Mastering Mean and Standard Deviation in Binomial Distributions

Unlock the power of binomial distributions! Learn to calculate mean and standard deviation, interpret results, and apply concepts to real-world scenarios. Elevate your statistical analysis skills today.

Table of Contents:

• Mean and standard deviation of binomial distribution
• Table of Discrete Probability Distributions
• What is a binomial distribution
• Mean and standard deviation of binomial distribution
• Binomial distribution examples
• Example 1
• Example 2
• Example 3

Mean and standard deviation of binomial distribution

After looking into what the binomial distribution is, let us start this lecture by saying, such a discrete distribution will be used throughout the remaining of chapter 4 in our course for statistics; take it as the base for all the remaining discrete distributions as it actually shares most of its characteristics with all of the remaining ones (look at the table below).

Table of Discrete Probability Distributions

Distribution definition:	Characteristics:	Probability formula:
Binomial distribution: Number of successes in a certain amount of trials (with replacement)	Fixed number of trials. Each trial has only two possible outcomes: a success or a failure. The probability of success in each trial is constant.	$P(x)= \; _{n}C_{x}p ^{x}(1-p)^{n-x}=$ $\large \frac{n!}{x!(n-x)!}$ $p^{x}(1-p)^{n-x}$ Where: $n$ = number of trials $x$ = number of successes in n trials $p$ = probability of success in each trial $_{n}C_{x}=$$\large \frac{n!}{x!(n-x)!}$ = number of success outcomes $P(x)$ = probability of getting $x$ successes out of n trials
Poisson distribution: Used as an approx. to the binomial distribution when the amount of trials in the experiment is very high in comparison with the amount of successes.	Fixed number of trials. Each trial has only two possible outcomes: a success or a failure. The probability of success in each trial is constant.	$P(x)=e^{-np}$$\large \frac{(np)^{x}}{x!}$ $\;$→$\; P(x)=e^{-\lambda}$ $\large \frac{(\lambda)^{x}}{x!}$ Where: $n$ = number of trials $x$ = number of successes in n trials $p$ = probability of success in each trial $P(x)$=probability of getting $x$ successes out of n trials $\mu = \lambda = np =$ average number of events per time interval
Geometric distribution: Number of trials until the first success.	Fixed number of trials. Each trial has only two possible outcomes: a success or a failure. The probability of success in each trial is constant.	$P(n)=(1-p)^{n-1}p$ Where: $n$ = number of trials until the first success $p$ = probability of success in each trial $P(n)$ = probability of getting your 1st success at nth trial
Negative binomial distribution: Number of trials needed for a certain amount of successes.	Fixed number of trials. Each trial has only two possible outcomes: a success or a failure. The probability of success in each trial is constant.	$P(x)=\; _{n}C_{x}p^{x}(1-p)^{n-x}=$ $\large \frac{n!}{x!(n-x)!}$$p^{x}(1-p)^{n-x}$ Where: $n$ = number of trials $x$ = number of successes in n trials $p$ = probability of success in each trial $_{n}C_{x}=$$\large \frac{n!}{x!(n-x)!}$ = number of success outcomes $P(x)$ = probability of getting x successes out of n trials
Hypergeometric distribution: Number of successes in a certain amount of trials (without replacement)	A randomly selected sample of fixed size is selected without replacement from a population. The population and the sample have only two possible outcomes: a success or a failure. The probability of success in each trial is not constant.	$P(x)=$$\large \frac{(_{m}C_{x})(_{N-m}C_{n-x})}{NCn}$ Where: $N$ = population size $m$ = number of successes in the population $n$ = sample size $x$ = number of successes in the sample $_{m}C_{x}=$ $\large \frac{m!}{x!(m-x)!}$ $_{N-m}C_{n-x}=$$\large \frac{(N-m)!}{(n-x)!((N-m)-(n-x))!}$ $_{N}C_{n}=$$\large \frac{N!}{n!(N-n)!}$

With all of this in mind, let us take this lecture just as an expansion of our latest one on the binomial distribution, since we will be remembering the concept of it in order to introduce its mean, standard deviation and variance; all important concepts to the full comprehension of the distribution itself. And with that, we can move onto the rest of the discrete probability distributions.

What is a binomial distribution

From our past lesson we learned that a binomial distribution is a discrete probability distribution which can have only two possible outcomes: either a success or a failure. This type of probability distribution shows the possible results to occur in a series of finite experiments where the answers are yes or no (did the expected outcome happened? Or did it not?), also referred as success (if yes) or failure (if no), true or false, and the values of one (for the occurrence to happen) or zero (for the occurence not happening).

In simple terms, when studying a binomial probability distribution we are looking for the number of successes in a certain amount of trials, and just as most of the rest discrete probability distributions, in order to be sure that our statistical experiment is based on a binomial distribution the experiment must have fixed trials to be followed, only two possible outcomes (success or failure) and be sampled with replacement which results on a constant probability of success in each trial.

