Mixing
What Can Median Tell Me About my Data?
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For a final assignment for math, I was required to take the "Yes!" side of 'Should Uniforms Be Implemented at School?' My job was to create a biased survey that would make people agree with our side of the topic, where one of the questions went like this:
Estimate the amount of time it takes you to plan, pick out, and accessorize your outfit for school every morning.
a) 5 minutes or less
b) 6 - 10 minutes
c) 11 - 15 minutes
d) 16 - 20 minutes
e) 21 - 25 minutes
f) 26 - 30 minutes
g) Other (please specify): __________
I had 37 people do my survey, and for this question, the majority stated they took either 5 minutes or less, or 6-10 minutes. I even found the midpoint for each of the ranges above and I organized the data in ascending order: 5, 8, 13, 18, 23, 28, 60 (where (16+20) / 2 = 18 is the median). 60 is an input for 1 hour by a respondent for the 'Other' option, so I had to include it as it was a part of the raw data.
There was a task where I had to pick a quantitative set of data (I picked this question above) and analyze it using percentiles for example. I calculated the percentile for 16 - 20 minutes (midpoint would be 18). I found this value to be in the 57th percentile and stated how a little over half of the respondents take 16 - 20 minutes to prepare their outfits for school in the morning and that this cuts off their time to shower and eat breakfast, and that they might be late for class due to taking a while preparing their outfits and not getting a proper breakfast due to rushing.
However, I want to justify properly why I chose to determine to find the percentile for 16-20 minutes. I said this originally: We will determine the percentile in which students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning. We know that 16 - 20 minutes is the median in this data, and the median is the point where fifty percent of the data is above and below it. The median in the data (18) can represent the maximum result.
Honestly speaking, I believe that my reasoning above does not make logical sense. 16-20 minutes isn`t the maxiumum chosen by the respondents.
Note that the graph for this question is a bar graph and that the data shows that the majority of the survey respondents said that they take either 5 minutes or less/6-10 minutes to prepare their outfits for school every morning. 
is the image for the graph.
data-analysis median percentile
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
$endgroup$
add a comment |
$begingroup$
For a final assignment for math, I was required to take the "Yes!" side of 'Should Uniforms Be Implemented at School?' My job was to create a biased survey that would make people agree with our side of the topic, where one of the questions went like this:
Estimate the amount of time it takes you to plan, pick out, and accessorize your outfit for school every morning.
a) 5 minutes or less
b) 6 - 10 minutes
c) 11 - 15 minutes
d) 16 - 20 minutes
e) 21 - 25 minutes
f) 26 - 30 minutes
g) Other (please specify): __________
I had 37 people do my survey, and for this question, the majority stated they took either 5 minutes or less, or 6-10 minutes. I even found the midpoint for each of the ranges above and I organized the data in ascending order: 5, 8, 13, 18, 23, 28, 60 (where (16+20) / 2 = 18 is the median). 60 is an input for 1 hour by a respondent for the 'Other' option, so I had to include it as it was a part of the raw data.
There was a task where I had to pick a quantitative set of data (I picked this question above) and analyze it using percentiles for example. I calculated the percentile for 16 - 20 minutes (midpoint would be 18). I found this value to be in the 57th percentile and stated how a little over half of the respondents take 16 - 20 minutes to prepare their outfits for school in the morning and that this cuts off their time to shower and eat breakfast, and that they might be late for class due to taking a while preparing their outfits and not getting a proper breakfast due to rushing.
However, I want to justify properly why I chose to determine to find the percentile for 16-20 minutes. I said this originally: We will determine the percentile in which students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning. We know that 16 - 20 minutes is the median in this data, and the median is the point where fifty percent of the data is above and below it. The median in the data (18) can represent the maximum result.
Honestly speaking, I believe that my reasoning above does not make logical sense. 16-20 minutes isn`t the maxiumum chosen by the respondents.
Note that the graph for this question is a bar graph and that the data shows that the majority of the survey respondents said that they take either 5 minutes or less/6-10 minutes to prepare their outfits for school every morning. is the image for the graph.
data-analysis median percentile
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
$endgroup$
add a comment |
$begingroup$
For a final assignment for math, I was required to take the "Yes!" side of 'Should Uniforms Be Implemented at School?' My job was to create a biased survey that would make people agree with our side of the topic, where one of the questions went like this:
Estimate the amount of time it takes you to plan, pick out, and accessorize your outfit for school every morning.
a) 5 minutes or less
b) 6 - 10 minutes
c) 11 - 15 minutes
d) 16 - 20 minutes
e) 21 - 25 minutes
f) 26 - 30 minutes
g) Other (please specify): __________
I had 37 people do my survey, and for this question, the majority stated they took either 5 minutes or less, or 6-10 minutes. I even found the midpoint for each of the ranges above and I organized the data in ascending order: 5, 8, 13, 18, 23, 28, 60 (where (16+20) / 2 = 18 is the median). 60 is an input for 1 hour by a respondent for the 'Other' option, so I had to include it as it was a part of the raw data.
