Question

If X denotes the number of heads in n tosses of a coin, what is the standard deviation of the random variable X? Does this standard deviation get larger or smaller when n gets larger? (Hint: this is a theoretical question, and is not asking about your simulation results. You should get an algebraic expression involving n.)

Answer 1

Answer:

$\sigma=\sqrt{np(1-p)}=\sqrt{n*0.5(1-0.5)}=\frac{\sqrt{n}}{2}$

As we can see the deviation is proportional to the value of n and if n increase then the deviation increases too. So then the deviation would be larger when n gets larger.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

Solution to the problem

We can define the following random variable X ="Number of heads in n tosses of a coin".

We assume that the coin is fair and then $P(H)= 0.5$ for any trial so then we can model X with the following distribution:

$X \sim Bin (n, p=0.5)$

For this distribution the mean and variance are given by:

$E(X)=np=n*0.5=\frac{n}{2}$

$Var(X)= np(1-p) = n*0.5*(1-0.5) = 0.25 n = \frac{n}{4}$

And the deviation would be just the square root of the variance and we got:

$\sigma=\sqrt{np(1-p)}=\sqrt{n*0.5(1-0.5)}=\frac{\sqrt{n}}{2}$

Does this standard deviation get larger or smaller when n gets larger?

As we can see the deviation is proportional to the value of n and if n increase then the deviation increases too. So then the deviation would be larger when n gets larger.