Do test scores really follow a normal distribution?
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13
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I've been trying to learn which distributions to use in GLMs, and I'm a little fuzzled on when to use the normal distribution. In one part of my textbook, it says that a normal distribution could be good for modeling exam scores. In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values. Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there? Doesn't the normal distribution allow for negative values?
normal-distribution generalized-linear-model gamma-distribution inverse-gaussian-distrib
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up vote
13
down vote
favorite
I've been trying to learn which distributions to use in GLMs, and I'm a little fuzzled on when to use the normal distribution. In one part of my textbook, it says that a normal distribution could be good for modeling exam scores. In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values. Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there? Doesn't the normal distribution allow for negative values?
normal-distribution generalized-linear-model gamma-distribution inverse-gaussian-distrib
1
If you're worried about the bounds on scores, you could try en.wikipedia.org/wiki/Truncated_normal_distribution
– J.G.
Nov 12 at 6:39
5
In the real world, of course, exam score distributions often don't look anything like a normal distribution anyway. As an example from my math undergrad days, I remember the Topology I class as having been notorious for its highly bimodal "dumbbell curve" grade distribution: you either understood the key concepts and got a nearly perfect score, or you didn't and were lucky to get any points at all. Very few people ended up scoring anywhere in the middle between those two extremes.
– Ilmari Karonen
Nov 12 at 10:44
2
No. Next question.
– Carl Witthoft
Nov 12 at 16:24
add a comment |
up vote
13
down vote
favorite
up vote
13
down vote
favorite
I've been trying to learn which distributions to use in GLMs, and I'm a little fuzzled on when to use the normal distribution. In one part of my textbook, it says that a normal distribution could be good for modeling exam scores. In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values. Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there? Doesn't the normal distribution allow for negative values?
normal-distribution generalized-linear-model gamma-distribution inverse-gaussian-distrib
I've been trying to learn which distributions to use in GLMs, and I'm a little fuzzled on when to use the normal distribution. In one part of my textbook, it says that a normal distribution could be good for modeling exam scores. In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values. Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there? Doesn't the normal distribution allow for negative values?
normal-distribution generalized-linear-model gamma-distribution inverse-gaussian-distrib
normal-distribution generalized-linear-model gamma-distribution inverse-gaussian-distrib
asked Nov 12 at 1:03
mistersunnyd
190110
190110
1
If you're worried about the bounds on scores, you could try en.wikipedia.org/wiki/Truncated_normal_distribution
– J.G.
Nov 12 at 6:39
5
In the real world, of course, exam score distributions often don't look anything like a normal distribution anyway. As an example from my math undergrad days, I remember the Topology I class as having been notorious for its highly bimodal "dumbbell curve" grade distribution: you either understood the key concepts and got a nearly perfect score, or you didn't and were lucky to get any points at all. Very few people ended up scoring anywhere in the middle between those two extremes.
– Ilmari Karonen
Nov 12 at 10:44
2
No. Next question.
– Carl Witthoft
Nov 12 at 16:24
add a comment |
1
If you're worried about the bounds on scores, you could try en.wikipedia.org/wiki/Truncated_normal_distribution
– J.G.
Nov 12 at 6:39
5
In the real world, of course, exam score distributions often don't look anything like a normal distribution anyway. As an example from my math undergrad days, I remember the Topology I class as having been notorious for its highly bimodal "dumbbell curve" grade distribution: you either understood the key concepts and got a nearly perfect score, or you didn't and were lucky to get any points at all. Very few people ended up scoring anywhere in the middle between those two extremes.
– Ilmari Karonen
Nov 12 at 10:44
2
No. Next question.
– Carl Witthoft
Nov 12 at 16:24
1
1
If you're worried about the bounds on scores, you could try en.wikipedia.org/wiki/Truncated_normal_distribution
– J.G.
Nov 12 at 6:39
If you're worried about the bounds on scores, you could try en.wikipedia.org/wiki/Truncated_normal_distribution
– J.G.
Nov 12 at 6:39
5
5
In the real world, of course, exam score distributions often don't look anything like a normal distribution anyway. As an example from my math undergrad days, I remember the Topology I class as having been notorious for its highly bimodal "dumbbell curve" grade distribution: you either understood the key concepts and got a nearly perfect score, or you didn't and were lucky to get any points at all. Very few people ended up scoring anywhere in the middle between those two extremes.