There was a task where I had to pick a quantitative set of data (I picked this question above) and analyze it using percentiles for example. I calculated the percentile for 16 - 20 minutes (midpoint would be 18). I found this value to be in the 57th percentile and stated how a little over half of the respondents take 16 - 20 minutes to prepare their outfits for school in the morning and that this cuts off their time to shower and eat breakfast, and that they might be late for class due to taking a while preparing their outfits and not getting a proper breakfast due to rushing.
However, I want to justify properly why I chose to determine to find the percentile for 16-20 minutes. I said this originally: We will determine the percentile in which students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning. We know that 16 - 20 minutes is the median in this data, and the median is the point where fifty percent of the data is above and below it. The median in the data (18) can represent the maximum result.
Honestly speaking, I believe that my reasoning above does not make logical sense. 16-20 minutes isn`t the maxiumum chosen by the respondents.
Note that the graph for this question is a bar graph and that the data shows that the majority of the survey respondents said that they take either 5 minutes or less/6-10 minutes to prepare their outfits for school every morning. is the image for the graph.
data-analysis median percentile
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
$endgroup$
For a final assignment for math, I was required to take the "Yes!" side of 'Should Uniforms Be Implemented at School?' My job was to create a biased survey that would make people agree with our side of the topic, where one of the questions went like this:
Estimate the amount of time it takes you to plan, pick out, and accessorize your outfit for school every morning.
a) 5 minutes or less
b) 6 - 10 minutes
c) 11 - 15 minutes
d) 16 - 20 minutes
e) 21 - 25 minutes
f) 26 - 30 minutes
g) Other (please specify): __________
I had 37 people do my survey, and for this question, the majority stated they took either 5 minutes or less, or 6-10 minutes. I even found the midpoint for each of the ranges above and I organized the data in ascending order: 5, 8, 13, 18, 23, 28, 60 (where (16+20) / 2 = 18 is the median). 60 is an input for 1 hour by a respondent for the 'Other' option, so I had to include it as it was a part of the raw data.
There was a task where I had to pick a quantitative set of data (I picked this question above) and analyze it using percentiles for example. I calculated the percentile for 16 - 20 minutes (midpoint would be 18). I found this value to be in the 57th percentile and stated how a little over half of the respondents take 16 - 20 minutes to prepare their outfits for school in the morning and that this cuts off their time to shower and eat breakfast, and that they might be late for class due to taking a while preparing their outfits and not getting a proper breakfast due to rushing.
However, I want to justify properly why I chose to determine to find the percentile for 16-20 minutes. I said this originally: We will determine the percentile in which students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning. We know that 16 - 20 minutes is the median in this data, and the median is the point where fifty percent of the data is above and below it. The median in the data (18) can represent the maximum result.
Honestly speaking, I believe that my reasoning above does not make logical sense. 16-20 minutes isn`t the maxiumum chosen by the respondents.
Note that the graph for this question is a bar graph and that the data shows that the majority of the survey respondents said that they take either 5 minutes or less/6-10 minutes to prepare their outfits for school every morning. is the image for the graph.
data-analysis median percentile
data-analysis median percentile
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
edited Jan 13 at 3:26
Rodrigo Zepeda
535415
535415
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
asked Jan 13 at 3:06
Yashvi ShahYashvi Shah
166
166
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The median is the value where half the data is above and half is below. As you stated, it is for the point which is $18$ values, from either top or bottom, when they are arranged in numeric order. According to your bar chart, this is for a value in the $6$-$10$ minutes range, not $16$-$20$ minutes, as the first $20$ values are less than $11$ minutes. I believe you made a mistake in calculating your percentiles, and they aren't even generally needed for determining a median. After looking again at your bar chart, I believe the problem is you are using each group by itself, instead of accumulating each set of values as you go from left to right. Thus, the percentiles will be $100 times frac{10}{37}$ for up to $5$ minutes, $100 times frac{10 + 10}{37}$ for up to $10$ minutes, etc.
Also, I'm unsure of what you mean by "The median in the data (18) can represent the maximum result". In most cases, the median and the maximum result of a set of data are different, including in your case here. As you say yourself, it doesn't make logical sense to you.
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
$endgroup$
$begingroup$
Looking at my organized data, since 18 is the midpoint between 16 and 20, I can say that 18 is the fourth element. P would represent the percentile where students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning P = (4 / 7) * 100 P = 0.5714 * 100 P = 57.14 P = 57, so this is how I determined the percentile for 16-20 minutes
$endgroup$
– Yashvi Shah
Jan 13 at 3:34
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@YashviShah The $18$ is for the middle of the number of responses, not the values of the responses. Also, I'm not sure how you got $4 / 7$ for determining the percentile.
