Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating skewed distribution?

  1. Aug 13, 2014 #1
    I am trying to calculate the distribution of a number of units between two points with a desired average not necessarily in the middle. In an even distribution I would normally find the difference between the two points and use the result to divide the number of units for distribution.

    100 units
    50 - 0 = 50
    100 / 50 = 2 per interval
    $50 / 2 = $25.00

    So in the above if I had a 100 units that I wanted sell evenly for between $50 and $1 with an average price of $25. I would have to sell 2 units at $50, 2 for $49 ... and so on until inventory was depleted.

    My question is lets say I wanted to sell the exact same number of units between the exact dollar values, but I wanted to skew my distribution so that my average final sale price was say $30 or $20. How would I go about calculating that ?
  2. jcsd
  3. Aug 13, 2014 #2


    User Avatar
    Homework Helper

    Actually, your average in this case would be $25.50 because you calculated 50-0 even though you don't include giving away 2 items for free.

    [tex]=\frac{1}{50}\cdot \frac{50(50+1)}{2}=\frac{51}{2}[/tex]

    I'm not exactly following what you want. You want to sell, say, 100 units and the prices you're restricted to is $1-$50 and you want to know what prices you should sell them at so that your average is close to some value $x? If so, what other restrictions are there, because you can do this in many ways. If not, could you please rephrase the question.
  4. Aug 13, 2014 #3
    Yes, you pretty much have it covered.

    My two main concerns are

    1. Have a final average price of $x

    2. Distribute the selling as evenly as possible between the starting and ending values.

    So that say at $43 I know to sell x number of units and so on until the average is met and the inventory is completely liquidated.

    Please feel free to ask any questions needed for clarity ...I don't do this type of math everyday so describing it doesn't come as second nature.
  5. Aug 13, 2014 #4


    User Avatar
    Homework Helper

    Ok I understand now. Like I said earlier, there are many ways to end up near the same average, but going by your "distribute evenly as possible" criterion, then here is one way I would suggest.

    We already know that given 100 units sold over $1 - $50 evenly gives us 2 units per dollar sale, so if we want to be able to move units around, we should cut our starting point down to 1 unit per dollar sale (so we have 50 units left to work with) and we already know that the average for this is $25.50, so say you want an average of $30 then we need to start placing around the $35 mark.

    The reason for $35 as opposed to $30 is because half of our values have an average of about $25, so when the other half then has an average of $35, the total average will be the average of $25 and $35 = $30.

    So we first sell 1 of each unit at each dollar, then we sell as many units around $35 till we hit the $50 wall. So we would be selling $21 - $50 (a total of 30 units). Then we finally sell the last 20 units around $35 again, so we will be selling at $26 - $45.

    In total, you will be selling
    1 unit at 1-20
    2 units at 21-25
    3 units at 26-45
    2 units at 46-50

    The average is calculated by

    [tex]\frac{1}{100}\left(\sum_{i=1}^{50}i +\sum_{j=21}^{50}j +\sum_{k=26}^{45}k\right) = 30.5 [/tex]

    Notice however that the average is 30.5 for the same reason that you calculated 25.5 earlier (we really should have been placing around the $34.50 mark), if you want this fixed then just move 1 unit a total of $50 down, or 50 units by $1 each, or any variation thereof.

    So if I shift the last 2 sums down by 1 value, we get the desired result:

    [tex]\frac{1}{100}\left(\sum_{i=1}^{50}i +\sum_{j=20}^{49}j +\sum_{k=25}^{44}k\right) = 30 [/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook