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Suppose a discrete random variable X takes on the values $0, 1, 2, ,n$ with frequencies proportional to binomial coefficients $\binom{n}{0}, \binom{n}{1},....,\binom{n}{n}$ Then the mean (m ) and the variance ($s_{2}$ ) of the distribution are?

I differentiated the binominal expansion and took $x=1$ which concluded to $\sum x_{i}= n2^{n-1}$ .Therefore the mean should be $ \frac{\sum x_{i}}{n}=2^{n-1} $. But, the mean is $\frac{n}{2}$.

Can anyone help me, thanks in advance.

user405925
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    You don't sum up the number of values... you sum up the frequencies of those values. Remember that $\sum\limits_{k=0}^n\binom{n}{k}=2^n$ as per the binomial theorem. Regardless, this is a binomial distribution where $p=q=\frac{1}{2}$. The mean is $np$ and the variance is $npq$ – JMoravitz Mar 08 '17 at 08:09
  • OH! Got it , Many thanks friend. – user405925 Mar 08 '17 at 08:20

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The total frequency isn't $n$. In fact, it's proportional to

$$\sum_{i = 0}^n {n\choose i} = 2^n$$

hence the mean should be

$$\frac{n\cdot 2^{n - 1}}{2^n} = \frac{n}{2}$$

Yiyuan Lee
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