What percentage of people scored between 100 and 110, for mean 100, s = 20?

klhoff?

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Hello,
I am having trouble answering a question from my statistics homework. I do not know how to get started. Please walk me through it like you were teaching a child.

Mean=100
s=20

What percentage of people scored between 100 and 110?

I have to draw a graph and calculate z scores. Unfortunately I do not know how to type the formula using the key pad. I am fully aware that your time is precious and that you are going out of your way to help me. i would like to thank you in advance.
 
Assuming that you were given literally no information other than what you included in your post, and thus you don't know anything about the variable's distribution, you won't be able to get an exact answer. The best you can do is Chebyshev’s Theoremhttp://www.statisticshowto.com/how-to-calculate-chebyshevs-theorem/, which gives a minimum proportion of observations that will fall within k standard deviations of the mean. Specifically:

\(\displaystyle P(|X - \mu| \le k\sigma) \ge 1 - \dfrac{1}{k^2}\)

Here, one of the endcaps of your interval is the mean, which even further complicates things because the interval you're interested in isn't symmetric about the mean. To deal with this, you can use the generalized form of Chebyshev’s Theoremhttp://www.ams.sunysb.edu/~jsbm/courses/311/cheby.pdf, which states:

\(\displaystyle P(a < X < b) \ge 1 - \dfrac{\sigma^2 + \left( \mu - \dfrac{a+b}{2} \right)^2}{\dfrac{b-a}{2}}\)

Here, plug in a = 100 and b = 110. However, I'm fairly certain this isn't what's expected of you, considering that I didn't even know that the generalized form of the theorem even existed until just now when I went looking to see if it did. It would be in pretty poor form for a homework exercise to expect you to use Google like that.

Hence, I can reasonably conclude that there is more to this exercise than what you've posted. Please share with us the full and exact text of the problem, ideally quoting word-for-word if possible. Thank you.
 
Hello,
I am having trouble answering a question from my statistics homework. I do not know how to get started. Please walk me through it like you were teaching a child.

Mean=100
s=20

What percentage of people scored between 100 and 110?

I have to draw a graph and calculate z scores. Unfortunately I do not know how to type the formula using the key pad.

I would not teach this to a child! We need to have some starting point that we can assume you know.

Your comments imply (a) that you know something about the normal distribution, and perhaps the formula z = (x - mean)/sd, and (b) that the problem justifies using it. But the problem as you state it says nothing about such assumptions. You also have not said whether you will be using a normal distribution table, technology, or something else.

Please state the problem completely and exactly as given to you, as we say in the Read Before Posting warning. Also, give us some idea of where you are stuck, and how much you do know. That is part of how you show respect to people who are willing to help you -- don't make it harder than it has to be.
 
I would not teach this to a child! We need to have some starting point that we can assume you know.

Your comments imply (a) that you know something about the normal distribution, and perhaps the formula z = (x - mean)/sd, and (b) that the problem justifies using it. But the problem as you state it says nothing about such assumptions. You also have not said whether you will be using a normal distribution table, technology, or something else.

Please state the problem completely and exactly as given to you, as we say in the Read Before Posting warning. Also, give us some idea of where you are stuck, and how much you do know. That is part of how you show respect to people who are willing to help you -- don't make it harder than it has to be.

It is normal distribution and I have stated the problem exactly how it was given. I do not know where to begin to answer this problem. I am stuck at is x 100 + 110 or are they two different equations. The only thing that is stated clearly is the mean and standard deviation. When I say teach me as you would a child I mean simplify the steps it takes to solve the problem. I would like to understand how to do this problem. I will be tested on similar equations. I am not disrespecting anyone. If you do not want to help please let another person respond to my post. The tone of your response comes across as hostile.
 
I would not teach this to a child! We need to have some starting point that we can assume you know.

Your comments imply (a) that you know something about the normal distribution, and perhaps the formula z = (x - mean)/sd, and (b) that the problem justifies using it. But the problem as you state it says nothing about such assumptions. You also have not said whether you will be using a normal distribution table, technology, or something else.

Please state the problem completely and exactly as given to you, as we say in the Read Before Posting warning. Also, give us some idea of where you are stuck, and how much you do know. That is part of how you show respect to people who are willing to help you -- don't make it harder than it has to be.

The formula is definitely z = (x - mean)/sd
 
Assuming that you were given literally no information other than what you included in your post, and thus you don't know anything about the variable's distribution, you won't be able to get an exact answer. The best you can do is Chebyshev’s Theorem, which gives a minimum proportion of observations that will fall within k standard deviations of the mean. Specifically:

\(\displaystyle P(|X - \mu| \le k\sigma) \ge 1 - \dfrac{1}{k^2}\)

Here, one of the endcaps of your interval is the mean, which even further complicates things because the interval you're interested in isn't symmetric about the mean. To deal with this, you can use the generalized form of Chebyshev’s Theorem, which states:

\(\displaystyle P(a < X < b) \ge 1 - \dfrac{\sigma^2 + \left( \mu - \dfrac{a+b}{2} \right)^2}{\dfrac{b-a}{2}}\)

Here, plug in a = 100 and b = 110. However, I'm fairly certain this isn't what's expected of you, considering that I didn't even know that the generalized form of the theorem even existed until just now when I went looking to see if it did. It would be in pretty poor form for a homework exercise to expect you to use Google like that.

