Mixing
Can You Have 2 Numerical, Identical Values To Represent the nth Percentile?
$begingroup$
For an assignment, I am required to find the value of a piece of data in the 70th
percentile using mean and standard deviation.
Here is my organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.
The image link attached below called 'Percentiles With Normal Distribution' is the work I have done to find the value. (For some reason I can`t seem to upload the actual image here...)
Percentiles with Normal Distribution
Just in case you cannot access the image, I will explain what I did.
I created the normal distribution graph and shaded approximately 70% of it, leaving a little part on the right side empty.
I am finding b, where P(X > b) = p, where p = 0.7. I reversed this and it is nowP(X > b) = 1 - p (where p = 0.7).This means to find the (1-p)Th percentile for X.Next, I found the corresponding percentile for Z by looking in the body of the Z-Score Table, and finding the probability that is closest to p = 0.7, to which I did this:
(1-0.7) = 0.3 so the closest value to this is 0.3015, which falls under row = -0.5 and column = 0.02.This means the70th percentile for Z is equal to -0.52.Lastly, I just changed the Z-Score value back into an x-value (original units).
x = Mean + Z(Standard Deviation).I substiuted my mean value of -0.55, Z-Score value of -0.52, and standard deviation of 2.4 and solved for x.x = -2.
Now, if you look at the organized data set I provided above, youd see that there are two -2s. This is confusing me because I`m not sure whether there can be two numerical values to represent the value of a certain percentile.
probability normal-distribution standard-deviation percentile
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
$endgroup$
add a comment |
$begingroup$
For an assignment, I am required to find the value of a piece of data in the 70th
percentile using mean and standard deviation.
Here is my organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.
The image link attached below called 'Percentiles With Normal Distribution' is the work I have done to find the value. (For some reason I can`t seem to upload the actual image here...)
Percentiles with Normal Distribution
Just in case you cannot access the image, I will explain what I did.
I created the normal distribution graph and shaded approximately 70% of it, leaving a little part on the right side empty.
I am finding b, where P(X > b) = p, where p = 0.7. I reversed this and it is nowP(X > b) = 1 - p (where p = 0.7).This means to find the (1-p)Th percentile for X.Next, I found the corresponding percentile for Z by looking in the body of the Z-Score Table, and finding the probability that is closest to p = 0.7, to which I did this:
(1-0.7) = 0.3 so the closest value to this is 0.3015, which falls under row = -0.5 and column = 0.02.This means the70th percentile for Z is equal to -0.52.Lastly, I just changed the Z-Score value back into an x-value (original units).
x = Mean + Z(Standard Deviation).I substiuted my mean value of -0.55, Z-Score value of -0.52, and standard deviation of 2.4 and solved for x.x = -2.
Now, if you look at the organized data set I provided above, youd see that there are two -2s. This is confusing me because I`m not sure whether there can be two numerical values to represent the value of a certain percentile.
probability normal-distribution standard-deviation percentile
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
$endgroup$
add a comment |
$begingroup$
For an assignment, I am required to find the value of a piece of data in the 70th
percentile using mean and standard deviation.
Here is my organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.
The image link attached below called 'Percentiles With Normal Distribution' is the work I have done to find the value. (For some reason I can`t seem to upload the actual image here...)
Percentiles with Normal Distribution
Just in case you cannot access the image, I will explain what I did.
I created the normal distribution graph and shaded approximately 70% of it, leaving a little part on the right side empty.
I am finding b, where P(X > b) = p, where p = 0.7. I reversed this and it is nowP(X > b) = 1 - p (where p = 0.7).This means to find the (1-p)Th percentile for X.Next, I found the corresponding percentile for Z by looking in the body of the Z-Score Table, and finding the probability that is closest to p = 0.7, to which I did this:
(1-0.7) = 0.3 so the closest value to this is 0.3015, which falls under row = -0.5 and column = 0.02.This means the70th percentile for Z is equal to -0.52.Lastly, I just changed the Z-Score value back into an x-value (original units).
x = Mean + Z(Standard Deviation).I substiuted my mean value of -0.55, Z-Score value of -0.52, and standard deviation of 2.4 and solved for x.x = -2.
Now, if you look at the organized data set I provided above, youd see that there are two -2s. This is confusing me because I`m not sure whether there can be two numerical values to represent the value of a certain percentile.
probability normal-distribution standard-deviation percentile
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
$endgroup$
For an assignment, I am required to find the value of a piece of data in the 70th
percentile using mean and standard deviation.
Here is my organized data in ascending order: 0, 1, -4, 0, -3, 0, -4, -2, 0, 1, -1, 2, 0, -4, 5, -2, 0, 3, -3, 0.
The image link attached below called 'Percentiles With Normal Distribution' is the work I have done to find the value. (For some reason I can`t seem to upload the actual image here...)
Percentiles with Normal Distribution
Just in case you cannot access the image, I will explain what I did.
I created the normal distribution graph and shaded approximately 70% of it, leaving a little part on the right side empty.
