Mean and Variance of Binomial Distribution is a topic in probability theory and statistics. Mean and Variance is properties of Binomial Distribution. In this article, we will study Mean and Variance of Binomial Distribution, how to find Mean and Variance of Binomial Distribution, formula, derivation with proof, solved examples, mean & variance of negative binomial distribution and FAQs

What is Mean and Variance of Binomial Distribution?

The binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes-no question, and each with its own Boolean-valued outcome: success or failure. Mean and Variance is the properties of Binomial Distribution. The concept of mean and variance is also seen in standard deviation. Binomial Distribution is a topic of statistics. Mean deviation is also a useful topic of probability.

Maths Notes Free PDFs

Mean

Mean is the expected value of Binomial Distribution. It is calculated by multiplying the number of trials (n) by the probability of successes (p), or n x p.

Variance

Variance is a measure of dispersion that takes into account the spread of all data points in a data set. The variance is the mean squared difference between each data point and the centre of the distribution measured by the mean.

The mean of the distribution\( (μ_x)\) is equal to np.

The variance \((σ^2_x)\) is \(n\times{p}\times( 1 – p )\). The standard deviation \((σ_x)\) is \(\sqrt{n\times{p}\times( 1 – p )}\)

When p > 0.5, the distribution is skewed to the left. When p < 0.5, the distribution is skewed to the right

What is the mean and variance of a Bernoulli binomial distribution?

Bernoulli distribution is a discrete probability distribution for a Bernoulli trial. It is a random experiment that has only two outcomes, usually either a Success or a Failure.

Example: The probability of getting a head i.e a success while flipping a coin is 0.5. The probability of “failure” is 1 – P (1 minus the probability of success, which also equals 0.5 for a coin toss). It is a special case of the binomial distribution for n = 1. In other words, it is a binomial distribution with a single trial (e.g. a single coin toss). The expected value for a random variable, X, for a Bernoulli distribution is E[X] = p.

For example, if p = 0.4, then E[X] = 0.4.

The variance of a Bernoulli random variable is:

Var[X] = p(1 – p).

Derivation of Mean and Variance of Binomial Distribution

Now let’s see the derivation of how the formulae of the Mean and Variance are derived.

Mean and Variance of Binomial Distribution Proof

\(\begin{equation}

\text{If } P(x)= \binom{n}{x} p^x(1-p)^{n-x} \text{then}\\

\mathop{\mathbb{E}[X] = \sum_{x=0}^{n} x \cdot \binom{n}{x} p^x(1-p)^{n-x}}\\

=\sum_{x=0}^{n}{n!\over{(n-x)!x!}}p^x(1-p)^{n-x}\\

=\sum_{x=1}^{n}{n!\over{(n-x)!(x-1)!}}p^x(1-p)^{n-x}\\

\mathop{\mathbb{E}[X] = \sum_{x=1}^{n}{n(n-1)!\over{(n-x)!(x-1)!}}p.p^{x-1}(1-p)^{n-x}}\\

=np\sum_{x=1}^{n}{(n-1)!\over{(n-x)!(x-1)!}}.p^{x-1}(1-p)^{n-x}\\

=np\sum_{x=1}^{n}{(n-1)!\over{[(n-1)-(x-1)]!(x-1)!}}.p^{x-1}(1-p)^{n-x}\\

=np\sum_{x=1}^{n} x \cdot \binom{n-1}{x-1}p^{x-1}(1-p)^{n-x}\\

\text{We put 1 – p = q}\\

=np\sum_{x=1}^{n} x \cdot \binom{n-1}{x-1}p^{x-1}q^{n-x}\\

=np[^{n-1}C_0q^{n-1}+^{n-1}C_1pq^{n-2}+^{n-1}C_2p^2q^{n-3}+ … + ^{n-1}C_{n-1}p^{n-1}]\\

[^{n-1}C_0q^{n-1}+^{n-1}C_1pq^{n-2}+^{n-1}C_2p^2q^{n-3}+ … + ^{n-1}C_{n-1}p^{n-1}] \\ =\text{Binomial Expansion of} (p+q)^{n-1}\\

