As Maths Question

[kratos009]

Hey guys,
This might sound like a very trivial question but i seem to have forgotten how to do it . A table is given with the intervals

   time spent in x mins               Frequency

0<   x < or equal to 10                 210
10< x < or equal to 20                  134
20< x < or equal to 30                  78
30< x < or equal to 40                   72
40< x < or equal to 60                  46

The question asked estimate the mean time in minutes of the data given above. So using the interval do you use 5.5*210 + 15.5*134 + 25.5*78 etc... or do you use 5*210 + 15*134 and so on. The correct answer was using 5*210 + 15*134 but i don't get why the midpoint of the intervals are 5,15,25 etc instead of 5.5,15.5 etc. Could someone explain to me why that is?
I hope that all made sense.

Thank you in advance  

[Light]

well, it is easy if you know how.to calculate midpoint, u take the sum of upper boundary and lower boundary divided by 2.lets say 0<x or equal to 10,the lower boundary=0.upper boundary =10.midpoint =(10+0)/2=5.there is no gap between the classes.if there is gap like 1-10,11-20,...then u take 5.5 as the first midpoint as the lower boundary=0.5 and upper=10.5.midpoint=(10.5+0.5)/2=5.5.

[kratos009]

Oh right i get most of it. I'm still a little confused why the lower boundary for 0 in 0< x < or equal to 10 is 0. wouldn't the lower boundary be 0.5< x, because it is not greater than or equal to 0? and the higher boundary for 10 is < or equal to 10.5 because the table states it is < or equal to 10? Sorry i know this is a very trivial question but i can't seem to get my head around it...

[Light]

if lower boundary is 0.5 u r saying that the value starts form 0.5 but it is wrong cause if x>0,it can be also 0.1,0.2,0.3,0.0002 and etc which doesnt start from 0.5.the values has to start from 0. if the higher boundary is 10.5 u r saying the max is 10.5 ,but the question is less or equal to 10.the values greater than 10 like 10.1,10.2 ... are not counted.the maximum value must be 10 and so the upper boundary.

[kratos009]

Oh right I got what you are saying. Thanks a lot , i know it wasn't a very hard question but i couldn't seem to get my head around it which was frustrating.

[Tyserius]

Actually there is a trick to this. You can use the following examples to check your midpoints.

i) When there is a case whereby :
   0 < x < or = 10
   10 < x < or = 20
   From the above you can see that the 1st class includes 10 while the 2nd class is right at the boundary of 10 so there is no  way that the 1st class can have an upper boundary of 10.5 because it would have exceeded into the 2nd class. In this case the upper boundary would be 10. Therefore, your midpoint is 5.

ii) When there is a case whereby :
    1 < or = x < or = 10
    11 < or = x < or = 20
    From the above you can see that the 1st class includes any number from 1 to 10 while the 2nd class from 11 to 20. Since there is no way that the two classes may interfere with one another as like how they would from the first case in (i), you can safely use the upper boundaries and lower boundaries to find your midpoint. Therefore, the midpoint for the 1st class would be 5 as well.

iii) When there is a case whereby :
     0 < or = x < 10
     10 < or = x < 20
     From the above you can see that the intervals may interfere with one another if you consider the upper boundary for the 1st class as it will then exceed 10.5 which is intruding upon the 2nd class. Hence, it is not viable for you to consider finding the midpoints using the upper boundaries and the lower boundaries. Thus, you can just consider the median number, which is 5.

Hope this helps! All the best for Paper 6 tomorrow