Originally Posted by

**dexteronline** I tried the following solution, not sure about the accuracy of the results though

Coin Toss

Number----------- Probability

of Heads--------- of Outcome

x --------------- P(x)

0 --------------- 1/2 = 0.5

1 --------------- 1/2 = 0.5

Total------------ 2/2 = 1

Mean of a probability distribution

$\displaystyle \mu = \sum [xP(x)]$

$\displaystyle \mu = 0(0.5)+1(0.5)$

$\displaystyle \mu = 0.5$

Variance

$\displaystyle \sigma^2 = \sum [(x-\mu)^2P(x)]$

$\displaystyle \sigma^2 = [(0-0.5)^2 (0.5) + (1-0.5)^2 (0.5)]$

$\displaystyle \sigma^2 = [(-0.5)^2 (0.5) + (0.5)^2 (0.5)]$

$\displaystyle \sigma^2 = [(0.25)(0.5) + (0.25)(0.5)]$

$\displaystyle \sigma^2 = [(0.125) + (0.125)]$

$\displaystyle \sigma^2 = 0.25$

Standard Deviation

$\displaystyle \sigma = 0.5$

In one toss of a fair coin, we expect ___0.5_____ heads, give or take ___0.5____ heads or so.

In n=100 tosses, we expect ___50______ heads, give or take ___5___ heads or so

In n=400 tosses, we expect ___200_____ heads, give or take ___10____ heads or so.