\mathop{\mathbb{E}[X]} = np(p+q)^{n-1}\\

\text{But we know that p + q = 1}\\

\mathop{\mathbb{E}[X]} = np(1)^{n-1} = np\\

\text{This the mean of the binomial distribution.}\\

\text{Now,} Var(X) = \mathop{\mathbb{E}[X^2] – [{\mathop{\mathbb{E}[X]}]}}^2\\

\mathop{\mathbb{E}[X^2]} = \sum_{x=0}^{n} x^2 \cdot \binom{n}{x} p^xq^{n-x}\\

= {\sum_{x=0}^{n} [x(x-1)+x] \cdot \binom{n}{x} p^xq^{n-x}} + \sum_{x=0}^{n} x \cdot \binom{n}{x} p^xq^{n-x}\\

= \sum_{x=2}^{n} {\frac{x(x-1)n!}{(n-x)!x(x-1)(x-2)!}} p^x q^{n-x} + np\\

= n(n-1)p^2 \sum_{x=2}^{n} {\frac{(n-2)!}{(n-x)!(x-2)!}} p^{x-2}q^{n-x}\\

= n(n-1)p^2 \sum_{x=2}^{n} {\frac{(n-2)!}{[(n-2)-(x-2)]!(x-2)!}} p^{x-2}q^{n-x}\\

= n(n-1)p^2\sum_{x=2}^{n} \binom{n-2}{x–2}p^{x-2}q^{n-x}\\

= n(n-1)p^2 [^{n-2}C_0q^{n-2}+^{n-2}C_1pq^{n-3}+^{n-2}C_2p^2q^{n-4}+ … + ^{n-2}C_{n-2}p^{n-2}] + np\\

= n(n-1)p^2[(p+q)^{n-2}]+np\\

\text{Since p + q =1, we have} \\

\mathop{\mathbb{E}[X^2]} = n(n-1)p^2+np\\

\text{Using this,} \\

Var(X) = n(n-1)p^2+np -(np)^2\\

= n^2p^2 – np^2 + np – n^2p^2\\

= np(1-p)\\

= npq\\

\text{Hence the variance of the binomial distribution is npq.}\\

\end{equation}\)

Binomial Distribution Formula

The binomial distribution helps us find the probability of getting a certain number of successes in a fixed number of experiments (or trials), where each experiment has only two outcomes: success or failure.

The formula to find this probability is:

P(x; n, p) = nCx × p^x × (1 - p)^(n - x)

Or you can also write it as:

P(x; n, p) = nCx × p^x × q^(n - x)

Where:

• n = total number of trials
• x = number of successes you want
• p = chance of success in one trial
• q = chance of failure = 1 - p
• nCx = number of combinations of n things taken x at a time
  (This is calculated using: nCx = n! / [x!(n - x)!])

Mean, Variance, and Standard Deviation of Binomial Distribution

To describe the binomial distribution, we can use three important values:

• Mean (μ): This tells us the average number of successes.
  μ = n × p
   
• Variance (σ²): This shows how much the values are spread out.
  σ² = n × p × q
   
• Standard Deviation (σ): This is the square root of the variance. It also tells us about the spread.
  σ = √(n × p × q)

Binomial Distribution vs Normal Distribution

Solved Example to find Mean and Variance of Binomial Distribution

1. What is the mean of a binomial random variable with n = 18 and p = 0.4?

Ans: The mean of a binomial random variable X is represented by the symbol \(\mu\)

\(\mu\)
\(\mu\) = np
Here, n = 18 and p = 0.4, so
\(\mu\) = 18 x 0.4 = 7.2

1. What is the standard deviation of a binomial distribution with n = 18 and p = 0.4? Round your answer to two decimal places.

Ans: The standard deviation of X is represented by \(\sigma\) and represents the square root of the variance. If X has a binomial distribution, the formula for the standard deviation is
\(\begin{matrix}
\sigma=\sqrt{npq}
\sigma=\sqrt{18\times0.4(1-0.4)}
\end{matrix}\)

Word Problems to find Mean and Variance of Binomial Distribution

Q. An agent sells life insurance policies to five equally aged, healthy people. According to recent data, the probability of a person living in these conditions for 30 years or more is 2/3. Calculate the probability that after 30 years:

Ans: Case 1: If all 5 people are living
\(\begin{matrix}
B(5, \frac{2}{3}) p = \frac{2}{3} 1 – p = \frac{1}{3}\\
p(X = 5) = \binom{5}{5} (\frac{2}{3})^5 = 0.132
\end{matrix}\)

Case 2: At least three people are still living.