$endgroup$
– John Omielan
Jan 13 at 3:37
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I got 4/7 because 18 is the fourth value in my organized data.
$endgroup$
– Yashvi Shah
Jan 13 at 3:39
$begingroup$
@YashviShah Thanks for your explaination. The basic problem is as I stated above, i.e., you're confusing the # of responses to determine the median with the values of the responses. For example, consider if you had $37$ values, as here, but the values ranged between $1000$ and $2000$. What you're doing, I hope you see, doesn't really make any sense then.
$endgroup$
– John Omielan
Jan 13 at 3:42
$begingroup$
So, what Im thinking I should be doing is looking at how many people chose each option in the graph, and find the mean from that data when I organize it in ascending order. Then, I should determine the percentile for 16-20 minutes (still trying to figure out a logical reason for this because I cant say "I want to.") Am I on the right track? Because my teacher said I would need to use the raw data from my survey question (the quantitative question listed above). I dont remember what she said after that but she said its better over using the data from the responses.
$endgroup$
– Yashvi Shah
Jan 13 at 3:58
|
show 6 more comments
$begingroup$
Out of $37$ responses, $20$ people ($54%$, more than half) said $10$ minutes or less.
I see also that in your last paragraph you recognized this fact.
Therefore the median is certainly not greater than $10$ minutes.
A total of $27$ people ($73%$) said $15$ minutes or less. So the $57$th percentile is definitely less than $15$ minutes.
Just $6$ respondents ($16%$) said $16$ to $20$ minutes. That's hardly "a little over half".
I thought possibly you had confused the rank of your "median" response ($18$th out of $37$ responses) with the value of the median.
They are not at all the same thing. But I see now it is worse than that.
You say you have $7$ "values" (actually groups of values) in your data.
The value $18$ is in the $4$th group, which would make it the median of the groups. But "median of the groups" is a useless concept; it mainly tells us how you wrote the survey, not how people answered it.
The standard method for the median response is always $50%$ of the individual responses, not $50%$ of the number of boxes the survey offered.
By the way, while it seems reasonable to assign $8$ minutes as the typical length of time for everyone who said between $6$ and $10$ minutes,
following the same reasoning you should not assign $5$ minutes as the typical time for everyone who said $5$ minutes or less. That would be saying, in effect, that nobody was in the "or less" category, which you certainly do not know for a fact.
answered Jan 13 at 3:34
David KDavid K
54.1k342116
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$begingroup$
You are given the tabulated grouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
0-5&10&10\
6-10&10&20\
11-15&7&27\
16-20&6&33\
21-25&1&34\
26-30&2&36\
31-60&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan $11-15$ minutes. $F=27$ students plan $0-15$ minutes.
The median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$, which falls in the group $6-10$ (it is the median group).
The median is:
$$m=l+frac{frac {n+1}2-F}{2}cdot w=5.5+frac{19-10}{10}cdot 5=color{red}{10},$$
which shows $50%$ of students plan $0-color{red}{10}$ minutes.
When you calculated the mid-points of groups, then you converted the data to tabulated ungrouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
2.5&10&10\
8&10&20\
13&7&27\
18&6&33\
23&1&34\
28&2&36\
45.5&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan on average $13$ minutes.
Again the median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$ and the median number is $m=color{red}8$, which shows $50%$ students plan on average $0-8$ minutes. Indeed, if you arrange the numbers from the above tabulated ungrouped data in increasing order, you will have $10$ numbers $2.5$, then $10$ numbers $8$, in particular, the $19$th number will also be $8$.
Finally, note that when you are converting the data, you are approximating or assuming the average planning time. So, the first result $(m=10)$ is more accurate than the second result $(m=8)$.
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
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3 Answers
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3 Answers
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$begingroup$
The median is the value where half the data is above and half is below. As you stated, it is for the point which is $18$ values, from either top or bottom, when they are arranged in numeric order. According to your bar chart, this is for a value in the $6$-$10$ minutes range, not $16$-$20$ minutes, as the first $20$ values are less than $11$ minutes. I believe you made a mistake in calculating your percentiles, and they aren't even generally needed for determining a median. After looking again at your bar chart, I believe the problem is you are using each group by itself, instead of accumulating each set of values as you go from left to right. Thus, the percentiles will be $100 times frac{10}{37}$ for up to $5$ minutes, $100 times frac{10 + 10}{37}$ for up to $10$ minutes, etc.
Also, I'm unsure of what you mean by "The median in the data (18) can represent the maximum result". In most cases, the median and the maximum result of a set of data are different, including in your case here. As you say yourself, it doesn't make logical sense to you.
