Mixing
Mean and median of a distribution
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One study mentions that they found over 90% of the nation's school districts reported that their students were scoring "above the national average." Explain why this report is incorrect.
Does the analysis change if the term "average" refers to mean? What about the median?
I think this is incorrect since it should be exactly 50% above average and 50% below average. I'm not sure if it changes when it's referring to the mean, because the mean can get skewed by outliers, right? I know that it's incorrect when referring to the median because median is the center again.
statistics descriptive-statistics
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add a comment |
$begingroup$
One study mentions that they found over 90% of the nation's school districts reported that their students were scoring "above the national average." Explain why this report is incorrect.
Does the analysis change if the term "average" refers to mean? What about the median?
I think this is incorrect since it should be exactly 50% above average and 50% below average. I'm not sure if it changes when it's referring to the mean, because the mean can get skewed by outliers, right? I know that it's incorrect when referring to the median because median is the center again.
statistics descriptive-statistics
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Technically, the statement could be correct if the number of students in the other 10% of districts was far more than the number in the 90%, that is, they were populated by a very large number of low-scoring students.
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– Naman Kumar
Jan 28 at 4:03
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It is only possible if average" means mean. To be possible requires the 90% to be just above average while the remaining 10% have average near 0.
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– herb steinberg
Jan 28 at 4:06
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@herb steinberg It could be possible for median if the population was highly skewed towards the 10% districts. I don't think there's enough information to solve this problem, unless you look at a highly realistic scenario where it makes it nearly impossible for 90% of districts score above the national average. I mean, at this point the districts are probably lying.
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– Naman Kumar
Jan 28 at 4:10
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Suppose $9$ districts had student scores of $71.0$ and $1$ district had student scores of $70.0$. The average score would be $70.9$ and $90%$ of the districts would have student scores above the average
$endgroup$
– Henry
Jan 28 at 8:08
add a comment |
$begingroup$
One study mentions that they found over 90% of the nation's school districts reported that their students were scoring "above the national average." Explain why this report is incorrect.
Does the analysis change if the term "average" refers to mean? What about the median?
I think this is incorrect since it should be exactly 50% above average and 50% below average. I'm not sure if it changes when it's referring to the mean, because the mean can get skewed by outliers, right? I know that it's incorrect when referring to the median because median is the center again.
statistics descriptive-statistics
$endgroup$
One study mentions that they found over 90% of the nation's school districts reported that their students were scoring "above the national average." Explain why this report is incorrect.
Does the analysis change if the term "average" refers to mean? What about the median?
I think this is incorrect since it should be exactly 50% above average and 50% below average. I'm not sure if it changes when it's referring to the mean, because the mean can get skewed by outliers, right? I know that it's incorrect when referring to the median because median is the center again.
statistics descriptive-statistics
statistics descriptive-statistics
asked Jan 28 at 3:44
user614735
$begingroup$
Technically, the statement could be correct if the number of students in the other 10% of districts was far more than the number in the 90%, that is, they were populated by a very large number of low-scoring students.
$endgroup$
– Naman Kumar
Jan 28 at 4:03
$begingroup$
It is only possible if average" means mean. To be possible requires the 90% to be just above average while the remaining 10% have average near 0.
$endgroup$
– herb steinberg
Jan 28 at 4:06
$begingroup$
@herb steinberg It could be possible for median if the population was highly skewed towards the 10% districts. I don't think there's enough information to solve this problem, unless you look at a highly realistic scenario where it makes it nearly impossible for 90% of districts score above the national average. I mean, at this point the districts are probably lying.
$endgroup$
– Naman Kumar
Jan 28 at 4:10
$begingroup$
Suppose $9$ districts had student scores of $71.0$ and $1$ district had student scores of $70.0$. The average score would be $70.9$ and $90%$ of the districts would have student scores above the average
$endgroup$
– Henry
Jan 28 at 8:08
add a comment |
$begingroup$
Technically, the statement could be correct if the number of students in the other 10% of districts was far more than the number in the 90%, that is, they were populated by a very large number of low-scoring students.
