Question on mean and standard deviation.

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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
 

Answers and Replies

  • #2
cepheid
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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]
 
  • #3
cepheid
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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).
 

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