– Ilmari Karonen
Nov 12 at 10:44
In the real world, of course, exam score distributions often don't look anything like a normal distribution anyway. As an example from my math undergrad days, I remember the Topology I class as having been notorious for its highly bimodal "dumbbell curve" grade distribution: you either understood the key concepts and got a nearly perfect score, or you didn't and were lucky to get any points at all. Very few people ended up scoring anywhere in the middle between those two extremes.
– Ilmari Karonen
Nov 12 at 10:44
2
2
No. Next question.
– Carl Witthoft
Nov 12 at 16:24
No. Next question.
– Carl Witthoft
Nov 12 at 16:24
add a comment |
2 Answers
2
active
oldest
votes
up vote
14
down vote
accepted
Height, for instance, is often modelled as being normal. Maybe the height of men is something like 5 foot 10 with a standard deviation of 2 inches. We know negative height is unphysical, but under this model, the probability of observing a negative height is essentially zero. We use the model anyway because it is a good enough approximation.
All models are wrong. The question is "can this model still be useful", and in instances where we are modelling things like height and test scores, modelling the phenomenon as normal is useful despite it technically allowing for unphysical things.
In this particular case, it's questionable whether the normal distribution is even a useful approximation. Nearly every grade distribution I've seen resembled the bimodal curve Ilmari Karonen mentioned in the comments to some degree. (Although usually with modes around 0.6 and 0.9, rather than on the extreme ends) However, a linear combination of two normal distributions with different means wouldn't be a bad approximation.
– Ray
Nov 13 at 4:13
I wasn't arguing that the normal is THE BEST approximation. The entire point of my comment is really made in that last paragraph. Comments about bimodality of actual grade distributions, at least at this level of abstraction, are really not helpful.
– Demetri Pananos
Nov 13 at 4:16
I agree with everything you said in your answer, but part of the question concerns whether the normal distribution is specifically applicable to modeling grade distributions. And the answer to that is usually "No". All models are wrong and some models are useful, but some are more wrong and less useful than others. The fact that the normal distribution in particular is an especially bad fit for this problem is important, and the answer as it is seems to suggest that the normal is only wrong because the tails go negative and infinite, when there are actually much deeper problems.
– Ray
Nov 13 at 4:23
You're being a little pedantic here. OP's problem was that the normal allows for negative scores. Bimodality wasn't the issue. You are not seeing the forest for the trees with respect to this question. Modelling details aren't relevant right now.
– Demetri Pananos
Nov 13 at 4:29
add a comment |
up vote
10
down vote
Doesn't the normal distribution allow for negative values?
Correct. It also has no upper bound.
In one part of my textbook, it says that a normal distribution could be good for modeling exam scores.
In spite of the previous statements, nevertheless this is sometimes the case. If you have many components to the test, not too strongly related (e.g. so you're not essentially the same question a dozen times, nor having each part requiring a correct answer to the previous part), and not very easy or very hard (so that most marks are somewhere near the middle), then marks may often be reasonably well approximated by a normal distribution; often well enough that typical analyses should cause little concern.
We know for sure that they aren't normal, but that's not automatically a problem -- as long as the behaviour of the procedures we use are close enough to what they should be for our purposes (e.g. standard errors, confidence intervals, significance levels and power - whichever are needed - do close to what we expect them to)
In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values.
Yes, but more than that -- they tend to be heavily right skew and the variability tends to increase when the mean gets larger.
Here's an example of a claim-size distribution for vehicle claims:
https://ars.els-cdn.com/content/image/1-s2.0-S0167668715303358-gr5.jpg
(Fig 5 from Garrido, Genest & Schulz (2016) "Generalized linear models for dependent frequency and severity of insurance claims", Insurance: Mathematics and Economics, Vol 70, Sept., p205-215. https://www.sciencedirect.com/science/article/pii/S0167668715303358)
This shows a typical right-skew and heavy right tail. However we must be very careful because this is a marginal distribution, and we are writing a model for the conditional distribution, which will typically be much less skew (the marginal distribution we look at if we just do a histogram of claim sizes being a mixture of these conditional distributions). Nevertheless it is typically the case that if we look at the claim size in subgroups of the predictors (perhaps categorizing continuous variables) that the distribution is still strongly right skew and quite heavy tailed on the right, suggesting that something like a gamma model* is likely to be much more suitable than a Gaussian model.