$endgroup$
– Naman Kumar
Jan 28 at 4:03
$begingroup$
It is only possible if average" means mean. To be possible requires the 90% to be just above average while the remaining 10% have average near 0.
$endgroup$
– herb steinberg
Jan 28 at 4:06
$begingroup$
@herb steinberg It could be possible for median if the population was highly skewed towards the 10% districts. I don't think there's enough information to solve this problem, unless you look at a highly realistic scenario where it makes it nearly impossible for 90% of districts score above the national average. I mean, at this point the districts are probably lying.
$endgroup$
– Naman Kumar
Jan 28 at 4:10
$begingroup$
Suppose $9$ districts had student scores of $71.0$ and $1$ district had student scores of $70.0$. The average score would be $70.9$ and $90%$ of the districts would have student scores above the average
$endgroup$
– Henry
Jan 28 at 8:08
$begingroup$
Technically, the statement could be correct if the number of students in the other 10% of districts was far more than the number in the 90%, that is, they were populated by a very large number of low-scoring students.
$endgroup$
– Naman Kumar
Jan 28 at 4:03
$begingroup$
Technically, the statement could be correct if the number of students in the other 10% of districts was far more than the number in the 90%, that is, they were populated by a very large number of low-scoring students.
$endgroup$
– Naman Kumar
Jan 28 at 4:03
$begingroup$
It is only possible if average" means mean. To be possible requires the 90% to be just above average while the remaining 10% have average near 0.
$endgroup$
– herb steinberg
Jan 28 at 4:06
$begingroup$
It is only possible if average" means mean. To be possible requires the 90% to be just above average while the remaining 10% have average near 0.
$endgroup$
– herb steinberg
Jan 28 at 4:06
$begingroup$
@herb steinberg It could be possible for median if the population was highly skewed towards the 10% districts. I don't think there's enough information to solve this problem, unless you look at a highly realistic scenario where it makes it nearly impossible for 90% of districts score above the national average. I mean, at this point the districts are probably lying.
$endgroup$
– Naman Kumar
Jan 28 at 4:10
$begingroup$
@herb steinberg It could be possible for median if the population was highly skewed towards the 10% districts. I don't think there's enough information to solve this problem, unless you look at a highly realistic scenario where it makes it nearly impossible for 90% of districts score above the national average. I mean, at this point the districts are probably lying.
$endgroup$
– Naman Kumar
Jan 28 at 4:10
$begingroup$
Suppose $9$ districts had student scores of $71.0$ and $1$ district had student scores of $70.0$. The average score would be $70.9$ and $90%$ of the districts would have student scores above the average
$endgroup$
– Henry
Jan 28 at 8:08
$begingroup$
Suppose $9$ districts had student scores of $71.0$ and $1$ district had student scores of $70.0$. The average score would be $70.9$ and $90%$ of the districts would have student scores above the average
$endgroup$
– Henry
Jan 28 at 8:08
add a comment |
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$begingroup$
Technically, the statement could be correct if the number of students in the other 10% of districts was far more than the number in the 90%, that is, they were populated by a very large number of low-scoring students.
$endgroup$
– Naman Kumar
Jan 28 at 4:03
$begingroup$
It is only possible if average" means mean. To be possible requires the 90% to be just above average while the remaining 10% have average near 0.
$endgroup$
– herb steinberg
Jan 28 at 4:06
$begingroup$
@herb steinberg It could be possible for median if the population was highly skewed towards the 10% districts. I don't think there's enough information to solve this problem, unless you look at a highly realistic scenario where it makes it nearly impossible for 90% of districts score above the national average. I mean, at this point the districts are probably lying.
$endgroup$
– Naman Kumar
Jan 28 at 4:10
$begingroup$
Suppose $9$ districts had student scores of $71.0$ and $1$ district had student scores of $70.0$. The average score would be $70.9$ and $90%$ of the districts would have student scores above the average
$endgroup$
– Henry
Jan 28 at 8:08