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

[karush]

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?

 

[I like Serena]

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.

 

[karush]

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

 

[I like Serena]

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.

 

[karush]

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.$\displaystyle \frac{(15\cdot11 + 16\cdot21 + 17\cdot33 + 18\cdot34 + 19\cdot18 + 20\cdot8)}{125} \approx 17$

 

[I like Serena]

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.

 

[karush]

so did I but I rounded off
hopefully i will step out of square one with this stuff...(Wasntme)

thnx much for your help...