* there may be any number of other distributions which would be more suitable than a Gaussian - the inverse Gaussian is another choice - though less common; lognormal or Weibull models, while not GLMs as they stand, may be quite useful also.
[It's rarely the case that any of these distributions are near-perfect descriptions; they're inexact approximations, but in many cases sufficiently good that the analysis is useful and has close to the desired properties.]
Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there?
Because (under the conditions I mentioned before -- lots of components, not too dependent, not to hard or easy) the distribution tends to be fairly close to symmetric, unimodal and not heavy-tailed.
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
Height, for instance, is often modelled as being normal. Maybe the height of men is something like 5 foot 10 with a standard deviation of 2 inches. We know negative height is unphysical, but under this model, the probability of observing a negative height is essentially zero. We use the model anyway because it is a good enough approximation.
All models are wrong. The question is "can this model still be useful", and in instances where we are modelling things like height and test scores, modelling the phenomenon as normal is useful despite it technically allowing for unphysical things.
In this particular case, it's questionable whether the normal distribution is even a useful approximation. Nearly every grade distribution I've seen resembled the bimodal curve Ilmari Karonen mentioned in the comments to some degree. (Although usually with modes around 0.6 and 0.9, rather than on the extreme ends) However, a linear combination of two normal distributions with different means wouldn't be a bad approximation.
– Ray
Nov 13 at 4:13
I wasn't arguing that the normal is THE BEST approximation. The entire point of my comment is really made in that last paragraph. Comments about bimodality of actual grade distributions, at least at this level of abstraction, are really not helpful.
– Demetri Pananos
Nov 13 at 4:16
I agree with everything you said in your answer, but part of the question concerns whether the normal distribution is specifically applicable to modeling grade distributions. And the answer to that is usually "No". All models are wrong and some models are useful, but some are more wrong and less useful than others. The fact that the normal distribution in particular is an especially bad fit for this problem is important, and the answer as it is seems to suggest that the normal is only wrong because the tails go negative and infinite, when there are actually much deeper problems.
– Ray
Nov 13 at 4:23
You're being a little pedantic here. OP's problem was that the normal allows for negative scores. Bimodality wasn't the issue. You are not seeing the forest for the trees with respect to this question. Modelling details aren't relevant right now.
– Demetri Pananos
Nov 13 at 4:29
add a comment |
up vote
14
down vote
accepted
Height, for instance, is often modelled as being normal. Maybe the height of men is something like 5 foot 10 with a standard deviation of 2 inches. We know negative height is unphysical, but under this model, the probability of observing a negative height is essentially zero. We use the model anyway because it is a good enough approximation.
All models are wrong. The question is "can this model still be useful", and in instances where we are modelling things like height and test scores, modelling the phenomenon as normal is useful despite it technically allowing for unphysical things.
In this particular case, it's questionable whether the normal distribution is even a useful approximation. Nearly every grade distribution I've seen resembled the bimodal curve Ilmari Karonen mentioned in the comments to some degree. (Although usually with modes around 0.6 and 0.9, rather than on the extreme ends) However, a linear combination of two normal distributions with different means wouldn't be a bad approximation.
– Ray
Nov 13 at 4:13
I wasn't arguing that the normal is THE BEST approximation. The entire point of my comment is really made in that last paragraph. Comments about bimodality of actual grade distributions, at least at this level of abstraction, are really not helpful.
– Demetri Pananos
Nov 13 at 4:16
I agree with everything you said in your answer, but part of the question concerns whether the normal distribution is specifically applicable to modeling grade distributions. And the answer to that is usually "No". All models are wrong and some models are useful, but some are more wrong and less useful than others. The fact that the normal distribution in particular is an especially bad fit for this problem is important, and the answer as it is seems to suggest that the normal is only wrong because the tails go negative and infinite, when there are actually much deeper problems.
– Ray
Nov 13 at 4:23
You're being a little pedantic here. OP's problem was that the normal allows for negative scores. Bimodality wasn't the issue. You are not seeing the forest for the trees with respect to this question. Modelling details aren't relevant right now.
