Need help to understand a density histogram
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I've a density histogram, I don't understand. I hope somebody can help me with that.
The histogram: 
I know that a density histogram is the bar areas sum to 1. But I don't understand the histogram I have. It's a histogram of a consumption. Is it positive? How is the density? Is it symmetric or skewed?
statistics
edited Oct 14 '15 at 21:59
tomi
6,18611132
asked Oct 14 '15 at 21:46
user4476151
14
add a comment |
up vote
0
down vote
favorite
I've a density histogram, I don't understand. I hope somebody can help me with that.
The histogram:
I know that a density histogram is the bar areas sum to 1. But I don't understand the histogram I have. It's a histogram of a consumption. Is it positive? How is the density? Is it symmetric or skewed?
statistics
edited Oct 14 '15 at 21:59
tomi
6,18611132
asked Oct 14 '15 at 21:46
user4476151
14
it looks like a mixture of an exponential and a gamma (or something similar). Might be a useful way to model it.
– user237392
Oct 15 '15 at 3:58
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've a density histogram, I don't understand. I hope somebody can help me with that.
The histogram:
I know that a density histogram is the bar areas sum to 1. But I don't understand the histogram I have. It's a histogram of a consumption. Is it positive? How is the density? Is it symmetric or skewed?
statistics
edited Oct 14 '15 at 21:59
tomi
6,18611132
asked Oct 14 '15 at 21:46
user4476151
14
I've a density histogram, I don't understand. I hope somebody can help me with that.
The histogram:
I know that a density histogram is the bar areas sum to 1. But I don't understand the histogram I have. It's a histogram of a consumption. Is it positive? How is the density? Is it symmetric or skewed?
statistics
statistics
edited Oct 14 '15 at 21:59
tomi
6,18611132
asked Oct 14 '15 at 21:46
user4476151
14
edited Oct 14 '15 at 21:59
tomi
6,18611132
asked Oct 14 '15 at 21:46
user4476151
14
edited Oct 14 '15 at 21:59
tomi
6,18611132
edited Oct 14 '15 at 21:59
tomi
6,18611132
edited Oct 14 '15 at 21:59
tomi
6,18611132
6,18611132
asked Oct 14 '15 at 21:46
user4476151
14
asked Oct 14 '15 at 21:46
user4476151
14
asked Oct 14 '15 at 21:46
user4476151
14
14
it looks like a mixture of an exponential and a gamma (or something similar). Might be a useful way to model it.
– user237392
Oct 15 '15 at 3:58
add a comment |
it looks like a mixture of an exponential and a gamma (or something similar). Might be a useful way to model it.
– user237392
Oct 15 '15 at 3:58
it looks like a mixture of an exponential and a gamma (or something similar). Might be a useful way to model it.
– user237392
Oct 15 '15 at 3:58
it looks like a mixture of an exponential and a gamma (or something similar). Might be a useful way to model it.
– user237392
Oct 15 '15 at 3:58
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
In an ordinary histogram the area of a bar is equal to the frequency. In this example the area of a bar is equal to the relative frequency = frequency divided by sum of frequencies.
So, for example, the very leftmost bar has height about 0.49 and width 0.5, so area = 0.245 - which means that about 24.5% of the observations are found to take a value of between 0 and 0.4 (whatever HE$Q is).
The distribution is not symmetrical. There is a tail stretching out on the right, so the distribution is positively skewed.
answered Oct 14 '15 at 22:07
tomi
6,18611132
Isn't the width 0.5? How did you find the area?
– user4476151
Oct 14 '15 at 22:16
Is it something you calculated?
– user4476151
Oct 14 '15 at 22:35
Corrected the width now
– tomi
Oct 15 '15 at 11:33
But how did you find the area?
– user4476151
Oct 15 '15 at 12:13
Area = width $times$ height $=0.5 times 0.49 =0.245$
– tomi
Oct 15 '15 at 22:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In an ordinary histogram the area of a bar is equal to the frequency. In this example the area of a bar is equal to the relative frequency = frequency divided by sum of frequencies.
So, for example, the very leftmost bar has height about 0.49 and width 0.5, so area = 0.245 - which means that about 24.5% of the observations are found to take a value of between 0 and 0.4 (whatever HE$Q is).
The distribution is not symmetrical. There is a tail stretching out on the right, so the distribution is positively skewed.
answered Oct 14 '15 at 22:07
tomi
6,18611132
Isn't the width 0.5? How did you find the area?
– user4476151
Oct 14 '15 at 22:16
Is it something you calculated?