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
$endgroup$
$begingroup$
Looking at my organized data, since 18 is the midpoint between 16 and 20, I can say that 18 is the fourth element. P would represent the percentile where students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning P = (4 / 7) * 100 P = 0.5714 * 100 P = 57.14 P = 57, so this is how I determined the percentile for 16-20 minutes
$endgroup$
– Yashvi Shah
Jan 13 at 3:34
$begingroup$
@YashviShah The $18$ is for the middle of the number of responses, not the values of the responses. Also, I'm not sure how you got $4 / 7$ for determining the percentile.
$endgroup$
– John Omielan
Jan 13 at 3:37
$begingroup$
I got 4/7 because 18 is the fourth value in my organized data.
$endgroup$
– Yashvi Shah
Jan 13 at 3:39
$begingroup$
@YashviShah Thanks for your explaination. The basic problem is as I stated above, i.e., you're confusing the # of responses to determine the median with the values of the responses. For example, consider if you had $37$ values, as here, but the values ranged between $1000$ and $2000$. What you're doing, I hope you see, doesn't really make any sense then.
$endgroup$
– John Omielan
Jan 13 at 3:42
$begingroup$
So, what Im thinking I should be doing is looking at how many people chose each option in the graph, and find the mean from that data when I organize it in ascending order. Then, I should determine the percentile for 16-20 minutes (still trying to figure out a logical reason for this because I cant say "I want to.") Am I on the right track? Because my teacher said I would need to use the raw data from my survey question (the quantitative question listed above). I dont remember what she said after that but she said its better over using the data from the responses.
$endgroup$
– Yashvi Shah
Jan 13 at 3:58
|
show 6 more comments
$begingroup$
The median is the value where half the data is above and half is below. As you stated, it is for the point which is $18$ values, from either top or bottom, when they are arranged in numeric order. According to your bar chart, this is for a value in the $6$-$10$ minutes range, not $16$-$20$ minutes, as the first $20$ values are less than $11$ minutes. I believe you made a mistake in calculating your percentiles, and they aren't even generally needed for determining a median. After looking again at your bar chart, I believe the problem is you are using each group by itself, instead of accumulating each set of values as you go from left to right. Thus, the percentiles will be $100 times frac{10}{37}$ for up to $5$ minutes, $100 times frac{10 + 10}{37}$ for up to $10$ minutes, etc.
Also, I'm unsure of what you mean by "The median in the data (18) can represent the maximum result". In most cases, the median and the maximum result of a set of data are different, including in your case here. As you say yourself, it doesn't make logical sense to you.
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
$endgroup$
$begingroup$
Looking at my organized data, since 18 is the midpoint between 16 and 20, I can say that 18 is the fourth element. P would represent the percentile where students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning P = (4 / 7) * 100 P = 0.5714 * 100 P = 57.14 P = 57, so this is how I determined the percentile for 16-20 minutes
$endgroup$
– Yashvi Shah
Jan 13 at 3:34
$begingroup$
@YashviShah The $18$ is for the middle of the number of responses, not the values of the responses. Also, I'm not sure how you got $4 / 7$ for determining the percentile.
$endgroup$
– John Omielan
Jan 13 at 3:37
$begingroup$
I got 4/7 because 18 is the fourth value in my organized data.
$endgroup$
– Yashvi Shah
Jan 13 at 3:39
$begingroup$
@YashviShah Thanks for your explaination. The basic problem is as I stated above, i.e., you're confusing the # of responses to determine the median with the values of the responses. For example, consider if you had $37$ values, as here, but the values ranged between $1000$ and $2000$. What you're doing, I hope you see, doesn't really make any sense then.
$endgroup$
– John Omielan
Jan 13 at 3:42
$begingroup$
So, what Im thinking I should be doing is looking at how many people chose each option in the graph, and find the mean from that data when I organize it in ascending order. Then, I should determine the percentile for 16-20 minutes (still trying to figure out a logical reason for this because I cant say "I want to.") Am I on the right track? Because my teacher said I would need to use the raw data from my survey question (the quantitative question listed above). I dont remember what she said after that but she said its better over using the data from the responses.
$endgroup$
– Yashvi Shah
Jan 13 at 3:58
|
show 6 more comments
$begingroup$
The median is the value where half the data is above and half is below. As you stated, it is for the point which is $18$ values, from either top or bottom, when they are arranged in numeric order. According to your bar chart, this is for a value in the $6$-$10$ minutes range, not $16$-$20$ minutes, as the first $20$ values are less than $11$ minutes. I believe you made a mistake in calculating your percentiles, and they aren't even generally needed for determining a median. After looking again at your bar chart, I believe the problem is you are using each group by itself, instead of accumulating each set of values as you go from left to right. Thus, the percentiles will be $100 times frac{10}{37}$ for up to $5$ minutes, $100 times frac{10 + 10}{37}$ for up to $10$ minutes, etc.