– Demetri Pananos
Nov 13 at 4:29
add a comment |
up vote
14
down vote
accepted
up vote
14
down vote
accepted
Height, for instance, is often modelled as being normal. Maybe the height of men is something like 5 foot 10 with a standard deviation of 2 inches. We know negative height is unphysical, but under this model, the probability of observing a negative height is essentially zero. We use the model anyway because it is a good enough approximation.
All models are wrong. The question is "can this model still be useful", and in instances where we are modelling things like height and test scores, modelling the phenomenon as normal is useful despite it technically allowing for unphysical things.
Height, for instance, is often modelled as being normal. Maybe the height of men is something like 5 foot 10 with a standard deviation of 2 inches. We know negative height is unphysical, but under this model, the probability of observing a negative height is essentially zero. We use the model anyway because it is a good enough approximation.
All models are wrong. The question is "can this model still be useful", and in instances where we are modelling things like height and test scores, modelling the phenomenon as normal is useful despite it technically allowing for unphysical things.
edited Nov 12 at 1:55
answered Nov 12 at 1:27
Demetri Pananos
678317
678317
In this particular case, it's questionable whether the normal distribution is even a useful approximation. Nearly every grade distribution I've seen resembled the bimodal curve Ilmari Karonen mentioned in the comments to some degree. (Although usually with modes around 0.6 and 0.9, rather than on the extreme ends) However, a linear combination of two normal distributions with different means wouldn't be a bad approximation.
– Ray
Nov 13 at 4:13
I wasn't arguing that the normal is THE BEST approximation. The entire point of my comment is really made in that last paragraph. Comments about bimodality of actual grade distributions, at least at this level of abstraction, are really not helpful.
– Demetri Pananos
Nov 13 at 4:16
I agree with everything you said in your answer, but part of the question concerns whether the normal distribution is specifically applicable to modeling grade distributions. And the answer to that is usually "No". All models are wrong and some models are useful, but some are more wrong and less useful than others. The fact that the normal distribution in particular is an especially bad fit for this problem is important, and the answer as it is seems to suggest that the normal is only wrong because the tails go negative and infinite, when there are actually much deeper problems.
– Ray
Nov 13 at 4:23
You're being a little pedantic here. OP's problem was that the normal allows for negative scores. Bimodality wasn't the issue. You are not seeing the forest for the trees with respect to this question. Modelling details aren't relevant right now.
– Demetri Pananos
Nov 13 at 4:29
add a comment |
In this particular case, it's questionable whether the normal distribution is even a useful approximation. Nearly every grade distribution I've seen resembled the bimodal curve Ilmari Karonen mentioned in the comments to some degree. (Although usually with modes around 0.6 and 0.9, rather than on the extreme ends) However, a linear combination of two normal distributions with different means wouldn't be a bad approximation.
– Ray
Nov 13 at 4:13
I wasn't arguing that the normal is THE BEST approximation. The entire point of my comment is really made in that last paragraph. Comments about bimodality of actual grade distributions, at least at this level of abstraction, are really not helpful.
– Demetri Pananos
Nov 13 at 4:16
I agree with everything you said in your answer, but part of the question concerns whether the normal distribution is specifically applicable to modeling grade distributions. And the answer to that is usually "No". All models are wrong and some models are useful, but some are more wrong and less useful than others. The fact that the normal distribution in particular is an especially bad fit for this problem is important, and the answer as it is seems to suggest that the normal is only wrong because the tails go negative and infinite, when there are actually much deeper problems.
– Ray
Nov 13 at 4:23
You're being a little pedantic here. OP's problem was that the normal allows for negative scores. Bimodality wasn't the issue. You are not seeing the forest for the trees with respect to this question. Modelling details aren't relevant right now.
– Demetri Pananos
Nov 13 at 4:29
In this particular case, it's questionable whether the normal distribution is even a useful approximation. Nearly every grade distribution I've seen resembled the bimodal curve Ilmari Karonen mentioned in the comments to some degree. (Although usually with modes around 0.6 and 0.9, rather than on the extreme ends) However, a linear combination of two normal distributions with different means wouldn't be a bad approximation.
