This problem is not easy

The current mean should do it..

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gurpreetsingh,
your answer is approximately correct only for large n. The exact answer is:
\(x_n_+_1=\mu_n\pm\sigma_n\sqrt{\frac{n+1}{n}}\)
The derivation of this formula is not straightforward and took me quite some time. I have the derivation, but unfortunately cannot post my website address yet because I just joined the forum today.

mainhoon,
Sorry, no cigar. the mean will not change the mean, but it will change the standard deviation.

Let d= standard deviation and a = Mean

\(d = \sqrt{\frac{{(a-x_1)^2 + (a-x_2)^2 + (a-x_2)^2 .....}}{n}}\)

New \(d_n\)= \(\sqrt{\frac{{(a-x_1)^2 + (a-x_2)^2 + (a-x_2)^2 .....(a-x_{n+1})^2}}{{n+1}}}\)

For \(d = d_n\)

\({d_n}^2 = \frac{{(a-x_1)^2 + (a-x_2)^2 + (a-x_2)^2 .....(a-x_{n+1})^2}}{{n+1}}\)

\({d}^2 = \frac{{ d^2 *n+ (a-x_{n+1})^2}}{{n+1}}\)

\({d}^2 * (n+1) = { d^2 *n+ (a-x_{n+1})^2}\)

=> \({d}^2 = {(a-x_{n+1})^2}\)

=> \(|d| = |a-x_{n+1}|\)

But why is the mean in the n and n+1 case the same? a? that should change...I think the idea is to equate both the standard deviation formula and calculate what the new value should look like.. But I doubt if this is GMAT type...

gurpreetsingh,
Your derivation is surprisingly similar to my first attempt, and it happens to be correct as an approximation only for large n. But there is a flaw in this approach, because you are assuming that the mean is the same in both lists, which is not true.

mainhoon,
You are right, the mean will change, and that is the flaw in gurpreetsingh derivation. The derivation of the exact formula is quite involved. Also, I am sure this is not a GMAT question.

The Problem:
Find a number, such that when added to a list of numbers will not change the standard deviation of the original list.

The solution:
Define the list of numbers as: \($x_1 ,x_2 , \cdots x_n ,x_{n + 1}$\) where \($x_{n + 1} $\) is the number added to the list.
The mean and the standard deviation of the original list are:
\(\bar x_n = \frac{1}\\
{n}\sum\limits_{i = 1}^n {x_i } \\
\\)
\(\sigma _n ^2 = \frac{1}\\
{{n - 1}}\sum\limits_{i = 1}^n {\left( {x_i - \bar x_n } \right)^2 } \\
\\)
After \($x_{n + 1} $\) is added to the original list, the mean and the standard deviation of the new list are:
\(\bar x_{n + 1} = \frac{1}\\
{{n + 1}}\sum\limits_{i = 1}^{n + 1} {x_i } \\
\\)
\(\sigma _{n + 1} ^2 = \frac{1}\\
{n}\sum\limits_{i = 1}^{n + 1} {\left( {x_i - \bar x_{n + 1} } \right)^2 } \\
\\)

We want to find \($x_{n + 1} $\)

such that \($\sigma _n ^2 = \sigma _{n + 1} ^2 $\)

therefore,
\(\frac{1}\\
{{n - 1}}\sum\limits_{i = 1}^n {\left( {x_i - \bar x_n } \right)^2 } = \frac{1}\\
{n}\sum\limits_{i = 1}^{n + 1} {\left( {x_i - \bar x_{n + 1} } \right)^2 } \\
\\)
The sum of terms up to n is
\(\sum\limits_{i = 1}^n {x_i = n\bar x_n } \\
\\)
and the sum of terms up to n+1 is
\(\sum\limits_{i = 1}^{n + 1} {x_i = (n + 1)\bar x_{n + 1} } \\
\\)
Subtracting \($x_{n + 1}$\)

\(\sum\limits_{i = 1}^{n + 1} {x_i - x_{n + 1} = \sum\limits_{i = 1}^n {x_i } } = n\bar x_n \\
\\)
therefore,

\(\(n + 1)\bar x_{n + 1} - x_{n + 1} = n\bar x_n \\
\\)

\(\bar x_{n + 1} = \frac{n}\\
{{n + 1}}\bar x_n + \frac{{x_{n + 1} }}\\
{{n + 1}}\\
\\) (1)

Some algebra:

