If we flip a coin following we have to note.

flip a coin 6 times.

if the last flip is heads record (-1)*#of heads in the first 5 flips

if the last flip is tail record  (1) * #of heads in the first 5 flips

1. Compute the expected standard deviation (S) for the sample mean if n=10 .

The process you describe will generate a binomial distribution with
mean = 0 and a standard deviation of the distribution equal to
sqrt(5).  To see this, let B_5(x) be the binomial distribution
describing the probability that x heads will occur in 5 throws of a
coin.  The probability distribution generated by the process in your
question is given by: P(y) = (probability of a head in the 5th toss) *
(probability of |y| heads in the previous 5 tosses) = 0.5 * B_5(|y|),
where y can take on all possible values integer values from -5 to 5,
and |y| is the absolute value of y.

P(y) is thus seen to be the standard binomial distribution B_5(x),
stretched out by a factor of two (i.e., the range of possible outcomes
goes from -5 to 5, instead of from 0 to 5).  The standard deviation of
this distribution is simply twice the standard deviation of the usual
binomial distribution, or 2 * sqrt(5*.5*.5) = sqrt(5).

The standard deviation of the mean is given by the standard deviation
of the population divided by the square root of the number of trials. 
If the number of trials = 10, then SD-mean = sqrt(5)/sqrt(10) =
1/sqrt(2).