Also, I'm unsure of what you mean by "The median in the data (18) can represent the maximum result". In most cases, the median and the maximum result of a set of data are different, including in your case here. As you say yourself, it doesn't make logical sense to you.
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
$endgroup$
The median is the value where half the data is above and half is below. As you stated, it is for the point which is $18$ values, from either top or bottom, when they are arranged in numeric order. According to your bar chart, this is for a value in the $6$-$10$ minutes range, not $16$-$20$ minutes, as the first $20$ values are less than $11$ minutes. I believe you made a mistake in calculating your percentiles, and they aren't even generally needed for determining a median. After looking again at your bar chart, I believe the problem is you are using each group by itself, instead of accumulating each set of values as you go from left to right. Thus, the percentiles will be $100 times frac{10}{37}$ for up to $5$ minutes, $100 times frac{10 + 10}{37}$ for up to $10$ minutes, etc.
Also, I'm unsure of what you mean by "The median in the data (18) can represent the maximum result". In most cases, the median and the maximum result of a set of data are different, including in your case here. As you say yourself, it doesn't make logical sense to you.
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
edited Jan 13 at 3:34
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
answered Jan 13 at 3:27
John OmielanJohn Omielan
2,626212
2,626212
$begingroup$
Looking at my organized data, since 18 is the midpoint between 16 and 20, I can say that 18 is the fourth element. P would represent the percentile where students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning P = (4 / 7) * 100 P = 0.5714 * 100 P = 57.14 P = 57, so this is how I determined the percentile for 16-20 minutes
$endgroup$
– Yashvi Shah
Jan 13 at 3:34
$begingroup$
@YashviShah The $18$ is for the middle of the number of responses, not the values of the responses. Also, I'm not sure how you got $4 / 7$ for determining the percentile.
$endgroup$
– John Omielan
Jan 13 at 3:37
$begingroup$
I got 4/7 because 18 is the fourth value in my organized data.
$endgroup$
– Yashvi Shah
Jan 13 at 3:39
$begingroup$
@YashviShah Thanks for your explaination. The basic problem is as I stated above, i.e., you're confusing the # of responses to determine the median with the values of the responses. For example, consider if you had $37$ values, as here, but the values ranged between $1000$ and $2000$. What you're doing, I hope you see, doesn't really make any sense then.
$endgroup$
– John Omielan
Jan 13 at 3:42
$begingroup$
So, what Im thinking I should be doing is looking at how many people chose each option in the graph, and find the mean from that data when I organize it in ascending order. Then, I should determine the percentile for 16-20 minutes (still trying to figure out a logical reason for this because I cant say "I want to.") Am I on the right track? Because my teacher said I would need to use the raw data from my survey question (the quantitative question listed above). I dont remember what she said after that but she said its better over using the data from the responses.
$endgroup$
– Yashvi Shah
Jan 13 at 3:58
|
show 6 more comments
$begingroup$
Looking at my organized data, since 18 is the midpoint between 16 and 20, I can say that 18 is the fourth element. P would represent the percentile where students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning P = (4 / 7) * 100 P = 0.5714 * 100 P = 57.14 P = 57, so this is how I determined the percentile for 16-20 minutes
$endgroup$
– Yashvi Shah
Jan 13 at 3:34
$begingroup$
@YashviShah The $18$ is for the middle of the number of responses, not the values of the responses. Also, I'm not sure how you got $4 / 7$ for determining the percentile.
$endgroup$
– John Omielan
Jan 13 at 3:37
$begingroup$
I got 4/7 because 18 is the fourth value in my organized data.
$endgroup$
– Yashvi Shah
Jan 13 at 3:39
$begingroup$
@YashviShah Thanks for your explaination. The basic problem is as I stated above, i.e., you're confusing the # of responses to determine the median with the values of the responses. For example, consider if you had $37$ values, as here, but the values ranged between $1000$ and $2000$. What you're doing, I hope you see, doesn't really make any sense then.
$endgroup$
– John Omielan
Jan 13 at 3:42
$begingroup$
So, what Im thinking I should be doing is looking at how many people chose each option in the graph, and find the mean from that data when I organize it in ascending order. Then, I should determine the percentile for 16-20 minutes (still trying to figure out a logical reason for this because I cant say "I want to.") Am I on the right track? Because my teacher said I would need to use the raw data from my survey question (the quantitative question listed above). I dont remember what she said after that but she said its better over using the data from the responses.
