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

Question on mean and standard deviation.

  1. Nov 24, 2011 #1
    The length of time, t minutes, taken to do the crossword in a certain newspaper was observed on
    12 occasions. The results are summarised below.
    Σ(t − 35) = −15
    Σ(t − 35)^2 = 82.23
    Calculate the mean and standard deviation of these times taken to do the crossword.

    My real problem is finding out the value of t. or can we find the answer without finding t??. I took t-35 as y, and hence got y's mean as -1.25. Then im stuck.
    I cant make head or tail of the answer provided for the mean:( why do we deduct 35 -1.25?)

    Answer
    mean = 35 – 15/12
    = 33.75 (33.8) minutes

    sd = 82.23/12 ( 15/12)
    = 2.3 minutes
     
  2. jcsd
  3. Nov 24, 2011 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You took 12 time measurements, and you'll label them:

    t1, t2, t3, ... ,t10, t11, t12.

    By definition, the arithmetic mean of these time measurements is just: [tex] \bar{t} = \frac{t_1 + t_2 + t_3 + \ldots + t_{10} + t_{11} + t_{12}}{12} = \frac{1}{12}\sum_{i=1}^{12} t_i [/tex]But you haven't been given [itex] \sum_{i=1}^{12} t_i [/itex]. In this problem, what you've been given is that[tex]\sum_{i=1}^{12} (t_i - 35) = -15[/tex]We can write this as:[tex] \sum_{i=1}^{12}t_i - \sum_{i=1}^{12} 35 = \sum_{i=1}^{12}t_i -(12)(35) = -15[/tex][tex]\sum_{i=1}^{12}t_i = (12)(35) - 15 [/tex]Therefore:[tex]\bar{t} = \frac{1}{12}\sum_{i=1}^{12}t_i = \frac{(12)(35)}{12} - \frac{15}{12} = 35 - \frac{15}{12}[/tex]
     
  4. Nov 24, 2011 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I should point out that your method was fine too (and totally equivalent). You said, let yi = ti - 35 so that:[tex]\bar{y} = \frac{1}{12}\sum_{i=0}^{12}y_i = -\frac{15}{12} [/tex]Now, if yi = ti - 35, then it follows that ti = 35 + yi. It also follows that [itex] \bar{t} = 35 + \bar{y}[/itex]. Hence[tex] \bar{t} = 35 - \frac{15}{12} [/tex]My only assumption here was that if you apply a constant offset to all your measurements, their mean will also differ from the mean of the original measurements by that constant offset. This is pretty easy to prove just using the definition of a mean (EDIT: in fact, you can take the steps in my previous post as being a proof of that).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question on mean and standard deviation.
  1. Standard Deviation (Replies: 4)

  2. Standard deviation (Replies: 3)

Loading...