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

Mean, Deviation (DEV), and Average (DEV)? question

  1. Aug 28, 2011 #1
    For the following data set of eight values, give the MEAN, DEVIATION (DEV), and AVERAGE DEVIATION (AD). Show a sample calculation for each: 0.77, 0.92, 1.12, 1.00, 0.96, 0.88, 1.16, 1.02

    2. Relevant equations
    Well before we began, can anyone take the courtesy and see if I'm getting the correct answer?

    MEAN: 0.97875
    DEV: .0126 (maybe?)
    AD: ???

    3. The attempt at a solution

    (-167/800)^2+(-47/800)^2+(113/800)^2+(-3/160)^2+(-79/808)^2+(29/160)^2+(33/800)^2 = 0.111635/(8-1) = 0.0159 =sqrt(0.0159)=0.126

    I'm confused now....I think I found the standard deviation. However, I'm unsure how to even calculate the average deviation cause I don't have enough values. what am i doing wrong? What should I do next?
  2. jcsd
  3. Aug 29, 2011 #2


    User Avatar
    Homework Helper

    mean looks good
    \mu = \frac{1}{n}\sum_i x_i
    terminology, generally I would consider:
    standard deviation (square root of variance)
    \sigma = \sqrt{\frac{1}{n}\sum_i (x_i-\mu)^2}
    sample standard deviation (square root of sample variance)
    s = \sqrt{\frac{1}{n-1}\sum_i (x_i-\mu)^2}

    can you give your formulas for each DEV and AD?

    looks like you are calculating the sample deviation (based on the (8-1) term)
    Last edited: Aug 29, 2011
  4. Aug 29, 2011 #3


    User Avatar
    Homework Helper

    Also though I cpiece together what you are attempting, writing it like this may confuse things for other people, in particular
    which doesn't make any sense

    it would be better to be break it up into

    DEV^2 = (-167/800)^2+(-47/800)^2+(113/800)^2+(-3/160)^2+(-79/808)^2+(29/160)^2+(33/800)^2 = 0.111635/(8-1) = 0.0159


    sorry if it comes across as pedantic, but if its clear what you're attempting you'll generally get a better and quicker answer (probably in test as well)
    Last edited: Aug 29, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook