What is the standard deviation of the following data set rounded to the nearest tenth? 60, 40, 35, 45, 39
What tools have you available? What method have you at your disposal?
first find the mean then use the formula 1/n-1((xi -u)^2 +.... +(xn - u)^2)
u = mean, xi is u each data for x til ur last one. n= total sample
8.7
then u sqrt it, I forgot to mention that part
no 8.7 isn't right
what did u get for ur mean?
s^2 = sum (value - mean)^2 /(n-1) where n = number of samples s = standard deviation
then i have been lied to http://www.mathsisfun.com/data/standard-deviation-calculator.html
@douglaswinslowcooper How do you know it's a sample? @alexgriffin If only we could get a response from the OP.
"data set" suggested sample to me rather than being the entire population of interest
Not a standard definition. No real objection, though. If that's the whole population, why in the name of reason would we care about the standard deviation?
@tkhunny you seem to be very wise. Help me on my physics of you don't mind
sorry for the wait I got 43.8
43.8 is the Mean, or Average. That is not the question.
i was replying to @fiberdust
Ah. Missed that. Very good. That is the Mean. Not sure what else fiberdust is calculating. Rounding to integer, maybe?
if u don't get these numbers right you won't get the sd right
it's a very tedious process
sum/n = mean
see i used 45 twice
219/5
my bad
haha its okay :)
i had 44.8 cuz i used 45 instead of 40 mb
so yes 43.8 is correct mean
what i do from there
I might be wrong how i view this but maybe others can confirm or correct me. Since you are given data sample it's going to be 1/n-1 for your formula the other part of the formula is same you will use your data sample value for example (60- mean)^2 +... continue to do each data value then add them up then use 1/n-1( ) this will be s^2 which is your variance. square root the variance to get standard deviation
We are ASSUMING it's a SAMPLE. State that and use n-1 in the denominator.
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