Hence, I can reasonably conclude that there is more to this exercise than what you've posted. Please share with us the full and exact text of the problem, ideally quoting word-for-word if possible. Thank you.

Hi,
Thank you so much for responding. The equation is z = (x - mean)/sd. The above information is all that was provided. I do not know if I am to add 100 and 110 for X or if I am suppose to use different equations. This is my first statistics class. I was told to calculate z scores. I know there are z tables.

Auna
 
The formula is definitely z = (x - mean)/sd
So try using the formula and let's see what you get from that. Also, the next time you say that a helper is hostile on this forum I promise you that will never happen again because no one will help you.
 
The formula is definitely z = (x - mean)/sd
You should state the whole problem which includes the section/formulas you think should be used. We don't mind helping people but our time is valuable and we rather not spend it asking for students to post the entire problem along with their work.
 
It is normal distribution and I have stated the problem exactly how it was given. I do not know where to begin to answer this problem. I am stuck at is x 100 + 110 or are they two different equations. The only thing that is stated clearly is the mean and standard deviation.

When I say teach me as you would a child I mean simplify the steps it takes to solve the problem. I would like to understand how to do this problem. I will be tested on similar equations. I am not disrespecting anyone. If you do not want to help please let another person respond to my post. The tone of your response comes across as hostile.

I'm sorry for the tone of my last sentence; my intention was only to let you know how to ask questions better in the future, and I was responding to your very nice expression of thanks and respect, not to any disrespect I perceived.

The more you tell us about the problem and your needs (as you have now), the better we can do. (If I were teaching a child, I would start out by trying to find out what she knew and encouraging her to try as much as she could, not by assuming she knew nothing). The most important thing in helping a student is knowledge of their particular needs.) I'll try to treat you as a beginner, which you are.

Okay, somewhere (maybe not in this problem, but in the header for a set of problems), it must have said to assume a normal distribution. That's an essential part, and is the reason the first response you got was totally unrelated to your actual need.

So the problem was,

Mean=100
s=20

What percentage of people scored between 100 and 110?

You are expected to sketch the normal distribution, where the middle (mean) will be 100. Then you would draw vertical lines at 100 (the middle) and at 110 (somewhere to the right of that) and shading in the region between. The percentage you want can be described as the area of the shaded region.

The first step is to find z, by plugging the given numbers into the formula. You know the mean and standard deviation; the x is any value you are interested in, in this case 110.

So the value of z corresponding to x=110 is (110 - 100)/20. Calculate that.

Now what to do next depends on some things I asked about that you have not answered, probably because you don't know why I had to ask. How are you finding the area (probability) in your course? In some courses, you are given a computer program to do this, or use a graphing calculator. In others, you are given a table to look it up in. But even those tables differ. Some give the answer to your question directly (the area between the middle and a given value); others give the entire area to the left, and for your problem you would have to subtract 50% (the left half of the graph). So this is one of those situations in which I would much rather be sitting next to you, so I could look over your book or notes and determine what method you need to use! You can see examples of three kinds of tables here. Please let me know what your class is using, so we can continue.
 
The help provided is appreciated

So try using the formula and let's see what you get from that. Also, the next time you say that a helper is hostile on this forum I promise you that will never happen again because no one will help you.

Thank you for your reply.
Please note my first post . "I am fully aware that your time is precious and that you are going out of your way to help me. i would like to thank you in advance."

Also note people with whom I associate will find the feedback I give pertaining to this forum helpful. I would like to continue giving positive feedback. The help provided is appreciated. Thank you again for your thoughts concerning this matter.

Warmly,
Auna
 
The help provided is appreciated

I'm sorry for the tone of my last sentence; my intention was only to let you know how to ask questions better in the future, and I was responding to your very nice expression of thanks and respect, not to any disrespect I perceived.

The more you tell us about the problem and your needs (as you have now), the better we can do. (If I were teaching a child, I would start out by trying to find out what she knew and encouraging her to try as much as she could, not by assuming she knew nothing). The most important thing in helping a student is knowledge of their particular needs.) I'll try to treat you as a beginner, which you are.

Okay, somewhere (maybe not in this problem, but in the header for a set of problems), it must have said to assume a normal distribution. That's an essential part, and is the reason the first response you got was totally unrelated to your actual need.

So the problem was,
Mean=100
s=20

What percentage of people scored between 100 and 110?

You are expected to sketch the normal distribution, where the middle (mean) will be 100. Then you would draw vertical lines at 100 (the middle) and at 110 (somewhere to the right of that) and shading in the region between. The percentage you want can be described as the area of the shaded region.

