Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By HallsofIvy

Math Help - precentage change and weighted mean question

  1. #1
    Senior Member
    Joined
    May 2012
    From
    Toronto
    Posts
    258
    Thanks
    1

    precentage change and weighted mean question

    A weighted mean for the following test scores

    4/10
    8/15
    19/31
    3/5

    2/5
    5/5
    21/28
    72/100

    is the sum of the values times their weights over the sum of the weights so

    (4*10+8*15+19*31+....+72*100)/(10+15+31+....+100)

    which comes to 8587/199 = 43.2 (to two decimals places)

    and if the first term changes from 4/10 to 6/10 then the new weighted mean would be

    (6*10+8*15+19*31+....+72*100)/(10+15+31+....+100) = 43.3 to two decimal places.

    The change between the two would be equal to 0.1

    The question I'm trying to answer is how much of a change will there be in the final result? My options are

    "2 percent", "2 percentage points", "0.6 percent" and "0.6 percentage points"

    I'm not sure what the difference is between "percentage points" and "percent". I also can't figure a way to get any of the above answers (2 percent, two percentage points etc).

    If I set all the fractions equal to 1 (ie 100 percent on every test) I get just over 61 = 100% on every test, which means the weighted mean is itself not a percentage. If I convert the scores to a percentage I get 43.2/61 which is 70.81 percent (71 if rounded to the nearest percentage) and 43.2/61 which is 70.98 percent (ALSO 71 if rounded to the nearest percentage). The percent difference is about .23 percent according to what I have.

    If I take 43.2 and scale it up by 2 percent I get 43.2*(1.02)=44.064 so that's not right but if I scale up 43.2 by 0.6 percent I get 43.2*1.006 = 43.54 which is also wrong. The score has been changed by 0.0023148 because 1.0023148*43.2= 43.29999936

    I must have misunderstood something about this, can anyone help? is my weighted mean of 43.2 wrong? or is it percentage change that I'm doing wrong?

    thanks



    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,444
    Thanks
    1863

    Re: precentage change and weighted mean question

    Since only the first value changes, and its weight doesn't, that's relatively simple to calculate.
    The first values give, as you say, a weighted average of (4*10+8*15+19*31+....+72*100)/(10+15+31+....+100)= (40+ 8547)/199= 8587/199= 43.2 (to one decimal place- it is 43.15 to two decimal places). If the "4" increases to 6 we have (60+ 8547)/199= 8607/199= 43.26 to two decimal places. In terms of percentages, the second number is 43.26/43.15= 1.003 or 103%
    Thanks from kingsolomonsgrave
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    May 2012
    From
    Toronto
    Posts
    258
    Thanks
    1

    Re: precentage change and weighted mean question

    Thanks much!

    One question though, at the end shouldn't that be an increase of 100.3%, or am I missing something there?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Weighted Moving Average Question
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 15th 2012, 05:24 PM
  2. Replies: 0
    Last Post: March 4th 2011, 12:55 AM
  3. Weighted Integral Question.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2010, 09:40 AM
  4. precentage
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 10th 2008, 07:43 AM
  5. precentage change
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 4th 2007, 01:46 PM

Search Tags


/mathhelpforum @mathhelpforum