– user4476151
Oct 14 '15 at 22:35
Corrected the width now
– tomi
Oct 15 '15 at 11:33
But how did you find the area?
– user4476151
Oct 15 '15 at 12:13
Area = width $times$ height $=0.5 times 0.49 =0.245$
– tomi
Oct 15 '15 at 22:52
add a comment |
up vote
0
down vote
In an ordinary histogram the area of a bar is equal to the frequency. In this example the area of a bar is equal to the relative frequency = frequency divided by sum of frequencies.
So, for example, the very leftmost bar has height about 0.49 and width 0.5, so area = 0.245 - which means that about 24.5% of the observations are found to take a value of between 0 and 0.4 (whatever HE$Q is).
The distribution is not symmetrical. There is a tail stretching out on the right, so the distribution is positively skewed.
answered Oct 14 '15 at 22:07
tomi
6,18611132
Isn't the width 0.5? How did you find the area?
– user4476151
Oct 14 '15 at 22:16
Is it something you calculated?
– user4476151
Oct 14 '15 at 22:35
Corrected the width now
– tomi
Oct 15 '15 at 11:33
But how did you find the area?
– user4476151
Oct 15 '15 at 12:13
Area = width $times$ height $=0.5 times 0.49 =0.245$
– tomi
Oct 15 '15 at 22:52
add a comment |
up vote
0
down vote
up vote
0
down vote
In an ordinary histogram the area of a bar is equal to the frequency. In this example the area of a bar is equal to the relative frequency = frequency divided by sum of frequencies.
So, for example, the very leftmost bar has height about 0.49 and width 0.5, so area = 0.245 - which means that about 24.5% of the observations are found to take a value of between 0 and 0.4 (whatever HE$Q is).
The distribution is not symmetrical. There is a tail stretching out on the right, so the distribution is positively skewed.
answered Oct 14 '15 at 22:07
tomi
6,18611132
In an ordinary histogram the area of a bar is equal to the frequency. In this example the area of a bar is equal to the relative frequency = frequency divided by sum of frequencies.
So, for example, the very leftmost bar has height about 0.49 and width 0.5, so area = 0.245 - which means that about 24.5% of the observations are found to take a value of between 0 and 0.4 (whatever HE$Q is).
The distribution is not symmetrical. There is a tail stretching out on the right, so the distribution is positively skewed.
answered Oct 14 '15 at 22:07
tomi
6,18611132
edited Oct 15 '15 at 11:33
answered Oct 14 '15 at 22:07
tomi
6,18611132
answered Oct 14 '15 at 22:07
tomi
6,18611132
answered Oct 14 '15 at 22:07
tomi
6,18611132
6,18611132
Isn't the width 0.5? How did you find the area?
– user4476151
Oct 14 '15 at 22:16
Is it something you calculated?
– user4476151
Oct 14 '15 at 22:35
Corrected the width now
– tomi
Oct 15 '15 at 11:33
But how did you find the area?
– user4476151
Oct 15 '15 at 12:13
Area = width $times$ height $=0.5 times 0.49 =0.245$
– tomi
Oct 15 '15 at 22:52
add a comment |
Isn't the width 0.5? How did you find the area?
– user4476151
Oct 14 '15 at 22:16
Is it something you calculated?
– user4476151
Oct 14 '15 at 22:35
Corrected the width now
– tomi
Oct 15 '15 at 11:33
But how did you find the area?
– user4476151
Oct 15 '15 at 12:13
Area = width $times$ height $=0.5 times 0.49 =0.245$
– tomi
Oct 15 '15 at 22:52
Isn't the width 0.5? How did you find the area?
– user4476151
Oct 14 '15 at 22:16
Isn't the width 0.5? How did you find the area?
– user4476151
Oct 14 '15 at 22:16
Is it something you calculated?
– user4476151
Oct 14 '15 at 22:35
Is it something you calculated?
– user4476151
Oct 14 '15 at 22:35
Corrected the width now
– tomi
Oct 15 '15 at 11:33
Corrected the width now
– tomi
Oct 15 '15 at 11:33
But how did you find the area?
– user4476151
Oct 15 '15 at 12:13
But how did you find the area?
– user4476151
Oct 15 '15 at 12:13
Area = width $times$ height $=0.5 times 0.49 =0.245$
– tomi
Oct 15 '15 at 22:52
Area = width $times$ height $=0.5 times 0.49 =0.245$
– tomi
Oct 15 '15 at 22:52
add a comment |
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it looks like a mixture of an exponential and a gamma (or something similar). Might be a useful way to model it.
– user237392
Oct 15 '15 at 3:58