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Standard Deviation Formula for Binomial Distribution

What is the standard deviation formula for a binomial distribution \(X \sim \mathrm{Bin}(n,p)\), and why does it equal \(\sqrt{np(1-p)}\)?

Subject: Statistics Chapter: Discrete Random Variables and Their Probability Distributions Topic: Mean and Standard Deviation of Binomial Distribution Answer included
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Accepted answer Answer included

Binomial model and the meaning of standard deviation

The phrase standard deviation formula for binomial distribution refers to the spread of a binomial random variable \(X\), where \(X\) counts the number of successes in \(n\) independent trials, each trial having success probability \(p\). This is written \(X \sim \mathrm{Bin}(n,p)\).

\[ \mu=\mathbb{E}[X]=np, \qquad \sigma=\sqrt{\mathrm{Var}(X)}. \]

Main result (formula)

\[ \mathrm{Var}(X)=np(1-p) \qquad\Longrightarrow\qquad \sigma=\sqrt{np(1-p)}. \]

The expression \(1-p\) is often denoted by \(q\), so the same standard deviation formula is also written \(\sigma=\sqrt{npq}\).

Step-by-step derivation from Bernoulli trials

A binomial count is a sum of independent indicator variables. Define \(X_i\) as the outcome of trial \(i\): \(X_i=1\) for success and \(X_i=0\) for failure. Then

\[ X=\sum_{i=1}^{n} X_i. \]

Step 1: Variance of one Bernoulli trial.

Each \(X_i\) has \(P(X_i=1)=p\) and \(P(X_i=0)=1-p\). Compute \(\mathbb{E}[X_i]\) and \(\mathbb{E}[X_i^2]\):

\[ \mathbb{E}[X_i]=1\cdot p+0\cdot(1-p)=p, \qquad \mathbb{E}[X_i^2]=1^2\cdot p+0^2\cdot(1-p)=p. \]

Therefore

\[ \mathrm{Var}(X_i)=\mathbb{E}[X_i^2]-\big(\mathbb{E}[X_i]\big)^2 =p-p^2 =p(1-p). \]

Step 2: Add variances using independence.

Independence implies covariances are zero, so the variance of a sum is the sum of variances:

\[ \mathrm{Var}(X)=\mathrm{Var}\!\left(\sum_{i=1}^{n}X_i\right) =\sum_{i=1}^{n}\mathrm{Var}(X_i) =\sum_{i=1}^{n}p(1-p) =np(1-p). \]

Step 3: Convert variance to standard deviation.

\[ \sigma=\sqrt{\mathrm{Var}(X)}=\sqrt{np(1-p)}. \]

Worked example

Suppose a quality-control process produces a defective item with probability \(p=0.20\) per item, independently across items, and a sample of \(n=25\) items is inspected. Then \(X\sim\mathrm{Bin}(25,0.20)\).

Quantity Computation Value
Mean \(\mu=np=25\cdot 0.20\) \(\mu=5\)
Variance \(\sigma^2=np(1-p)=25\cdot 0.20\cdot 0.80\) \(\sigma^2=4\)
Standard deviation \(\sigma=\sqrt{np(1-p)}=\sqrt{4}\) \(\sigma=2\)

Interpretation: the expected number of defectives is \(5\), and the typical fluctuation around that center is about \(2\) defectives (subject to the discreteness of counts).

Visualization: how \(\sigma=\sqrt{np(1-p)}\) changes with \(p\) (fixed \(n\))

Binomial standard deviation versus p for n equals 25 A curve of sigma as a function of p from 0 to 1 for fixed n=25, peaking at p=0.5 and symmetric about 0.5. Markers highlight p=0.2 and p=0.5. 0 1 2 2.5 0 0.2 0.5 1 p σ p = 0.2 p = 0.5 (maximum) Binomial standard deviation vs p for fixed n = 25
For fixed \(n\), the standard deviation \(\sigma=\sqrt{np(1-p)}\) is \(0\) at \(p=0\) and \(p=1\), and it reaches its maximum at \(p=0.50\). The curve is symmetric about \(p=0.50\), reflecting that success and failure swap roles when \(p\) is replaced by \(1-p\).

Key takeaways

The standard deviation formula for binomial distribution is \(\sigma=\sqrt{np(1-p)}\). It follows from writing the binomial count as a sum of independent Bernoulli trials and using variance additivity. Variability is largest near \(p=0.50\) and smaller when successes are very rare or very common.

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