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Five balls numbered $0,2,4,6,8$ are placed in a bag. After the balls are mixed, one of them is selected, its number is noted, and then it is replaced. If this experiment is repeated many times, find the variance and standard deviation of the numbers on the balls.

I choose $X=0,2,4,6,8$, and hence $f(0)= f(2)=f(4)=f(6)=f(8)= \frac15$. So I think to use the formula $\sigma^2=\mu_2-\mu^2$ to find variance where $\mu=\sum{xf(x)}$ and $\mu_2=\sum{x^2f(x)}$. This is what i think for this question!

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    I find this question a bit puzzling (which presumably is the question-setter's fault, not neemy's). Would the mean and variance be any different if the experiment is _not_ repeated many times? The question _does_ ask for the variance and standard deviation of the _numbers_ which could well mean: In $n$ experiments, the numbers $0,2,4,6,8$ were found to have occurred $n_0, n_2, n_4, n_6,$ and $n_8$ times where $n_0+n_2+n_4+n_6+n_8=n$. What is the variance and standard deviation? etc.2011-12-27

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Yes, this is the way to do it.

But the problem is stated a little imprecisely. You should write something like: "Let the random variable $X$ be the number that appears on the selected ball. Find the variance and standard deviation of $X$".

The variance of a random variable $X$, denoted by $\sigma^2_{ X}$, can be calculated using the formula $ \sigma^2_X=\Bbb E(X^2)-[\Bbb E(X)]^2;$

which, in the discrete case, gives $\sigma^2_X=\sum_i x_i^2 p_X(x_i) -\bigl[\,\sum_i x_i p_X(x_i) \,\bigr]^2 $ where $p_X$ is the probability mass function of $X$ and the $x_i$ are the distinct values that $X$ takes.

In your problem, as you state, $X$ takes the values $0, 2, 4, 6, 8$ with equal probabilities $1/5$. So, $p_X(i)=1/5$ for $i=0,2,4,6, 8$, and: $ \sigma_X^2= \sum_{i\in\{0,2,4,6,8\}} i^2\cdot{\textstyle{1\over 5}} - \Bigl[ \sum_{i\in\{0,2,4,6,8\}} i \cdot{\textstyle{1\over 5}}\Bigr]^2. $

I'll leave the actual computation to you.

Of course, the standard deviation of $X$ is just the square root of the variance.

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    David, @Kannappan 's edit left some more undesirable traces: for instance, the variances are now denoted both by $\sigma_{X}^2$ and $\sigma^2$.2011-12-27