1. Interpret the data

The values of $x$ are $0,2,4,6,\dots,2n$ with corresponding frequencies $\binom n0,\binom n1,\binom n2,\dots,\binom nn.$

So for the value $x=2r$, the frequency is $\binom nr$, where $r=0,1,2,\dots,n$.

2. Formula for mean

Mean of a frequency distribution is $\bar x=\frac{\sum f_i x_i}{\sum f_i}.$

Here, $\sum f_i=\sum_{r=0}^n \binom nr = 2^n.$

Also, $\sum f_i x_i=\sum_{r=0}^n \binom nr (2r)=2\sum_{r=0}^n r\binom nr.$

We use the standard identity $\sum_{r=0}^n r\binom nr = n2^{n-1}.$

Therefore, $\sum f_i x_i = 2\cdot n2^{n-1}=n2^n.$

Hence the mean is $\bar x=\frac{n2^n}{2^n}=n.$

3. Compare with the given mean

Given mean is $\frac{728}{2^n}.$

So, $n=\frac{728}{2^n}.$

Thus, $n2^n=728.$

Now check powers of $2$: $728=8\cdot 91=2^3\cdot 91.$

Try integer values of $n$:

• For $n=3$, $n2^n=3\cdot 8=24$
• For $n=4$, $4\cdot 16=64$
• For $n=5$, $5\cdot 32=160$
• For $n=6$, $6\cdot 64=384$
• For $n=7$, $7\cdot 128=896$

There is no integer $n$ such that $n2^n=728.$

4. Check the stored answer $6$

If $n=6$, then the mean should be $\bar x=n=6.$

But the given expression becomes $\frac{728}{2^6}=\frac{728}{64}=11.375,$ which is not equal to $6$.

So the stored answer $6$ does not satisfy the given question.

5. Conclusion

From the data, the mean is always $n$. Equating with $\dfrac{728}{2^n}$ gives $n2^n=728,$ which has no integer solution.

Therefore, the question as stated appears to contain an error, and the stored answer $6$ is inconsistent with the given mean.