Mean and Standard deviation question!
you know the formulas, right?
No not for this
um what is the formula called?
mean is just the average, sum all the values and divide by the number of values. first find mean, you'll need it for std deviation
std deviation is the square root of variance and variance is \(\Large {Var}(X) = \frac{1}{n} \sum \limits_{i=1}^n (x_i - \mu)^2. \)
problem, Im not sure where to put some of the values
\(\Large \mu = \dfrac{1.13+1.02+1.23+1.06+1.16}{5}\)
it is linear so 1.12u
yes, thats mean now calculate variance using the formula
n = 5 \(\mu\) =1.12
i got 1.25u
\(mean = \mu = 1.12\) \(\Large {Var}(X) = \dfrac{1}{n} \sum \limits_{i=1}^n (x_i - \mu)^2 \\ \Large = \dfrac{1}{5} [(1.13 - 1.12)^2+(1.02 - 1.12)^2 \\ \Large +(1.23 - 1.12)^2+(1.06 - 1.12)^2 +(1.16 - 1.12)^2] \)
calculating that has nothing to do with statistics you can use calculator or web to do that calculation for you
ok i got 8.43
ok this time i got 0.00685
see the example here, https://www.easycalculation.com/statistics/standard-deviation.php
Question: is standard deviation and variance considered the same thing? or do I find it now?
they are conceptually different, but one formula to find variance, and square root of variance is std deviation
Now, my last question is basically asking if mean and median are or are not conflicting between which is the central value.. right?
When the values of items in the data set are not important BUT the measure which divides the distribution into two equal parts is required, Median is considered as the measure of central tendency otherwise Mean is considered
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