SOLUTION: Consider a data set S with 51 signals of value 1 and 49 signals of value −1. Calculate the mean x ̄ of S and the standard deviation σ_x. What would happen to the

49 of the 1's would cancel all the -1's and leave two 1's, whose sum is 2.
Divide that by 100 and the mean is 2/100 or 0.02

Subtracting 0.02 from each of the 51 1's makes 51 0.98's. Subtracting 0.02 from
each of the 49 -1's makes 49 -1.02's.  Incidentally, if you added those
together you'd just get 0.  

So we square the 51 0.98's and that gives us 51 0.9604's.  We square the 49
-1.02's and that gives us 49 1.0404's.  Now we add 51*0.9604+49*1.0404 and get
99.96.  

If the data set is that of a population we divide by 100 and get 0.9996 for the
variance.

If the data set is that of a sample we divide by 99 and get 1.00969697 for the
variance.    

So if the data set is a population, we take the square root of 0.9996 and get
0.99979998 for the standard deviation.

If the data set is a sample, we take the square root of 1.00969697 and get
1.004836788 for the standard deviation. 

Suppose we add, say, 100 0's:

49 of the 1's would still cancel all the -1's and leave two 1's, whose sum is
2.  The 0's would not change that. Divide that by 200 and the mean is 2/200 or
0.01

Subtracting 0.01 from each of the 51 1's makes 51 0.99's. Subtracting 0.01 from
each of the 49 -1's makes 49 -1.01's. Subtracting 0.01 from each of the 100 0's
makes 100 -0.01's.   Incidentally, as before, and as in all such situations, if
you added those together you'd just get 0.  

So we square the 51 0.99's and that gives us 51 0.9801's.  We square the 49
-1.01's and that gives us 49 1.0201's. We square the 100 -0.01's and that gives
us 100 0.0001's.   

Now we add 51*0.9801+49*1.0201+100*(0.0001) and get 99.98.  

If the data set is that of a population we divide by 200 and get 0.9998 for the
variance.

If the data set is that of a sample we divide by 200 and get 0.4999 for the
variance.    

So if the data set is a population, we take the square root of 0.9998 and get
0.999899995 for the standard deviation.

If the data set is a sample, we take the square root of 0.4999 and get
0.707036067 for the standard deviation.

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My conclusion is that if we add as many 0's as we have data values, the mean
would be about half as much and the standard deviation would be reduced by
about 70%.

You could try adding different amounts of zeros and seeing what you get.

Edwin