$endgroup$
– Yashvi Shah
Jan 13 at 3:58
$begingroup$
Looking at my organized data, since 18 is the midpoint between 16 and 20, I can say that 18 is the fourth element. P would represent the percentile where students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning P = (4 / 7) * 100 P = 0.5714 * 100 P = 57.14 P = 57, so this is how I determined the percentile for 16-20 minutes
$endgroup$
– Yashvi Shah
Jan 13 at 3:34
$begingroup$
Looking at my organized data, since 18 is the midpoint between 16 and 20, I can say that 18 is the fourth element. P would represent the percentile where students take 16 - 20 minutes to plan, pick out, and accessorize their outfits for school in the morning P = (4 / 7) * 100 P = 0.5714 * 100 P = 57.14 P = 57, so this is how I determined the percentile for 16-20 minutes
$endgroup$
– Yashvi Shah
Jan 13 at 3:34
$begingroup$
@YashviShah The $18$ is for the middle of the number of responses, not the values of the responses. Also, I'm not sure how you got $4 / 7$ for determining the percentile.
$endgroup$
– John Omielan
Jan 13 at 3:37
$begingroup$
@YashviShah The $18$ is for the middle of the number of responses, not the values of the responses. Also, I'm not sure how you got $4 / 7$ for determining the percentile.
$endgroup$
– John Omielan
Jan 13 at 3:37
$begingroup$
I got 4/7 because 18 is the fourth value in my organized data.
$endgroup$
– Yashvi Shah
Jan 13 at 3:39
$begingroup$
I got 4/7 because 18 is the fourth value in my organized data.
$endgroup$
– Yashvi Shah
Jan 13 at 3:39
$begingroup$
@YashviShah Thanks for your explaination. The basic problem is as I stated above, i.e., you're confusing the # of responses to determine the median with the values of the responses. For example, consider if you had $37$ values, as here, but the values ranged between $1000$ and $2000$. What you're doing, I hope you see, doesn't really make any sense then.
$endgroup$
– John Omielan
Jan 13 at 3:42
$begingroup$
@YashviShah Thanks for your explaination. The basic problem is as I stated above, i.e., you're confusing the # of responses to determine the median with the values of the responses. For example, consider if you had $37$ values, as here, but the values ranged between $1000$ and $2000$. What you're doing, I hope you see, doesn't really make any sense then.
$endgroup$
– John Omielan
Jan 13 at 3:42
$begingroup$
So, what I
m thinking I should be doing is looking at how many people chose each option in the graph, and find the mean from that data when I organize it in ascending order. Then, I should determine the percentile for 16-20 minutes (still trying to figure out a logical reason for this because I cant say "I want to.") Am I on the right track? Because my teacher said I would need to use the raw data from my survey question (the quantitative question listed above). I dont remember what she said after that but she said its better over using the data from the responses.$endgroup$
– Yashvi Shah
Jan 13 at 3:58
$begingroup$
So, what I
m thinking I should be doing is looking at how many people chose each option in the graph, and find the mean from that data when I organize it in ascending order. Then, I should determine the percentile for 16-20 minutes (still trying to figure out a logical reason for this because I cant say "I want to.") Am I on the right track? Because my teacher said I would need to use the raw data from my survey question (the quantitative question listed above). I dont remember what she said after that but she said its better over using the data from the responses.$endgroup$
– Yashvi Shah
Jan 13 at 3:58
|
show 6 more comments
$begingroup$
Out of $37$ responses, $20$ people ($54%$, more than half) said $10$ minutes or less.
I see also that in your last paragraph you recognized this fact.
Therefore the median is certainly not greater than $10$ minutes.
A total of $27$ people ($73%$) said $15$ minutes or less. So the $57$th percentile is definitely less than $15$ minutes.
Just $6$ respondents ($16%$) said $16$ to $20$ minutes. That's hardly "a little over half".
I thought possibly you had confused the rank of your "median" response ($18$th out of $37$ responses) with the value of the median.
They are not at all the same thing. But I see now it is worse than that.
You say you have $7$ "values" (actually groups of values) in your data.
The value $18$ is in the $4$th group, which would make it the median of the groups. But "median of the groups" is a useless concept; it mainly tells us how you wrote the survey, not how people answered it.
The standard method for the median response is always $50%$ of the individual responses, not $50%$ of the number of boxes the survey offered.
By the way, while it seems reasonable to assign $8$ minutes as the typical length of time for everyone who said between $6$ and $10$ minutes,
following the same reasoning you should not assign $5$ minutes as the typical time for everyone who said $5$ minutes or less. That would be saying, in effect, that nobody was in the "or less" category, which you certainly do not know for a fact.
answered Jan 13 at 3:34
David KDavid K
54.1k342116
$endgroup$
add a comment |
$begingroup$
Out of $37$ responses, $20$ people ($54%$, more than half) said $10$ minutes or less.
I see also that in your last paragraph you recognized this fact.
Therefore the median is certainly not greater than $10$ minutes.
A total of $27$ people ($73%$) said $15$ minutes or less. So the $57$th percentile is definitely less than $15$ minutes.
