequency 1/10 (since it doesn't appear that any values actually repeat, but we wouldn't really expect that with a continuous variable). If you need to, redraw the graph with relative frequency on the y-axis and the serum creatinine values on the x-axis. This will make it easier to see that this sample has a positive skew (a few larger values in the higher end).

Any way you look at this question, it involves skew, and there's only one correct answer. It's important to keep in mind that skew is a property of the data (distribution), and it isn't changed by how someone chooses to present the data (although, certain presentations can more clearly demonstrate the shape of a distribution).

A is correct.

 

[zhopv10]

^not really sure who this is directed to, but just to be clear on mine. I was not saying that this graph was left/negative skew. I was cautioning not to read this graph at face value or else fall into the trap of thinking that it was left skew. I made it confusing by saying if it was a frequency graph and looked this way then it would be left skew but didn't mean to imply this data set was literally left skew. That was unclear, my bad. Anyway my point was you can't look at a graph and read the skew from the shape unless it is a frequency based graph (histogram, box plot etc.) so you have to pay attention the axis.

Now totally skew is inherent to the data, I was just meaning to say that: if you want to visualize skew you can't do it straight across with this graph without thinking about what the data is saying and mentally "redrawing" it mentally of you will, or just noticing the distribution. Anyway, spot on post I did say it wasn't a skew question which isn't right, I was thinking of it as you don't have to think about it that way to get it right but that really isn't the best way to do it, above posts edited accordingly to avoid confusion.

*edit: wow I take that back I totally did say it would be left skew at the first of my post, that's terrible :0 wow well it really was too late last night to be posting apparently and the it was totally contradictory to the rest of the post haha sorry guys that would explain why it was confusing lol. I edited it to fix that that thanks dempty for pointing that out! Sent from my iPhone using SDN mobile app

 

[dempty]

^not really sure who this is directed to, but just to be clear on mine. I was not saying that this graph was left/negative skew. I was cautioning not to read this graph at face value or else fall into the trap of thinking that it was left skew. I made it confusing by saying if it was a frequency graph and looked this way then it would be left skew but didn't mean to imply this data set was literally left skew. That was unclear, my bad. Anyway my point was you can't look at a graph and read the skew from the shape unless it is a frequency based graph (histogram, box plot etc.) so you have to pay attention the axis.

Now totally skew is inherent to the data, I was just meaning to say that: if you want to visualize skew you can't do it straight across with this graph without thinking about what the data is saying and mentally "redrawing" it mentally of you will, or just noticing the distribution. Anyway, spot on post I did say it wasn't a skew question which isn't right, I was thinking of it as you don't have to think about it that way to get it right but that really isn't the best way to do it, above posts edited accordingly to avoid confusion.

*edit: wow I take that back I totally did say it would be left skew at the first of my post, that terrible :0 wow well it really was late last night and totally contradictory to the rest of the post haha sorry guys that would explain why it was confusing lol. I edited it to fix that that thanks dempty for pointing that out! Sent from my iPhone using SDN mobile app

It looks like you got a good handle on it at that point. I agree that the shape of this graph would be trickier to infer skew from instead of a more appropriate method. I definitely think the data were presented in this manner in an attempt to trap someone moving quickly.

To be clear, I was referring to a few posts that weren't necessarily accurate (or possibly could have been worded better, from the way I read it). My intention was to hopefully clear up some of the concepts for anyone reading (and I think that's been accomplished, more or less).

 

[zhopv10]

Totally, I'm glad you did so too, otherwise I wouldn't have looked at that terrible and contradictory post I made originally haha


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