Mean and standard deviation of binomial distribution

We have learnt before that the binomial distribution definition for the probability of obtaining x amount of successes in n amount of trials is called the general probability formula of a binomial distribution, and its mathematical expression goes as follows:

$P(x) = \; _nC_xP^{x}(1-p)^{n-x} \; = \;$ $\large \frac{n!}{x!(n-x)!}$ $p^{x}(1-p^{n-x})$

Where:
$n$ = number of trial
$x$ = number of successes in $n$ trials
$p$ = probability of success in each trial
$_nC_x =$ $\large \frac{n!}{x!(n-x)!}$ = number of success outcomes
$P(x)$ = probability of getting x successes out of $n$ trials

The mean of a binomial distribution is defined as the multiplication of the number of trials in the experiments, times the probability of success in each trial, and is written as:

$\mu = np$
The mean provides the best approximation of the amount of times you can expect a certain outcome to occur per a certain number of trials. In other words, the mean of a binomial distribution is the expectation of success: given that it multiplies the probability of obtaining a success with the trials that will be worked through, the mean of a binomial distribution can be defined as the number of expected successes per n trials.

With this in mind, we can identify when a problem is asking us to obtain the mean of the distribution, since it will usually ask for the amount of times a certain outcome (which is taken as the success outcome for the case) is expected to occur (look at example 1 on its first question).

We move on to the next important concepts from a probability distribution and these are the standard deviation definition and the variance of binomial distribution (which is just the standard deviation squared) defined as:

$\sigma = \sqrt{np (1-p)} =$ standard dev.
$\sigma^{2} = np(1-p) =$ variance

From equations 2 and 3 we can observe that once we know the number of trials an experiment will consist of, and the probability of success each trial has, the mean and standard deviation calculations can be done in a simple and rapid manner; with that in mind, let us go to the next section to solve some example problems.

Binomial distribution examples

Now let us work through a few examples for the calculation of the mean and standard deviation of a binomial distribution. Notice that we will be focusing on equations 2 and 3, and their computations rather than using the binomial distribution equation for probability (equation 1).
And so, our first example is dedicated to describe the calculations of the mean and the standard deviation of a binomial distribution as an introduction, example 2 explains how to deal with a problem which has a non-integer mean, and example 3 teaches how to interpret and draw conclusions from all of these calculations.

Example 1

If you roll a fair die 12 times,
Therefore: $n=\frac{1}{2}$ and $p=\frac{1}{6}$

When the problem asks you by the amount of successes you expect in an experiment, is actually asking about the average since it provides a good estimation of the successes in the statistical experiment. Therefore, we will calculate the mean of this binomial distribution in order to see how many times do you expect to roll the die and obtain a 6.

So following equation 2 for the mean of a binomial distribution, we calculate:

$\mu = np = 12(\frac{1}{6}) = 2$
Therefore, if you roll the die 12 times, you can expect to roll a 6 two times.

Using equation 3 to calculate the standard deviation of a binomial distribution, we obtain:

$\sigma = \sqrt{np(1-p)} = \sqrt{12(\frac{1}{6})(1-\frac{1}{6})} = \sqrt{2(\frac{5}{6})} = \sqrt{\frac{10}{6}} =$

Example 2

How many times would you expect to roll a 6, if you rolled the die 10 times?

On this problem, we have that: $n=10$ and $p=\frac{1}{6}$
Once more, we use equation 2 which is the binomial distribution formula for the mean:

$\mu = np = 10(\frac{1}{6}) = 5/3 =$ 1.666
Therefore, you can expect to roll a 6 from 1 to 2 times when rolling a fair die 10 times. Given that the mean is closer to the value of 2, than to the value of 1, it can be said that you have a higher chance to rolling a 6 twice, than just once, but still, the chances are not complete for it.

Example 3

On this example we will be interpreting the mean and standard deviation of binomial distribution.
It is said that 10% of the accidents occuring while rock climbing are due to rockfall. In Squamish there are 280 climbing accidents a year. With this in mind, answer the next three questions, noting that on this case we have: $n=280$ and $p=\frac{1}{10}=0.1.$

Calculating the mean using equation 2:

$\mu = np = 280(\frac{1}{10} = 28)$
We can expect that out of 280 climbing accidents that occur in a year, 28 of them will be due rockfall.

Now using the standard deviation formula from equation 3 we calculate

$\sigma = \sqrt{np(1-p)} = \sqrt{280(\frac{1}{10})(1-\frac{1}{10})} = \sqrt{28(\frac{9}{10})} = \sqrt{25.2} =$ 5.02
In order to answer this question we use the range rule of thumb we learnt in our lesson for the probability distribution. Remember that:

$Maximum \; usual\; value = \mu + 2\sigma$
$Minimum \; usual\; value = \mu - 2\sigma \$
Calculating those maximum and minimum usual values in order to obtain the scope of the range of usual outcomes for this case, we have that:

$Maximum \; usual\; value = \mu + 2\sigma = 28 + 2(5.02) = 38.04$
$Minimum \; usual\; value = \mu - 2\sigma = 28 2 2(5.02) = 17.96$
And so, the range of usual values goes from 17.96 to 38.04, which means that a value of 34 accidents in Squamish due to rockfall is still within the usual amount of accidents one should expect from the area.

***
As you can see, finding the standard deviation, mean and variance of a binomial distribution is a simple endeavour. So we finalize our lesson of today with a few recommendations on other places where you can continue to study our topic of today:
On this handout for the mean and variance of binomial random variables you will have the process used in order to derive the formulas for the mean and variance, and thus, the standard deviation equation, associated to the binomial distribution. Also, on the next link you will see examples on how to find the standard deviation of a binomial distribution without a table.

So we have finished our lesson for today, see you in the next one!