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# Question: If a variable takes values with $0,1,2,3,...,n$ frequencies $1,C\left( {n,1} \right),C\left( {n,2} \right),C\left( {n,3} \right),...,C\left( {n,n} \right)$ respectively, then the arithmetic mean is A. $2n$B. $n + 1$ C. $n$D. $\dfrac{n}{2}$ Verified
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Hint: Arithmetic mean is the sum of a collection of numbers divided by the count of numbers in the collection. This definition will be useful in solving the problem.
$mean = \dfrac{{0 \times 1 + 1 \times C\left( {n,1} \right) + 2 \times C\left( {n,2} \right) + ... + n \times C\left( {n,n} \right)}}{{1 + C\left( {n,1} \right) + C\left( {n,2} \right) + ... + C\left( {n,n} \right)}}$
We can write $1$as$C\left( {n,0} \right)$, so that we can get a pattern and solve it easily.
$= \dfrac{{1 \times C\left( {n,1} \right) + 2 \times C\left( {n,2} \right) + ... + n \times C\left( {n,n} \right)}}{{C\left( {n,0} \right) + C\left( {n,1} \right) + C\left( {n,2} \right) + ... + C\left( {n,n} \right)}}$
$= \dfrac{{\sum\limits_{r = 1}^n {r \times C\left( {n,r} \right)} }}{{{2^n}}}$
Answer $= \dfrac{{n \times {2^{n - 1}}}}{{{2^n}}} = \dfrac{n}{2}$

Note: Make sure that you take the limits correctly otherwise you are likely to make mistakes in reaching the answer.

Last updated date: 20th Sep 2023
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