Let \(x=\) number of heads in one toss of a coin. What are the possible values of \(x\) and their probabilities? What is \(\left.\mu_{x} ? \text { Hence show that } \operatorname{Var}(x)=\text { [average of }\left(x-\mu_{x}\right)^{2}\right]\) \(=\frac{1}{4},\) so the standard deviation is \(\frac{1}{2} .\) Now use the result from Problem 6.15 "variance of a sum of independent random variables \(=\) sum of their variances" to show that if \(x=\) number of heads in \(n\) tosses of a coin, \(\operatorname{Var}(x)=\frac{1}{4} n\) and the standard deviation \(\sigma_{x}=\frac{1}{2} \sqrt{n}\).

The expected value, often denoted \(\mu\) or \(E(x)\), represents the long-run average outcome of a random variable after many trials.
The expected value gives you an idea of the center of the distribution for that variable.
In the context of a coin toss where \(x\) is the number of heads, the possible values are 0 for tails and 1 for heads. Since each outcome has a probability of \(\frac{1}{2}\), the expected value is calculated as follows:
\[ \mu_x = E(x) = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} = \frac{1}{2} \]
This means that, on average, out of many tosses, the number of heads you expect is 0.5 per toss.

Variance, denoted as \operatorname{Var}(x)\, quantifies the spread or dispersion of a set of values around their mean.
It represents how much the values differ from the expected value.
For our coin toss example, we need to find how values vary from \(\mu_x\):
\[ \operatorname{Var}(x) = E((x - \mu_x)^2) = (0 - \frac{1}{2})^2 \cdot \frac{1}{2} + (1 - \frac{1}{2})^2 \cdot \frac{1}{2} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \]
As shown, the variance for one coin toss is \(\frac{1}{4}\). This value tells us that there is a small, uniform spread around the mean value of 0.5.

Standard deviation is the square root of variance and provides a measure of how much values typically deviate from the mean.
It is denoted by \sigma_x\.
In our example, the variance was calculated as \(\frac{1}{4}\), so the standard deviation \(\sigma_x\) is:
\[ \sigma_x = \sqrt{\operatorname{Var}(x)} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]
This tells us that in one coin toss, the number of heads will typically deviate from the mean by 0.5.

Independent random variables are those whose outcomes do not affect each other.
In our case, each toss of a coin is independent of the others.
Whether you get heads or tails on one toss has no influence on the result of the next toss.
When dealing with sums of independent random variables, their variances add up.
For \(n\) tosses of the coin, the general variance would be:
\[ \operatorname{Var}(x) = \frac{1}{4}n \]
Consequently, the standard deviation for \(n\) tosses is:
\[ \sigma_x = \sqrt{\operatorname{Var}(x)} = \sqrt{\frac{1}{4}n} = \frac{1}{2}\sqrt{n} \]
This implies that with more tosses, the standard deviation increases, reflecting higher variability but still under the principle that each event is independent.