The first step is to find z, by plugging the given numbers into the formula. You know the mean and standard deviation; the x is any value you are interested in, in this case 110.

So the value of z corresponding to x=110 is (110 - 100)/20. Calculate that.

Now what to do next depends on some things I asked about that you have not answered, probably because you don't know why I had to ask. How are you finding the area (probability) in your course? In some courses, you are given a computer program to do this, or use a graphing calculator. In others, you are given a table to look it up in. But even those tables differ. Some give the answer to your question directly (the area between the middle and a given value); others give the entire area to the left, and for your problem you would have to subtract 50% (the left half of the graph). So this is one of those situations in which I would much rather be sitting next to you, so I could look over your book or notes and determine what method you need to use! You can see examples of three kinds of tables here. Please let me know what your class is using, so we can continue.

Hi,
Thank you for the step by step approach. Unfortunately none of the tables from the link is the table my professor is using. Could you try pasting this link in the browser file:///C:/Users/Keyau/OneDrive/Documents/Distribution%20Chart.pdf please let me know if it works. The pdf was to big for an attachment unless you have a different email address that I could send it to.

110-100/20=10/20=0.5
So far so good.

Respectfully,

Auna
 
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You should state the whole problem which includes the section/formulas you think should be used. We don't mind helping people but our time is valuable and we rather not spend it asking for students to post the entire problem along with their work.

Hello Jomo,
I am new to forums and unfamiliar with having to ask for help with my assignments in general. Please understand I am a beginner in my statistics class. I can only hope that those who help me have the virtue of patience. Their help will be appreciated. The feedback I give to my peers will be based on my experience. I would like to continue giving them positive information. Thank you again.

Warmly,
Auna
 
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So try using the formula and let's see what you get from that. Also, the next time you say that a helper is hostile on this forum I promise you that will never happen again because no one will help you.

Hey Jomo!
I am so transparent. Note that I give positive feedback too. Both serve a purpose, to make services better. I am sure that each helper is fabulous and brings a lot to the table. Thank you.

Best Regards,
Auna
 
Mind blown

Hi,
Thank you for the step by step approach. Unfortunately none of the tables from the link is the table my professor is using. Could you try pasting this link in the browser file:///C:/Users/Keyau/OneDrive/Documents/Distribution%20Chart.pdf please let me know if it works. The pdf was to big for an attachment unless you have a different email address that I could send it to.

110-100/20=10/20=0.5
So far so good.

Respectfully,

Auna

Hey,
In column A of Appendix D:Areas Under the Normal Curve Chart and it reads that the z score is 0.50 and next to it is Column B (the area between the mean and z) .1915. So I guess the next step would be to .1915 X 100=19.15%

I think Im right.

Auna
 
Hey,
In column A of Appendix D:Areas Under the Normal Curve Chart and it reads that the z score is 0.50 and next to it is Column B (the area between the mean and z) .1915. So I guess the next step would be to .1915 X 100=19.15%

I think Im right.

Auna

I can't open the file.

But if the table gives the area between the mean and z, then it is equivalent to the first table in Wikipedia. You have found that 19.15% are between 100 (the mean) and 110 (your x), so that is the answer.
 
I can't open the file.

But if the table gives the area between the mean and z, then it is equivalent to the first table in Wikipedia. You have found that 19.15% are between 100 (the mean) and 110 (your x), so that is the answer.

Hello Dr. P,
Thank you for your help. Next steps would include using the numbers I plugged in. Let's say I shade the area from 100 to 110. Do I put a line from 100 and 110 and write 19.15%? Where do I put 0.5 on the graph or Is there an app/website that can graph data? I'm excited that we are so close to the completion. Not sure what I would have done with out your help.

Auna
 
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Hello Dr. P,
Thank you for your help. Next steps would include using the numbers I plugged in. Let's say I shade the area from 100 to 110. Do I put a line from 100 and 110 and write 19.15%? Where do I put 0.5 on the graph or Is there an app/website that can graph data? I'm excited that we are so close to the completion. Not sure what I would have done with out your help.

Auna
Let's say I shade the area from 100 to 110. Do I put a line from 100 and 110 and write 19.15% Yes
Where do I put 0.5 on the graph To the left of the mean is 50% and to the right of the mean is 50%. Since you were asked for the area between 100 and 110 there is no need to place 50% anywhere. If you did I would put it to the left of the mean. Then you should now know the probability of a score being less than 110. That result would be 50% + 19.15% = 69.15% or .6915
 
Let's say I shade the area from 100 to 110. Do I put a line from 100 and 110 and write 19.15% Yes
Where do I put 0.5 on the graph To the left of the mean is 50% and to the right of the mean is 50%. Since you were asked for the area between 100 and 110 there is no need to place 50% anywhere. If you did I would put it to the left of the mean. Then you should now know the probability of a score being less than 110. That result would be 50% + 19.15% = 69.15% or .6915

Thanks a million

Auna
 
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