How many students are needed for the sit-up exercise problem?

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Discussion Overview

The discussion revolves around a problem related to calculating the number of students needed for a sit-up exercise, focusing on the concepts of median and mean in the context of student performance. Participants explore the implications of these statistical measures on the interpretation of the data provided.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants calculate values for p and q, noting that p=32+33=65 and q=99-p=34.
  • There is a discussion about whether the number of students needs to be considered when calculating the median and mean.
  • One participant suggests that with 125 total students, half being 62.5 complicates the interpretation of the median, questioning how many students can achieve a certain number of sit-ups.
  • Another participant explains that if students are ordered by performance, the median can be determined as 17, based on the scores of the 62nd and 63rd students.
  • There is a query about the mean being different from the median, with calculations provided for the mean based on weighted scores of students.
  • One participant calculates a mean of approximately 17, while another claims a mean of 17.41, indicating a discrepancy in results.

Areas of Agreement / Disagreement

Participants generally agree on the calculations for p and q, but there is disagreement regarding the interpretation of the median and mean, as well as the implications of the number of students on these calculations. The discussion remains unresolved regarding the exact values and their interpretations.

Contextual Notes

Participants express uncertainty about how to handle non-integer values when discussing the number of students achieving certain sit-up counts. There is also a lack of consensus on the correct mean value due to differing calculations.

karush
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image of given problem
https://www.physicsforums.com/attachments/1140

(a)
. (i)well looking at the pattern $p=32+33=65$

. (ii) $q=99-p=99-65=34$

(b) $\frac{35}{2}$

(c) $\frac{35}{2}$

. on (b) and (c) do we have to consider the # of students on these?
 
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karush said:
image of given problem
https://www.physicsforums.com/attachments/1140

(a)
. (i)well looking at the pattern $p=32+33=65$

. (ii) $q=99-p=99-65=34$

(b) $\frac{35}{2}$

(c) $\frac{35}{2}$

. on (b) and (c) do we have to consider the # of students on these?

Hi karush!

(a) is correct, but (b) and (c) are not.
And yes, you have to consider the # of students on those.

The median is such that half the students score lower and the other half score higher.
The mean is actually the weighted mean, where each score is weighted with the number of students.
 
well there are 125 total students half of that is 62.5, since p=65 does that mean that half the students did 17 or less? Its hard get an integer for half the students doing so many sit ups
 
karush said:
well there are 125 total students half of that is 62.5, since p=65 does that mean that half the students did 17 or less? Its hard get an integer for half the students doing so many sit ups

Suppose you put all the students in a row ordered on how many sit-ups they can do.
Then the 62nd student can do 17 and the 63rd student can also do 17.
In this case the numbers are the same, so yes, the median is 17.
 
I like Serena said:
Suppose you put all the students in a row ordered on how many sit-ups they can do.
Then the 62nd student can do 17 and the 63rd student can also do 17.
In this case the numbers are the same, so yes, the median is 17.

but is the mean is different?, assume the average of all situps by all the students divided by the # of students the ans seems to be the same.$\displaystyle \frac{(15\cdot11 + 16\cdot21 + 17\cdot33 + 18\cdot34 + 19\cdot18 + 20\cdot8)}{125} \approx 17$
 
karush said:
but is the mean is different?, assume the average of all situps by all the students divided by the # of students the ans seems to be the same.$\displaystyle \frac{(15\cdot11 + 16\cdot21 + 17\cdot33 + 18\cdot34 + 19\cdot18 + 20\cdot8)}{125} \approx 17$

Yep. The mean is different.
I actually get a mean of 17.41.
 
so did I but I rounded off
hopefully i will step out of square one with this stuff...(Wasntme)

thnx much for your help...
 

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