\(\sum\limits_{i = 1}^{n + 1} {\left( {x_i - \bar x_{n + 1} } \right)^2 } = \left( {x_{n + 1} - \bar x_{n + 1} } \right)^2 + \sum\limits_{i = 1}^n {\left( {x_i - \bar x_{n + 1} } \right)^2 } \\
\\)
Substituting \(\bar x_{n + 1} \\
\\) with (1)
\(\sum\limits_{i = 1}^{n + 1} {\left( {x_i - \bar x_{n + 1} } \right)^2 } = \left( {x_{n + 1} - \frac{n}\\
{{n + 1}}\bar x_n - \frac{{x_{n + 1} }}\\
{{n + 1}}} \right)^2 + \sum\limits_{i = 1}^n {\left( {x_i - \frac{n}\\
{{n + 1}}\bar x_n - \frac{{x_{n + 1} }}\\
{{n + 1}}} \right)^2 } \\
\\)
Rearranging,
\(\ = \left( {\frac{n}\\
{{n + 1}}x_{n + 1} - \frac{n}\\
{{n + 1}}\bar x_n } \right)^2 + \sum\limits_{i = 1}^n {\left( {x_i - \frac{n}\\
{{n + 1}}\bar x_n - \frac{{x_{n + 1} }}\\
{{n + 1}}} \right)^2 } \\
\\)
\(\ = \left( {\frac{n}\\
{{n + 1}}(x_{n + 1} - \bar x_n )} \right)^2 + \sum\limits_{i = 1}^n {\left( {x_i - \bar x_n + \frac{{\bar x_n }}\\
{{n + 1}} - \frac{{x_{n + 1} }}\\
{{n + 1}}} \right)^2 } \\
\\)
\(\ = \left( {\frac{n}\\
{{n + 1}}} \right)^2 (x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {\left( {x_i - \bar x_n + \frac{1}\\
{{n + 1}}(\bar x_n - x_{n + 1} )} \right)^2 } \\
\\)
Expanding the square in the summation,
\(\ = \left( {\frac{n}\\
{{n + 1}}} \right)^2 (x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {\left( {(x_i - \bar x_n )^2 + \frac{{2(x_i - \bar x_n )}}\\
{{n + 1}}(\bar x_n - x_{n + 1} ) + \frac{1}\\
{{(n + 1)^2 }}(\bar x_n - x_{n + 1} )^2 } \right)} \\
\\)
Note that
\(\\\
(x_{n + 1} - \bar x_n )^2 = (\bar x_n - x_{n + 1} )^2 \\
\\)
Therefore
\(\ = \left( {\frac{n}\\
{{n + 1}}} \right)^2 (x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {\left( {(x_i - \bar x_n )^2 + \frac{{2(x_i - \bar x_n )}}\\
{{n + 1}}(\bar x_n - x_{n + 1} ) + \frac{1}\\
{{(n + 1)^2 }}(x_{n + 1} - \bar x_n )^2 } \right)} \\
\\)
If C is a constant
\(\sum\limits_{i = 1}^n {(f(i) + C) = } \sum\limits_{i = 1}^n {f(i) + nC} \\
\\)
\(\sum\limits_{i = 1}^n {f(i) + Cg(i) = } \sum\limits_{i = 1}^n {f(i) + C\sum\limits_{i = 1}^n {g(i)} } \\
\\)
Therefore
\(\ = \left( {\frac{n}\\
{{n + 1}}} \right)^2 (x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {(x_i - \bar x_n )^2 + \sum\limits_{i + 1}^n {\left( {\frac{{2(x_i - \bar x_n )}}\\
{{n + 1}}(\bar x_n - x_{n + 1} ) + \frac{1}\\
{{(n + 1)^2 }}(x_{n + 1} - \bar x_n )^2 } \right)} } \\
\\)
\(\ = \left( {\frac{n}\\
{{n + 1}}} \right)^2 (x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {(x_i - \bar x_n )^2 + \frac{2}\\
{{n + 1}}(\bar x_n - x_{n + 1} )\sum\limits_{i + 1}^n {(x_i - \bar x_n ) + \frac{n}\\
{{(n + 1)^2 }}(x_{n + 1} - \bar x_n )^2 } } \\
\\)
Observe that
\(\sum\limits_{i = 1}^n {(x_i - \bar x_n )} = \sum\limits_{i = 1}^n {x_i } - n\bar x_n = n\bar x_n - n\bar x_n = 0\\
\\)
Therefore
\(\ = \left( {\frac{n}\\
{{n + 1}}} \right)^2 (x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {(x_i - \bar x_n )^2 + } \frac{n}\\
{{(n + 1)^2 }}(x_{n + 1} - \bar x_n )^2 \\
\\)
\(\ = \frac{n}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {(x_i - \bar x_n )^2 } \\
\\)
From the original problem,
\(\frac{1}\\
{{n - 1}}\sum\limits_{i = 1}^n {\left( {x_i - \bar x_n } \right)^2 } = \frac{1}\\
{n}\sum\limits_{i = 1}^{n + 1} {\left( {x_i - \bar x_{n + 1} } \right)^2 } \\
\\)
Therefore, we need \(x_{n + 1} \\
\\) to satisfy
\(\frac{1}\\
{{n - 1}}\sum\limits_{i = 1}^n {\left( {x_i - \bar x_n } \right)^2 } = \frac{1}\\
{n}\left( {\frac{n}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 + \sum\limits_{i = 1}^n {(x_i - \bar x_n )^2 } } \right) = \frac{1}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 + \frac{1}\\
{n}\sum\limits_{i = 1}^n {(x_i - \bar x_n )^2 } \\
\\)
\(\sigma _n ^2 = \frac{1}\\
{{n - 1}}\sum\limits_{i = 1}^n {\left( {x_i - \bar x_n } \right)^2 } = \frac{1}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 + \frac{1}\\
{n}\sum\limits_{i = 1}^n {(x_i - \bar x_n )^2 } \\
\\)
\(\sigma _n ^2 = \frac{1}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 + \frac{{n - 1}}\\
{n}\sigma _n ^2 \\
\\)
\(\sigma _n ^2 - \frac{{n - 1}}\\
{n}\sigma _n ^2 = \frac{1}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 \\
\\)
\(\frac{1}\\
{n}\sigma _n ^2 = \frac{1}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 \\
\\)
\(\sigma _n ^2 = \frac{n}\\
{{(n + 1)}}(x_{n + 1} - \bar x_n )^2 \\
\\)
\(x_{n + 1} = \bar x_n \pm \sigma _n \sqrt {\frac{{n + 1}}\\
{n}}\\)
For large n
\(x_{n + 1} \approx \bar x_n \pm \sigma _n \\
\\)
For large n, adding an observation one standard deviation on either side of the mean will not change the standard deviation significantly.