– Ray
Nov 13 at 4:13
In this particular case, it's questionable whether the normal distribution is even a useful approximation. Nearly every grade distribution I've seen resembled the bimodal curve Ilmari Karonen mentioned in the comments to some degree. (Although usually with modes around 0.6 and 0.9, rather than on the extreme ends) However, a linear combination of two normal distributions with different means wouldn't be a bad approximation.
– Ray
Nov 13 at 4:13
I wasn't arguing that the normal is THE BEST approximation. The entire point of my comment is really made in that last paragraph. Comments about bimodality of actual grade distributions, at least at this level of abstraction, are really not helpful.
– Demetri Pananos
Nov 13 at 4:16
I wasn't arguing that the normal is THE BEST approximation. The entire point of my comment is really made in that last paragraph. Comments about bimodality of actual grade distributions, at least at this level of abstraction, are really not helpful.
– Demetri Pananos
Nov 13 at 4:16
I agree with everything you said in your answer, but part of the question concerns whether the normal distribution is specifically applicable to modeling grade distributions. And the answer to that is usually "No". All models are wrong and some models are useful, but some are more wrong and less useful than others. The fact that the normal distribution in particular is an especially bad fit for this problem is important, and the answer as it is seems to suggest that the normal is only wrong because the tails go negative and infinite, when there are actually much deeper problems.
– Ray
Nov 13 at 4:23
I agree with everything you said in your answer, but part of the question concerns whether the normal distribution is specifically applicable to modeling grade distributions. And the answer to that is usually "No". All models are wrong and some models are useful, but some are more wrong and less useful than others. The fact that the normal distribution in particular is an especially bad fit for this problem is important, and the answer as it is seems to suggest that the normal is only wrong because the tails go negative and infinite, when there are actually much deeper problems.
– Ray
Nov 13 at 4:23
You're being a little pedantic here. OP's problem was that the normal allows for negative scores. Bimodality wasn't the issue. You are not seeing the forest for the trees with respect to this question. Modelling details aren't relevant right now.
– Demetri Pananos
Nov 13 at 4:29
You're being a little pedantic here. OP's problem was that the normal allows for negative scores. Bimodality wasn't the issue. You are not seeing the forest for the trees with respect to this question. Modelling details aren't relevant right now.
– Demetri Pananos
Nov 13 at 4:29
add a comment |
up vote
10
down vote
Doesn't the normal distribution allow for negative values?
Correct. It also has no upper bound.
In one part of my textbook, it says that a normal distribution could be good for modeling exam scores.
In spite of the previous statements, nevertheless this is sometimes the case. If you have many components to the test, not too strongly related (e.g. so you're not essentially the same question a dozen times, nor having each part requiring a correct answer to the previous part), and not very easy or very hard (so that most marks are somewhere near the middle), then marks may often be reasonably well approximated by a normal distribution; often well enough that typical analyses should cause little concern.
We know for sure that they aren't normal, but that's not automatically a problem -- as long as the behaviour of the procedures we use are close enough to what they should be for our purposes (e.g. standard errors, confidence intervals, significance levels and power - whichever are needed - do close to what we expect them to)
In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values.
Yes, but more than that -- they tend to be heavily right skew and the variability tends to increase when the mean gets larger.
Here's an example of a claim-size distribution for vehicle claims:
https://ars.els-cdn.com/content/image/1-s2.0-S0167668715303358-gr5.jpg
(Fig 5 from Garrido, Genest & Schulz (2016) "Generalized linear models for dependent frequency and severity of insurance claims", Insurance: Mathematics and Economics, Vol 70, Sept., p205-215. https://www.sciencedirect.com/science/article/pii/S0167668715303358)
This shows a typical right-skew and heavy right tail. However we must be very careful because this is a marginal distribution, and we are writing a model for the conditional distribution, which will typically be much less skew (the marginal distribution we look at if we just do a histogram of claim sizes being a mixture of these conditional distributions). Nevertheless it is typically the case that if we look at the claim size in subgroups of the predictors (perhaps categorizing continuous variables) that the distribution is still strongly right skew and quite heavy tailed on the right, suggesting that something like a gamma model* is likely to be much more suitable than a Gaussian model.
* there may be any number of other distributions which would be more suitable than a Gaussian - the inverse Gaussian is another choice - though less common; lognormal or Weibull models, while not GLMs as they stand, may be quite useful also.