\(\begin{matrix}
p(X \geq 3) = p (X = 3) + p (X = 4) + p (X = 5)\\
= \binom {5}{3} (\frac{2}{3})^3 (\frac{1}{3})^2 + \binom {5}{4} (\frac{2}{3})^4 (\frac{1}{3}) + \binom{5}{5} (\frac{2}{3})^5 = 0.791
\end{matrix}\)

Case 3: Exactly two people are still living.

\(p(X = 2) = \binom{5}{2} (\frac{2}{3})^2 (\frac{1}{3})^3 = 0.164\)

Q. If from six to seven in the evening one telephone line in every five is engaged in a conversation: what is the probability that when 10 telephone numbers are chosen at random, only two are in use?

Ans. : B(10, 1/5)p = 1/51 − p = 4/5
\(\begin{matrix}
B(10, \frac{1}{5}) p = \frac{1}{5} 1 – p = \frac{4}{5}\\
p(X = 2) = \binom {10}{2} (\frac{1}{5})^2 \cdot (\frac{4}{5})^8 = 0.3020
\end{matrix}\)

Q. A pharmaceutical lab states that a drug causes negative side effects in 3 of every 100 patients. To confirm this affirmation, another laboratory chooses 5 people at random who have consumed the drug. What is the probability of the following events?

Ans: Case: 1 None of the five patients experience side effects.

B(100, 0.03) p = 0.03 q = 0.97
p(X = 0) = \binom {5}{0} 0.97^5 = 0.8687[/latex]᠎᠎

Case 2: At least two experience side effects.

\(\begin{matrix}
p(X \geq 2) = 1 – p (X <2) = 1 – [p (X = 0) + p (X = 1)]\\
= 1 – [\binom {5}{0} 0.97^5 + \binom {5}{1} 0.03 \cdot 0.97^4] = 0.00847
\end{matrix}\)

What is the average number of patients that the laboratory should expect to experience side effects if they choose 100 patients at random?
\(\mu = 100 \cdot 0.03 = 3\)

Mean and Variance of Negative Binomial Distribution

The negative binomial distribution is a discrete probability distribution that models the number of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs. For example, we can define rolling a 6 on a dice as a failure, and rolling any other number as a success, and ask how many successful rolls will occur before we see the third failure (r = 3). In such a case, the probability distribution of the number of non-6s that appears will be a negative binomial distribution. We could similarly use the negative binomial distribution to model the number of days a certain machine works before it breaks down (r=1).
In a sequence of independent Bernoulli(p) trials, let the random variable X denote the trial at which the rth success occurs, where r is a fixed integer. Then

P(X = x|r, p) = \binom {x − 1}{r − 1}p^r(1 − p)^{x−r}, x = r, r + 1,…
\end{matrix}\)

and we say that X has a negative binomial(r, p) distribution.
Mean of Negative Binomial Distribution is given by,
\(= r({1 − p\over{p}})\)
Variance of Negative Binomial Distribution is given by,
\(VarY = {r(1 − p)\over{p^2}}\)

Special Case: The Mean and Variance of Binomial Distribution are same if

If the mean and the variance of the binomial distribution are same,

\(\begin{matrix}

\mu = Var(X)\\

np = \sqrt{npq}\\

\text{Squaring both the sides,}\\

n^2p^2 = npq\\

\therefore,np = q\\

np = (1-p)\\

np + p = 1\\

(n + 1)p = 1\\

p = {1\over{n+1}}\\

\end{matrix}\)

Properties of Mean and Variance of Binomial Distribution

The properties of mean and variance of binomial distribution

• Since p and q are numerically less than or equal to 1, npq < np
• The variance of a binomial variable is always less than its mean.
• The variance of binomial variable X attains its maximum value at p = q = 0.5 and this maximum value is n/4.
• The variance of a distribution measures how “spread out” the data is. Related is the standard deviation, the square root of the variance, useful due to being in the same units as the data.
• The mean, or expected value, of a distribution, gives useful information about what average one would expect from a large number of repeated trials.

Applications of Mean and Variance of Binomial Distribution

• The manufacturing company uses binomial distribution to detect defective goods or items.
• In a clinical trial binomial trial is used to detect the effectiveness of the drug.
• Moreover, the binomial trial is used in various fields such as market research.
• In a manufacturing context, the number of faulty items in a batch of products might follow a binomial distribution, if the probability of failures is constant.

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