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Homework Help: Can anyone help

  1. Oct 20, 2008 #1
    The table shows results for precipitation (rainfall) obtained with a home-made rain gauge over 7 consecutive days. The measurements were made at 7.30 am on each day, and the values given indicate the depth of water collected in the previous 24 hours. The original measurements were made to the nearest 0.5 mm (i.e. half a division on a ruler), and thus you may assume that the precision of each value presented here is also ±0.5 mm.

    Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7
    Precipitation in previous 24 hours/mm 2.0 0.0 3.5 2.5 4.5 3.0 4.5
    Calculate the mean daily precipitation for this week. Write this mean value and its uncertainty to an appropriate number of decimal places, and with appropriate units.
    The mean daily precipitation = 2.6 ± 0.5
    is this correct please
  2. jcsd
  3. Oct 20, 2008 #2


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    No, it is not correct. Since you don't say how you got it, I can't say more.
  4. Oct 20, 2008 #3
    This is how i got my answer. In order to get my mean, i added all those figures together and divided it by 7 and got the answer of 2.6. Then i chose 0.5 for my uncertainty value. Whilst i honestly thought my mean figure was correct, im very unsure of what this uncertainty value means. Hope that makes sense
  5. Oct 20, 2008 #4


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    (2.0+0.0+3.5+2.5+4.5+3.0+4.5)/7 is more than 2.6.

    Why did you choose 0.5 for your uncertainty? I'm not sure how your class is doing it, but I would expect more like 1.3 (or the 'conservative' 3.5).
  6. Oct 20, 2008 #5
    Hi CRGreathouse
    Please give me 5 mins and im going to start working it all out again and get back with what i think is correct.
  7. Oct 20, 2008 #6
    I get 2.85714 and another load of figures afterwards. Now im more confused than ever. Surely i now need to round off to nearest decimal figure.
    Can you also please explain uncertainty to me as i have absolutely no idea whatsoever.
  8. Oct 20, 2008 #7


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    The numbers add to 20. The mean is 20/7. That looks like what you now have.

    CRGreathouse, I believe that the uncertainty is 0.5. To find the uncertainty of a sum, you add the uncertainties of each. To find the uncertainty of an average you average the uncertainties- since the uncertainty of each measurement is the same, that is the same as that common uncertainty.

    Another way you could do this is to find the largest possible value for each measurement: that measurement + 0.5 here, and find that average, find the smallest possible value for each measurement, -0.5, and find that average. The error is the larger of "largest average minus the average you have" and "average you have minus smallest average". Since all 7 measurements have an error of 0.5, using the highest value gives you 20+ 7(0.5) which averages to 20/7+ 0.5. Using the least value gives you 20- 7(0.5) which averages to 20/7- 0.5. The error is plus or minus 0.5.
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