[It's rarely the case that any of these distributions are near-perfect descriptions; they're inexact approximations, but in many cases sufficiently good that the analysis is useful and has close to the desired properties.]
Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there?
Because (under the conditions I mentioned before -- lots of components, not too dependent, not to hard or easy) the distribution tends to be fairly close to symmetric, unimodal and not heavy-tailed.
add a comment |
up vote
10
down vote
Doesn't the normal distribution allow for negative values?
Correct. It also has no upper bound.
In one part of my textbook, it says that a normal distribution could be good for modeling exam scores.
In spite of the previous statements, nevertheless this is sometimes the case. If you have many components to the test, not too strongly related (e.g. so you're not essentially the same question a dozen times, nor having each part requiring a correct answer to the previous part), and not very easy or very hard (so that most marks are somewhere near the middle), then marks may often be reasonably well approximated by a normal distribution; often well enough that typical analyses should cause little concern.
We know for sure that they aren't normal, but that's not automatically a problem -- as long as the behaviour of the procedures we use are close enough to what they should be for our purposes (e.g. standard errors, confidence intervals, significance levels and power - whichever are needed - do close to what we expect them to)
In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values.
Yes, but more than that -- they tend to be heavily right skew and the variability tends to increase when the mean gets larger.
Here's an example of a claim-size distribution for vehicle claims:
https://ars.els-cdn.com/content/image/1-s2.0-S0167668715303358-gr5.jpg
(Fig 5 from Garrido, Genest & Schulz (2016) "Generalized linear models for dependent frequency and severity of insurance claims", Insurance: Mathematics and Economics, Vol 70, Sept., p205-215. https://www.sciencedirect.com/science/article/pii/S0167668715303358)
This shows a typical right-skew and heavy right tail. However we must be very careful because this is a marginal distribution, and we are writing a model for the conditional distribution, which will typically be much less skew (the marginal distribution we look at if we just do a histogram of claim sizes being a mixture of these conditional distributions). Nevertheless it is typically the case that if we look at the claim size in subgroups of the predictors (perhaps categorizing continuous variables) that the distribution is still strongly right skew and quite heavy tailed on the right, suggesting that something like a gamma model* is likely to be much more suitable than a Gaussian model.
* there may be any number of other distributions which would be more suitable than a Gaussian - the inverse Gaussian is another choice - though less common; lognormal or Weibull models, while not GLMs as they stand, may be quite useful also.
[It's rarely the case that any of these distributions are near-perfect descriptions; they're inexact approximations, but in many cases sufficiently good that the analysis is useful and has close to the desired properties.]
Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there?
Because (under the conditions I mentioned before -- lots of components, not too dependent, not to hard or easy) the distribution tends to be fairly close to symmetric, unimodal and not heavy-tailed.
add a comment |
up vote
10
down vote
up vote
10
down vote
Doesn't the normal distribution allow for negative values?
Correct. It also has no upper bound.
In one part of my textbook, it says that a normal distribution could be good for modeling exam scores.
In spite of the previous statements, nevertheless this is sometimes the case. If you have many components to the test, not too strongly related (e.g. so you're not essentially the same question a dozen times, nor having each part requiring a correct answer to the previous part), and not very easy or very hard (so that most marks are somewhere near the middle), then marks may often be reasonably well approximated by a normal distribution; often well enough that typical analyses should cause little concern.
We know for sure that they aren't normal, but that's not automatically a problem -- as long as the behaviour of the procedures we use are close enough to what they should be for our purposes (e.g. standard errors, confidence intervals, significance levels and power - whichever are needed - do close to what we expect them to)
In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values.
Yes, but more than that -- they tend to be heavily right skew and the variability tends to increase when the mean gets larger.
Here's an example of a claim-size distribution for vehicle claims:
https://ars.els-cdn.com/content/image/1-s2.0-S0167668715303358-gr5.jpg
(Fig 5 from Garrido, Genest & Schulz (2016) "Generalized linear models for dependent frequency and severity of insurance claims", Insurance: Mathematics and Economics, Vol 70, Sept., p205-215. https://www.sciencedirect.com/science/article/pii/S0167668715303358)
This shows a typical right-skew and heavy right tail. However we must be very careful because this is a marginal distribution, and we are writing a model for the conditional distribution, which will typically be much less skew (the marginal distribution we look at if we just do a histogram of claim sizes being a mixture of these conditional distributions). Nevertheless it is typically the case that if we look at the claim size in subgroups of the predictors (perhaps categorizing continuous variables) that the distribution is still strongly right skew and quite heavy tailed on the right, suggesting that something like a gamma model* is likely to be much more suitable than a Gaussian model.
