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Statistics 1 - Histogram help !

Why doesnt the frequency density * the width of the bar in question give me the answer ?

Also, I dont follow what the mark scheme is saying I should do....

Could someone please explain this.

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Reply 1
As far as I can see it, the argument in the mark scheme is essentially "The area under the histogram is 70, but there are 140 runners, therefore the density needs to be multiplied by 2 to reflect the actual number of runners".

I cannot say I think this makes any sense - it seems wrong to me for the y-axis to be labelled frequency density here.

However, I have next to no knowledge of the actual contents of S1.

Mr M
..
Comments? If typical of S1, I find this question more than a little depressing, to be honest.
Original post by DFranklin
As far as I can see it, the argument in the mark scheme is essentially "The area under the histogram is 70, but there are 140 runners, therefore the density needs to be multiplied by 2 to reflect the actual number of runners".

I cannot say I think this makes any sense - it seems wrong to me for the y-axis to be labelled frequency density here.

However, I have next to no knowledge of the actual contents of S1.

Comments? If typical of S1, I find this question more than a little depressing, to be honest.


S1 can be more than depressing... I assume you've quoted Mr M in order to get a second opinion ?
Reply 3
Original post by Ari Ben Canaan
S1 can be more than depressing... I assume you've quoted Mr M in order to get a second opinion ?
Yes, I'm after an opinion from someone who knows the actual material in the course.
Original post by DFranklin
Yes, I'm after an opinion from someone who knows the actual material in the course.


Thank you. :smile:
Total area = 70
total frequency = 140

Area = 0.5*frequency


0.5*12 = 6

area = 6, 6*2 = 12.


S1 edexcel is becoming such a joke in terms of the obscure questions they ask that I have opted to take mechanics instead.
(edited 12 years ago)
Original post by DFranklin
However, I have next to no knowledge of the actual contents of S1.

Comments? If typical of S1, I find this question more than a little depressing, to be honest.


I don't teach S1 but histograms are an A* topic at GCSE. This is not the first question I have seen with what I consider to be an incorrect frequency density scale. They want candidates to work out the areas of all the bars and see that you need to multiply by 2 to make 140. This would be fine, in my opinion, if they had not bothered to add a scale to the frequency density column.

Effectively they are saying

Frequency Density โˆ\propto Frequency / Class Width

It just sounds rubbish to me but I will ask my Head of Dept (a Fellow of the Royal Statistical Society) what he makes of it tomorrow and report back.
Original post by Mr M
I don't teach S1 but histograms are an A* topic at GCSE. This is not the first question I have seen with what I consider to be an incorrect frequency density scale. They want candidates to work out the areas of all the bars and see that you need to multiply by 2 to make 140. This would be fine, in my opinion, if they had not bothered to add a scale to the frequency density column.

Effectively they are saying

Frequency Density โˆ\propto Frequency / Class Width

It just sounds rubbish to me but I will ask my Head of Dept (a Fellow of the Royal Statistical Society) what he makes of it tomorrow and report back.


Thank you very much for taking such interest in the matter. I am much obliged.
Reply 8
Original post by Mr M
It just sounds rubbish to me but I will ask my Head of Dept (a Fellow of the Royal Statistical Society) what he makes of it tomorrow and report back.
Why am I expecting a post from you tomorrow along the lines of:

Mr M
I asked my Head of Dept (a Fellow of the Royal Statistical Society) and he said:

":facepalm:"


?
Original post by Ari Ben Canaan
Thank you very much for taking such interest in the matter. I am much obliged.


Could you tell me the source of the question - Awarding Body, session and year?
Original post by Mr M
Could you tell me the source of the question - Awarding Body, session and year?


Edexcel, 2008 January, Statistics 1.

I dont know whether I should feel like an absolute idiot for not being able to solve such a question or whether I should accept the fact that its just a stupid question.
Original post by Ari Ben Canaan
Edexcel, 2008 January, Statistics 1.

I dont know whether I should feel like an absolute idiot for not being able to solve such a question or whether I should accept the fact that its just a stupid question.


