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"I'm really sorry that I'm asking a basic question, I tried to understand but couldn't understand completely"

These are the 3 parameter results of rolling a single die. Expectation E(x)=3.5

Variance Var(x)=2.92

Standard deviation SD=1.709

Expectation is nothing but the average of outcomes of die when we perform infinite trials, that means if we roll a die infinite times and take out the mean of outcome , its value is close to 3.5. that's what I understood. I searched in Google and read about variance ,standard deviation. I didn't understand anything about those two, but left with this sentence "The variance measures how far each number in the set is from the mean" Can anyone please explain what do those two values indicate regarding to my die problem?? Thanks in advance

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  • $\begingroup$ Standard deviation is really just the same as variance, it's just the square root of the variance, so you don't really have two things to understand here, just one. Basically if you graph a histogram of the outcomes, the graph will be wider if the variance is greater and thinner if the variance is lower. If it is thin then values tend to be close to the mean and if it's really wide then they will be more spread out. $\endgroup$ Nov 11, 2015 at 18:04
  • $\begingroup$ Thank you. Can you please explain what can I conclude from those values with respect to above problem ,Var(x)=2.92 and SD=1.709 $\endgroup$
    – Ajay shifu
    Nov 11, 2015 at 18:16
  • $\begingroup$ There's not much you can conclude except that the SD is $1.709$. It would be easier to answer your question if you were comparing two distributions with different variances because you could then conclude that one random variable has more uncertainty than the other. But given just one RV with SD=$1.709$ all you can really conclude is that the SD is $1.709$. Unless maybe they want you to use Chebychev's inequality to say something. Do you know Chebychev yet? $\endgroup$ Nov 11, 2015 at 18:19

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The variance is the expected value of the square of the distance from the mean.

So with your dice, if the mean is $1\times \frac16+2\times \frac16+3\times \frac16+4\times \frac16+5\times \frac16+6\times \frac16 = \frac{7}{2}= 3.5$ then a value of for example $2$ contributes $(2-3.5)^2 = 2.25$, so overall the variance for the dice is $$(1-3.5)^2\times \tfrac16 +(2-3.5)^2\times \tfrac16+(3-3.5)^2\times \tfrac16+(4-3.5)^2\times \tfrac16+(5-3.5)^2\times \tfrac16+(6-3.5)^2\times \tfrac16 = \frac{35}{12} \approx 2.91667.$$ The variance has many useful mathematical properties, for example when adding or subtracting independent random variables or applying the Central Limit Theorem, though it needs to be used with some care as its units are the square of the units of the random variable and of the mean.

The standard deviation is simply the square root of the variance so in the the case of dice throws $\sqrt{\dfrac{35}{12}} \approx 1.7078$. This gives an indication of the scale of typical dispersion from the mean, measured in the same units as the random variable and its mean.

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