* there may be any number of other distributions which would be more suitable than a Gaussian - the inverse Gaussian is another choice - though less common; lognormal or Weibull models, while not GLMs as they stand, may be quite useful also.
[It's rarely the case that any of these distributions are near-perfect descriptions; they're inexact approximations, but in many cases sufficiently good that the analysis is useful and has close to the desired properties.]
Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there?
Because (under the conditions I mentioned before -- lots of components, not too dependent, not to hard or easy) the distribution tends to be fairly close to symmetric, unimodal and not heavy-tailed.
Doesn't the normal distribution allow for negative values?
Correct. It also has no upper bound.
In one part of my textbook, it says that a normal distribution could be good for modeling exam scores.
In spite of the previous statements, nevertheless this is sometimes the case. If you have many components to the test, not too strongly related (e.g. so you're not essentially the same question a dozen times, nor having each part requiring a correct answer to the previous part), and not very easy or very hard (so that most marks are somewhere near the middle), then marks may often be reasonably well approximated by a normal distribution; often well enough that typical analyses should cause little concern.
We know for sure that they aren't normal, but that's not automatically a problem -- as long as the behaviour of the procedures we use are close enough to what they should be for our purposes (e.g. standard errors, confidence intervals, significance levels and power - whichever are needed - do close to what we expect them to)
In the next part, it asks what distribution would be appropriate to model a car insurance claim. This time, it said that the appropriate distributions would be Gamma or Inverse Gaussian because they're continuous with only positive values.
Yes, but more than that -- they tend to be heavily right skew and the variability tends to increase when the mean gets larger.
Here's an example of a claim-size distribution for vehicle claims:
https://ars.els-cdn.com/content/image/1-s2.0-S0167668715303358-gr5.jpg
(Fig 5 from Garrido, Genest & Schulz (2016) "Generalized linear models for dependent frequency and severity of insurance claims", Insurance: Mathematics and Economics, Vol 70, Sept., p205-215. https://www.sciencedirect.com/science/article/pii/S0167668715303358)
This shows a typical right-skew and heavy right tail. However we must be very careful because this is a marginal distribution, and we are writing a model for the conditional distribution, which will typically be much less skew (the marginal distribution we look at if we just do a histogram of claim sizes being a mixture of these conditional distributions). Nevertheless it is typically the case that if we look at the claim size in subgroups of the predictors (perhaps categorizing continuous variables) that the distribution is still strongly right skew and quite heavy tailed on the right, suggesting that something like a gamma model* is likely to be much more suitable than a Gaussian model.
* there may be any number of other distributions which would be more suitable than a Gaussian - the inverse Gaussian is another choice - though less common; lognormal or Weibull models, while not GLMs as they stand, may be quite useful also.
[It's rarely the case that any of these distributions are near-perfect descriptions; they're inexact approximations, but in many cases sufficiently good that the analysis is useful and has close to the desired properties.]
Well, I believe that exam scores would also be continuous with only positive values, so why would we use a normal distribution there?
Because (under the conditions I mentioned before -- lots of components, not too dependent, not to hard or easy) the distribution tends to be fairly close to symmetric, unimodal and not heavy-tailed.
edited Nov 12 at 4:03
answered Nov 12 at 1:27
Glen_b♦
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If you're worried about the bounds on scores, you could try en.wikipedia.org/wiki/Truncated_normal_distribution
– J.G.
Nov 12 at 6:39
5
In the real world, of course, exam score distributions often don't look anything like a normal distribution anyway. As an example from my math undergrad days, I remember the Topology I class as having been notorious for its highly bimodal "dumbbell curve" grade distribution: you either understood the key concepts and got a nearly perfect score, or you didn't and were lucky to get any points at all. Very few people ended up scoring anywhere in the middle between those two extremes.
– Ilmari Karonen
Nov 12 at 10:44
2
No. Next question.
– Carl Witthoft
Nov 12 at 16:24