I think you probably should have known that just calculating the area of one bar would not be worth 5 marks but I am not arguing about the stupidity of the question .
(edited 12 years ago)
The Chief Examiner said "The common error here was to assume that frequency equals the area under a bar, rather than using the relationship that the frequency is proportional to the area under the bar. Many candidates therefore ignored the statement in line 1 of the question about the histogram representing 140 runners and simply gave an answer of 12ร—0.5 = 6 . A few candidates calculated the areas of the first 7 bars and subtracted this from 140, sadly they didnโ€™t think to look at the histogram and see if their answer seemed reasonable. Those who did find that the total area was 70 usually went on to score full marks. A small number of candidates had difficulty reading the scales on the graph and the examiners will endeavor to ensure that in any future questions of this type such difficulties are avoided."
(edited 12 years ago)
Original post by Mr M
The Chief Examiner says "The common error here was to assume that frequency equals the area under a bar, rather than using the relationship that the frequency is proportional to the area under the bar. Many candidates therefore ignored the statement in line 1 of the question about the histogram representing 140 runners and simply gave an answer of 12ร—0.5 = 6 . A few candidates calculated the areas of the first 7 bars and subtracted this from 140, sadly they didnโ€™t think to look at the histogram and see if their answer seemed reasonable. Those who did find that the total area was 70 usually went on to score full marks. A small number of candidates had difficulty reading the scales on the graph and the examiners will endeavor to ensure that in any future questions of this type such difficulties are avoided. A small number of candidates had difficulty reading the scales on the graph and the examiners will endeavor to ensure that in any future questions of this type such difficulties are avoided."


Hmmmm... Thank you for that. *SIGH* I'll have to accept it and hope I dont get such a vague question in my paper.

+rep to all.
Reply 14
Original post by Mr M
The Chief Examiner said "The common error here was to assume that frequency equals the area under a bar, rather than using the relationship that the frequency is proportional to the area under the bar. Many candidates therefore ignored the statement in line 1 of the question about the histogram representing 140 runners and simply gave an answer of 12ร—0.5 = 6 . A few candidates calculated the areas of the first 7 bars and subtracted this from 140, sadly they didnโ€™t think to look at the histogram and see if their answer seemed reasonable. Those who did find that the total area was 70 usually went on to score full marks. A small number of candidates had difficulty reading the scales on the graph and the examiners will endeavor to ensure that in any future questions of this type such difficulties are avoided."


Interesting. I'm pretty sure that, in that case, the y-axis is mislabelled -- it's not the frequency density. A histogram needn't have the frequency density on the y-axis (as long as it is proportional to the frequency density), but if it is labelled as the frequency density then one would assume that it is the frequency density! Perhaps I'm wrong; I probably am considering this wasn't plastered all over the news as one of those times where 'evil exam board makes errors'
Original post by nuodai
Interesting. I'm pretty sure that, in that case, the y-axis is mislabelled -- it's not the frequency density. A histogram needn't have the frequency density on the y-axis (as long as it is proportional to the frequency density), but if it is labelled as the frequency density then one would assume that it is the frequency density!Indeed.

I note also that the graph is drawn with an explicitly truncated range; looking at that graph I would not be at all inclined to deduce that it represented all the runners in the race (I'd expect to find an unknown number of people who took less than 60 minutes). And yet you have to assume that it does represent all the runners in order to use the suggested approach.

If I was the chief examiner, I would be more than a little embarrassed.
Original post by DFranklin
Indeed.

I note also that the graph is drawn with an explicitly truncated range; looking at that graph I would not be at all inclined to deduce that it represented all the runners in the race (I'd expect to find an unknown number of people who took less than 60 minutes). And yet you have to assume that it does represent all the runners in order to use the suggested approach.

If I was the chief examiner, I would be more than a little embarrassed.


I have emailed the question to my HoD. He has stomped his foot about things like this before - probably a bit late now after three years have elapsed though!
Original post by Mr M
. A small number of candidates had difficulty reading the scales on the graph and the examiners will endeavor to ensure that in any future questions of this type such difficulties are avoided."


I see that you generously deleted the second copy of the repeated final sentence which is in the official report.

Does make one wonder about the quality of Edexcel.
Original post by ghostwalker
I see that you generously deleted the second copy of the repeated final sentence which is in the official report.A common error is to think that the frequency of that sentence in the report has to equal its frequency in what the Chief Examiner actually meant to write, rather than only being proportional to its frequency in what the Chief Examiner actually meant to write.
Original post by DFranklin
A common error is to think that the frequency of that sentence in the report has to equal its frequency in what the Chief Examiner actually meant to write, rather than only being proportional to its frequency in what the Chief Examiner actually meant to write.


Ha!

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