Just $6$ respondents ($16%$) said $16$ to $20$ minutes. That's hardly "a little over half".
I thought possibly you had confused the rank of your "median" response ($18$th out of $37$ responses) with the value of the median.
They are not at all the same thing. But I see now it is worse than that.
You say you have $7$ "values" (actually groups of values) in your data.
The value $18$ is in the $4$th group, which would make it the median of the groups. But "median of the groups" is a useless concept; it mainly tells us how you wrote the survey, not how people answered it.
The standard method for the median response is always $50%$ of the individual responses, not $50%$ of the number of boxes the survey offered.
By the way, while it seems reasonable to assign $8$ minutes as the typical length of time for everyone who said between $6$ and $10$ minutes,
following the same reasoning you should not assign $5$ minutes as the typical time for everyone who said $5$ minutes or less. That would be saying, in effect, that nobody was in the "or less" category, which you certainly do not know for a fact.
answered Jan 13 at 3:34
David KDavid K
54.1k342116
$endgroup$
add a comment |
$begingroup$
Out of $37$ responses, $20$ people ($54%$, more than half) said $10$ minutes or less.
I see also that in your last paragraph you recognized this fact.
Therefore the median is certainly not greater than $10$ minutes.
A total of $27$ people ($73%$) said $15$ minutes or less. So the $57$th percentile is definitely less than $15$ minutes.
Just $6$ respondents ($16%$) said $16$ to $20$ minutes. That's hardly "a little over half".
I thought possibly you had confused the rank of your "median" response ($18$th out of $37$ responses) with the value of the median.
They are not at all the same thing. But I see now it is worse than that.
You say you have $7$ "values" (actually groups of values) in your data.
The value $18$ is in the $4$th group, which would make it the median of the groups. But "median of the groups" is a useless concept; it mainly tells us how you wrote the survey, not how people answered it.
The standard method for the median response is always $50%$ of the individual responses, not $50%$ of the number of boxes the survey offered.
By the way, while it seems reasonable to assign $8$ minutes as the typical length of time for everyone who said between $6$ and $10$ minutes,
following the same reasoning you should not assign $5$ minutes as the typical time for everyone who said $5$ minutes or less. That would be saying, in effect, that nobody was in the "or less" category, which you certainly do not know for a fact.
answered Jan 13 at 3:34
David KDavid K
54.1k342116
$endgroup$
Out of $37$ responses, $20$ people ($54%$, more than half) said $10$ minutes or less.
I see also that in your last paragraph you recognized this fact.
Therefore the median is certainly not greater than $10$ minutes.
A total of $27$ people ($73%$) said $15$ minutes or less. So the $57$th percentile is definitely less than $15$ minutes.
Just $6$ respondents ($16%$) said $16$ to $20$ minutes. That's hardly "a little over half".
I thought possibly you had confused the rank of your "median" response ($18$th out of $37$ responses) with the value of the median.
They are not at all the same thing. But I see now it is worse than that.
You say you have $7$ "values" (actually groups of values) in your data.
The value $18$ is in the $4$th group, which would make it the median of the groups. But "median of the groups" is a useless concept; it mainly tells us how you wrote the survey, not how people answered it.
The standard method for the median response is always $50%$ of the individual responses, not $50%$ of the number of boxes the survey offered.
By the way, while it seems reasonable to assign $8$ minutes as the typical length of time for everyone who said between $6$ and $10$ minutes,
following the same reasoning you should not assign $5$ minutes as the typical time for everyone who said $5$ minutes or less. That would be saying, in effect, that nobody was in the "or less" category, which you certainly do not know for a fact.
answered Jan 13 at 3:34
David KDavid K
54.1k342116
edited Jan 13 at 3:53
answered Jan 13 at 3:34
David KDavid K
54.1k342116
answered Jan 13 at 3:34
David KDavid K
54.1k342116
answered Jan 13 at 3:34
David KDavid K
54.1k342116
54.1k342116
add a comment |
add a comment |
$begingroup$
You are given the tabulated grouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
0-5&10&10\
6-10&10&20\
11-15&7&27\
16-20&6&33\
21-25&1&34\
26-30&2&36\
31-60&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan $11-15$ minutes. $F=27$ students plan $0-15$ minutes.
The median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$, which falls in the group $6-10$ (it is the median group).
The median is:
$$m=l+frac{frac {n+1}2-F}{2}cdot w=5.5+frac{19-10}{10}cdot 5=color{red}{10},$$
which shows $50%$ of students plan $0-color{red}{10}$ minutes.
When you calculated the mid-points of groups, then you converted the data to tabulated ungrouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
2.5&10&10\
8&10&20\
13&7&27\
18&6&33\
23&1&34\
28&2&36\
45.5&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan on average $13$ minutes.