I am finding b, where P(X > b) = p, where p = 0.7. I reversed this and it is nowP(X > b) = 1 - p (where p = 0.7).This means to find the (1-p)Th percentile for X.Next, I found the corresponding percentile for Z by looking in the body of the Z-Score Table, and finding the probability that is closest to p = 0.7, to which I did this:
(1-0.7) = 0.3 so the closest value to this is 0.3015, which falls under row = -0.5 and column = 0.02.This means the70th percentile for Z is equal to -0.52.Lastly, I just changed the Z-Score value back into an x-value (original units).
x = Mean + Z(Standard Deviation).I substiuted my mean value of -0.55, Z-Score value of -0.52, and standard deviation of 2.4 and solved for x.x = -2.
Now, if you look at the organized data set I provided above, youd see that there are two -2s. This is confusing me because I`m not sure whether there can be two numerical values to represent the value of a certain percentile.
probability normal-distribution standard-deviation percentile
probability normal-distribution standard-deviation percentile
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
asked Jan 3 at 21:34
Yashvi ShahYashvi Shah
165
165
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Three points:
If you look at your picture, you want $+0.52$ for the $70$th percentile of a standard normal distribution, but you used $-0.52$ which in fact corresponds to the $30$th percentile. Correcting this would have given you about $0.7$ rather than $-1.8$
There is no problem with a percentile corresponding to a value from the sample which happens multiple times. For example with the data
3, 5, 5, 5, 7it is obvious that the $50$th percentile or median is $5$I am not sure why you need to fit a normal distribution unless it is to fit the question's "using mean and standard deviation". As an alternative, sorting your $20$ data points to
-4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, you have $25%$ of the values below $-2$ and $65%$ of the values above $-2$ so you could say $-2$ is the $30$th percentile. Similarly you have $40%$ of the values below $0$ and $25%$ above $0$ so you could say $0$ is the $70$th percentile.
answered Jan 3 at 22:32
HenryHenry
99.2k478164
$endgroup$
$begingroup$
This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2.
$endgroup$
– Yashvi Shah
Jan 4 at 3:40
$begingroup$
@YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations.
$endgroup$
– Henry
Jan 4 at 8:03
$begingroup$
To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer
$endgroup$
– Henry
Jan 4 at 8:31
$begingroup$
Im not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something?
$endgroup$
– Yashvi Shah
Jan 4 at 17:47
$begingroup$
Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/…
$endgroup$
– Yashvi Shah
Jan 4 at 20:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Three points:
If you look at your picture, you want $+0.52$ for the $70$th percentile of a standard normal distribution, but you used $-0.52$ which in fact corresponds to the $30$th percentile. Correcting this would have given you about $0.7$ rather than $-1.8$
There is no problem with a percentile corresponding to a value from the sample which happens multiple times. For example with the data
3, 5, 5, 5, 7it is obvious that the $50$th percentile or median is $5$I am not sure why you need to fit a normal distribution unless it is to fit the question's "using mean and standard deviation". As an alternative, sorting your $20$ data points to
-4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, you have $25%$ of the values below $-2$ and $65%$ of the values above $-2$ so you could say $-2$ is the $30$th percentile. Similarly you have $40%$ of the values below $0$ and $25%$ above $0$ so you could say $0$ is the $70$th percentile.
answered Jan 3 at 22:32
HenryHenry
99.2k478164
$endgroup$
$begingroup$
This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2.
$endgroup$
– Yashvi Shah
Jan 4 at 3:40
$begingroup$
@YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations.
$endgroup$
– Henry
Jan 4 at 8:03
$begingroup$
To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer
$endgroup$
– Henry
Jan 4 at 8:31
$begingroup$
Im not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something?
$endgroup$
– Yashvi Shah
Jan 4 at 17:47
$begingroup$
Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/…
$endgroup$
– Yashvi Shah
Jan 4 at 20:31
add a comment |
$begingroup$
Three points:
If you look at your picture, you want $+0.52$ for the $70$th percentile of a standard normal distribution, but you used $-0.52$ which in fact corresponds to the $30$th percentile. Correcting this would have given you about $0.7$ rather than $-1.8$
There is no problem with a percentile corresponding to a value from the sample which happens multiple times. For example with the data
3, 5, 5, 5, 7it is obvious that the $50$th percentile or median is $5$I am not sure why you need to fit a normal distribution unless it is to fit the question's "using mean and standard deviation". As an alternative, sorting your $20$ data points to
-4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, you have $25%$ of the values below $-2$ and $65%$ of the values above $-2$ so you could say $-2$ is the $30$th percentile. Similarly you have $40%$ of the values below $0$ and $25%$ above $0$ so you could say $0$ is the $70$th percentile.
answered Jan 3 at 22:32
HenryHenry
99.2k478164
$endgroup$
$begingroup$
This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2.
$endgroup$
– Yashvi Shah
Jan 4 at 3:40
$begingroup$
@YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations.
$endgroup$
– Henry
Jan 4 at 8:03
$begingroup$
To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer
$endgroup$
– Henry
Jan 4 at 8:31
$begingroup$
Im not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something?
