Mean and standard deviation after a coin is tossed

Question

Could someone help solve this?

Let X equal the number of heads when $3600$ coins are tossed.

Find the Mean and Standard Deviation of X

Answer 1

I assume each coin toss is independent. Let each of the 3600 outcomes be numbered, so that you have 3600 independent Bernoulli random variables $X_1,\dots, X_{3600}$ (where $X_i$ equals 1 if the $i$-th coin toss is Heads, and $0$ otherwise).

In particular, for all $1\leq i\leq 3600$, we have $\mathbb{P}\{X_i=1\} = \frac{1}{2}$ (the coins are fair), and $$X = \sum_{i=1}^{3600} X_i.$$

You have that $X$ follows a Binomial$(3600,1/2)$ distribution. If you do not know what it is, let's use the following two facts:

• $\mathbb{E}[X_i]=\mathbb{P}\{X_i=1\} = \frac{1}{2}$ for all $i$, and expectation is linear: $$ \mathbb{E}[X]=\mathbb{E}\left[\sum_{i=1}^{3600} X_i\right]=\sum_{i=1}^{3600} \mathbb{E}[X_i] = 3600\cdot \frac{1}{2} = 1800.$$

• Variance is not linear, but the variance of a sum of independent random variables is the sum of their variances: $$ \operatorname{Var}[X]=\operatorname{Var}\left[\sum_{i=1}^{3600} X_i\right]=\sum_{i=1}^{3600} \operatorname{Var}[X_i] = 3600\cdot \frac{1}{4} = 900.$$ Where we used that the $X_i$'s are independent, and that the variance of a Bernoulli(1/2) ("a fair coin") random variable $Z$ is $$ \operatorname{Var}[Z]= \mathbb{E}[Z^2]-\mathbb{E}[Z]^2 = \mathbb{E}[Z] - \frac{1}{4} = \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$ where in the middle we relied on the fact that since $Z$ only can take values either $0$ or $1$, we have $Z^2 = Z$.

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Takeaway: this problem can be modeled by a Binomial$(n,p)$ distribution, where $n=3600$ and $p=1/2$: since you sum 3600 independent coin tosses with same probability of Heads (here, 1/2). In general, if $Y$ follows a Binomial$(n,p)$ distribution, we have $\mathbb{E}[Y] = np$ and $\operatorname{Var}Y = np(1-p)$.