Again the median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$ and the median number is $m=color{red}8$, which shows $50%$ students plan on average $0-8$ minutes. Indeed, if you arrange the numbers from the above tabulated ungrouped data in increasing order, you will have $10$ numbers $2.5$, then $10$ numbers $8$, in particular, the $19$th number will also be $8$.
Finally, note that when you are converting the data, you are approximating or assuming the average planning time. So, the first result $(m=10)$ is more accurate than the second result $(m=8)$.
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
You are given the tabulated grouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
0-5&10&10\
6-10&10&20\
11-15&7&27\
16-20&6&33\
21-25&1&34\
26-30&2&36\
31-60&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan $11-15$ minutes. $F=27$ students plan $0-15$ minutes.
The median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$, which falls in the group $6-10$ (it is the median group).
The median is:
$$m=l+frac{frac {n+1}2-F}{2}cdot w=5.5+frac{19-10}{10}cdot 5=color{red}{10},$$
which shows $50%$ of students plan $0-color{red}{10}$ minutes.
When you calculated the mid-points of groups, then you converted the data to tabulated ungrouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
2.5&10&10\
8&10&20\
13&7&27\
18&6&33\
23&1&34\
28&2&36\
45.5&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan on average $13$ minutes.
Again the median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$ and the median number is $m=color{red}8$, which shows $50%$ students plan on average $0-8$ minutes. Indeed, if you arrange the numbers from the above tabulated ungrouped data in increasing order, you will have $10$ numbers $2.5$, then $10$ numbers $8$, in particular, the $19$th number will also be $8$.
Finally, note that when you are converting the data, you are approximating or assuming the average planning time. So, the first result $(m=10)$ is more accurate than the second result $(m=8)$.
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
You are given the tabulated grouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
0-5&10&10\
6-10&10&20\
11-15&7&27\
16-20&6&33\
21-25&1&34\
26-30&2&36\
31-60&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan $11-15$ minutes. $F=27$ students plan $0-15$ minutes.
The median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$, which falls in the group $6-10$ (it is the median group).
The median is:
$$m=l+frac{frac {n+1}2-F}{2}cdot w=5.5+frac{19-10}{10}cdot 5=color{red}{10},$$
which shows $50%$ of students plan $0-color{red}{10}$ minutes.
When you calculated the mid-points of groups, then you converted the data to tabulated ungrouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
2.5&10&10\
8&10&20\
13&7&27\
18&6&33\
23&1&34\
28&2&36\
45.5&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan on average $13$ minutes.
Again the median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$ and the median number is $m=color{red}8$, which shows $50%$ students plan on average $0-8$ minutes. Indeed, if you arrange the numbers from the above tabulated ungrouped data in increasing order, you will have $10$ numbers $2.5$, then $10$ numbers $8$, in particular, the $19$th number will also be $8$.
Finally, note that when you are converting the data, you are approximating or assuming the average planning time. So, the first result $(m=10)$ is more accurate than the second result $(m=8)$.
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
$endgroup$
You are given the tabulated grouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
0-5&10&10\
6-10&10&20\
11-15&7&27\
16-20&6&33\
21-25&1&34\
26-30&2&36\
31-60&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan $11-15$ minutes. $F=27$ students plan $0-15$ minutes.
The median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$, which falls in the group $6-10$ (it is the median group).
The median is:
$$m=l+frac{frac {n+1}2-F}{2}cdot w=5.5+frac{19-10}{10}cdot 5=color{red}{10},$$
which shows $50%$ of students plan $0-color{red}{10}$ minutes.
When you calculated the mid-points of groups, then you converted the data to tabulated ungrouped data:
$$begin{array}{c|c|c|c}
text{Planning} (x)&text{Number of students} (f)&text{Cumulative frequency} (F)\
hline
2.5&10&10\
8&10&20\
13&7&27\
18&6&33\
23&1&34\
28&2&36\
45.5&1&37\
hline
text{Total}&37 end{array}$$
Interpretation: $f=7$ students plan on average $13$ minutes.
Again the median is the middle number when all numbers are arranged in increasing (or decreasing) order. There are $37$ students, so the median position is $frac{n+1}{2}=frac{37+1}{2}=19$ and the median number is $m=color{red}8$, which shows $50%$ students plan on average $0-8$ minutes. Indeed, if you arrange the numbers from the above tabulated ungrouped data in increasing order, you will have $10$ numbers $2.5$, then $10$ numbers $8$, in particular, the $19$th number will also be $8$.
Finally, note that when you are converting the data, you are approximating or assuming the average planning time. So, the first result $(m=10)$ is more accurate than the second result $(m=8)$.
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
answered Jan 13 at 8:33
farruhotafarruhota
20.2k2738
20.2k2738
add a comment |
add a comment |
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