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Binomial Standard Deviation Formula

What is the binomial standard deviation formula for \(X \sim \text{Bin}(n,p)\), and how is it derived from the variance of independent Bernoulli trials?

Subject: Statistics Chapter: Discrete Random Variables and Their Probability Distributions Topic: Mean and Standard Deviation of Binomial Distribution Answer included
binomial standard deviation formula binomial variance formula standard deviation of binomial distribution mean of binomial distribution X~Bin(n p) Bernoulli trials np(1-p)
Accepted answer Answer included

Binomial setting and what “standard deviation” measures

The binomial standard deviation formula applies when a random variable \(X\) counts the number of successes in \(n\) independent trials, each with success probability \(p\). This is written \(X \sim \mathrm{Bin}(n,p)\). The standard deviation quantifies the typical spread of \(X\) around its mean.

Quantity Formula for \(X \sim \mathrm{Bin}(n,p)\) Interpretation
Mean \(\mu = \mathbb{E}[X] = np\) Expected number of successes
Variance \(\sigma^2 = \mathrm{Var}(X) = np(1-p)\) Spread measured in squared units
Standard deviation \(\sigma = \sqrt{np(1-p)}\) Spread measured in the same units as \(X\)

Step-by-step derivation of the binomial standard deviation formula

A binomial count can be expressed as a sum of Bernoulli indicator variables. Let \(X_i\) be the indicator of success on trial \(i\), so \(X_i=1\) for success and \(X_i=0\) for failure. Then

\[ X = X_1 + X_2 + \cdots + X_n. \]

Step 1: Mean and variance of a single Bernoulli trial.

For \(X_i \sim \mathrm{Bernoulli}(p)\),

\[ \mathbb{E}[X_i]=p, \qquad \mathrm{Var}(X_i)=p(1-p). \]

Step 2: Use independence to add variances.

Because the trials are independent, the covariance terms 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: Take the square root to obtain the standard deviation.

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

Notation note: \(1-p\) is often written \(q\), so the same result appears as \(\sigma=\sqrt{npq}\).

Worked example

Suppose \(n=10\) independent trials with \(p=0.50\) (success is equally likely as failure). Then the binomial mean and the binomial standard deviation formula give:

\[ \mu=np=10\cdot 0.50=5, \qquad \sigma=\sqrt{np(1-p)}=\sqrt{10\cdot 0.50\cdot 0.50}=\sqrt{2.5}\approx 1.581. \]

Interpretation: counts near \(5\) are most typical, and values roughly within about \(1.6\) of \(5\) are comparatively common (subject to discreteness and the distribution’s shape).

Visualization: binomial probabilities with mean and ±1 standard deviation

Binomial distribution for n equals 10 and p equals 0.5 with mean and standard deviation A bar chart of the binomial probability mass function for k from 0 to 10. A solid vertical line marks the mean at k=5, and dashed vertical lines mark mean minus one standard deviation and mean plus one standard deviation. 0 0.10 0.20 0.25 0 1 2 3 4 5 6 7 8 9 10 k (number of successes) P(X = k) mean \(\mu=5\) \(\mu-\sigma\approx 3.42\) \(\mu+\sigma\approx 6.58\) Binomial PMF for \(n=10\), \(p=0.50\) with \(\mu=np\) and \(\sigma=\sqrt{np(1-p)}\)
Bars show \(P(X=k)\) for \(X\sim\mathrm{Bin}(10,0.50)\). The solid line marks the mean \(\mu=5\); dashed lines mark \(\mu\pm\sigma\) with \(\sigma\approx 1.581\), illustrating how the binomial standard deviation formula locates the typical spread around the center.

Common checks and practical notes

The formula \(\sigma=\sqrt{np(1-p)}\) requires independent trials with constant success probability \(p\). If the binomial model is appropriate, \(\sigma\) increases with \(n\) but is largest near \(p=0.50\) and decreases as \(p\) approaches \(0\) or \(1\).

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