$endgroup$
– Yashvi Shah
Jan 4 at 17:47
$begingroup$
Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/…
$endgroup$
– Yashvi Shah
Jan 4 at 20:31
add a comment |
$begingroup$
Three points:
If you look at your picture, you want $+0.52$ for the $70$th percentile of a standard normal distribution, but you used $-0.52$ which in fact corresponds to the $30$th percentile. Correcting this would have given you about $0.7$ rather than $-1.8$
There is no problem with a percentile corresponding to a value from the sample which happens multiple times. For example with the data
3, 5, 5, 5, 7it is obvious that the $50$th percentile or median is $5$I am not sure why you need to fit a normal distribution unless it is to fit the question's "using mean and standard deviation". As an alternative, sorting your $20$ data points to
-4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, you have $25%$ of the values below $-2$ and $65%$ of the values above $-2$ so you could say $-2$ is the $30$th percentile. Similarly you have $40%$ of the values below $0$ and $25%$ above $0$ so you could say $0$ is the $70$th percentile.
answered Jan 3 at 22:32
HenryHenry
99.2k478164
$endgroup$
Three points:
If you look at your picture, you want $+0.52$ for the $70$th percentile of a standard normal distribution, but you used $-0.52$ which in fact corresponds to the $30$th percentile. Correcting this would have given you about $0.7$ rather than $-1.8$
There is no problem with a percentile corresponding to a value from the sample which happens multiple times. For example with the data
3, 5, 5, 5, 7it is obvious that the $50$th percentile or median is $5$I am not sure why you need to fit a normal distribution unless it is to fit the question's "using mean and standard deviation". As an alternative, sorting your $20$ data points to
-4, -4, -4, -3, -3, -2, -2, -1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, you have $25%$ of the values below $-2$ and $65%$ of the values above $-2$ so you could say $-2$ is the $30$th percentile. Similarly you have $40%$ of the values below $0$ and $25%$ above $0$ so you could say $0$ is the $70$th percentile.
answered Jan 3 at 22:32
HenryHenry
99.2k478164
edited Jan 4 at 8:28
answered Jan 3 at 22:32
HenryHenry
99.2k478164
answered Jan 3 at 22:32
HenryHenry
99.2k478164
answered Jan 3 at 22:32
HenryHenry
99.2k478164
99.2k478164
$begingroup$
This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2.
$endgroup$
– Yashvi Shah
Jan 4 at 3:40
$begingroup$
@YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations.
$endgroup$
– Henry
Jan 4 at 8:03
$begingroup$
To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer
$endgroup$
– Henry
Jan 4 at 8:31
$begingroup$
Im not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something?
$endgroup$
– Yashvi Shah
Jan 4 at 17:47
$begingroup$
Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/…
$endgroup$
– Yashvi Shah
Jan 4 at 20:31
add a comment |
$begingroup$
This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2.
$endgroup$
– Yashvi Shah
Jan 4 at 3:40
$begingroup$
@YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations.
$endgroup$
– Henry
Jan 4 at 8:03
$begingroup$
To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer
$endgroup$
– Henry
Jan 4 at 8:31
$begingroup$
Im not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something?
$endgroup$
– Yashvi Shah
Jan 4 at 17:47
$begingroup$
Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/…
$endgroup$
– Yashvi Shah
Jan 4 at 20:31
$begingroup$
This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2.
$endgroup$
– Yashvi Shah
Jan 4 at 3:40
$begingroup$
This really helps! Thank you! But, is there a mathematical calculation way to prove that 0 is the 70th percentile? My teacher demands mathematical calculations. The way I’ve done it, I’m getting -2.
$endgroup$
– Yashvi Shah
Jan 4 at 3:40
$begingroup$
@YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations.
$endgroup$
– Henry
Jan 4 at 8:03
$begingroup$
@YashviShah: all my comments were mathematical. You should not be getting $-2$ but about $0.7$ from your calculations.
$endgroup$
– Henry
Jan 4 at 8:03
$begingroup$
To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer
$endgroup$
– Henry
Jan 4 at 8:31
$begingroup$
To justify $0$ you could draw an empirical cumulative distribution graph perhaps like the one I have added to my answer
$endgroup$
– Henry
Jan 4 at 8:31
$begingroup$
I
m not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something?$endgroup$
– Yashvi Shah
Jan 4 at 17:47
$begingroup$
I
m not quite sure how youre getting 0 as the 70th percentile through the calculations. Did you calculate the same way I did up there? What did you do? Im trying to get the result of 0 being the 70th percentile. And as for the empirical cumulative distribution graph, I dont think we learnt that. Is there a way to justify using a normal distribution graph or something?$endgroup$
– Yashvi Shah
Jan 4 at 17:47
$begingroup$
Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/…
$endgroup$
– Yashvi Shah
Jan 4 at 20:31
$begingroup$
Also, I used this link to help me with getting the percentile value using mean and standard deviation - dummies.com/education/math/statistics/…
$endgroup$
– Yashvi Shah
Jan 4 at 20:31
add a comment |
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