Standard Deviation of a Binomial Distribution

From Expected Count to Typical Variation

In Mean of a Binomial Distribution, you learned that if $X\sim B(n,p)$, then the expected number of successes is $\mu_X=np$. The mean describes the distribution’s center. This tutorial adds a measure of spread: the standard deviation describes how far the count of successes typically varies from that center.

Consider a manufacturing batch in which each item either meets a specified quality standard or does not. If the number of items is fixed and the probability of meeting the standard is the same for every item, $X$ can count the items that meet the standard. The binomial standard deviation helps describe how much that count tends to vary from batch to batch under the model.

Formula: If $X\sim B(n,p)$, then its variance is $\sigma_X^2=np(1-p)$, and its standard deviation is $$ \sigma_X=\sqrt{np(1-p)}. $$ Here, $n$ is the fixed number of trials and $p$ is the probability of success on one trial.

The quantity $p(1-p)$ is the variance of the success-or-failure outcome on one trial. Multiplying by $n$ gives the variance of the total count when the trials are independent; taking the square root gives the standard deviation. The result is in the original units of $X$, which for a manufacturing count means items.

Why the Formula Uses $np(1-p)$

For each item $i$, define an indicator variable $I_i$ that is 1 if the item is a success and 0 if it is a failure. The total success count is $X=I_1+I_2+\cdots+I_n$. Each indicator has mean $p$ and variance $p(1-p)$: it differs from its mean according to the two possible outcomes, success and failure.

The earlier tutorial Standard Deviation of the Sum of Independent Variables established that variances add for independent random variables. In a binomial setting, the indicators are independent and each has variance $p(1-p)$. Adding the $n$ equal variances gives $np(1-p)$; taking its nonnegative square root gives the binomial standard deviation.

\begin{aligned} X&=I_1+I_2+\cdots+I_n\\ \sigma_X^2&=\sigma_{I_1}^2+\sigma_{I_2}^2+\cdots+\sigma_{I_n}^2\\ &=p(1-p)+p(1-p)+\cdots+p(1-p)\\ &=np(1-p)\\ \sigma_X&=\sqrt{np(1-p)}. \end{aligned}

The expression $np(1-p)$ is the variance, not the standard deviation. Variance is measured in squared units, such as items squared. The square root returns to the original units, items. The earlier tutorials on variance and standard deviation of a random variable developed this distinction for general distributions; this formula makes the calculation shorter for a binomial count.

Interpretation: For a binomial count $X$, $\sigma_X$ describes the typical distance of the number of successes in a batch from its mean, $np$. It is not a maximum distance, a guaranteed range, or a prediction of the exact count in one batch.

Check the Manufacturing Model

The formula applies when the count is reasonably modeled as binomial. As in the earlier tutorials Recognizing a Binomial Setting and The BINS Checklist for Binomial Conditions, identify the trial, success, and count, then check the four features: Binary outcomes, Independent trials, a fixed Number of trials, and the Same probability of success on each trial.

In a manufacturing example, the trial is usually the inspection of one item, and success must be clearly defined—for example, “meets the quality standard.” If the model assumes independent production outcomes with a constant success probability, those assumptions support the binomial calculation. If items are sampled without replacement from a finite batch or inventory, consider whether dependence matters; when sampling from a larger population, the 10% condition can support treating the selections as approximately independent.

Conditions: Before using $\sigma_X=\sqrt{np(1-p)}$, define what one trial is and what counts as success. Check binary outcomes, independence, a fixed number $n$ of trials, and the same success probability $p$ on every trial. If sampling without replacement from a population of size $N$, check the 10% condition, $n\leq0.10N$, when using it to justify approximate independence.

Worked Example: Variation in a Batch of Components

A manufacturer models each component in a batch as having probability 0.02 of meeting a particular performance standard. A batch contains 200 components. Find and interpret the standard deviation of the number that meet the standard.

State. Let $X$ be the number of components in the batch that meet the performance standard. Under the stated model, $X\sim B(200,0.02)$.

Plan. The trial is inspecting one component, and success means the component meets the standard. Each component either meets the standard or does not, so outcomes are binary. The batch size is fixed at $n=200$. The model assigns the same success probability, $p=0.02$, to each component and treats outcomes as independent. Thus, the binomial conditions are satisfied by the assumptions in the model.

Do. Use the binomial standard deviation formula, retaining the failure probability $1-p=0.98$:

\begin{aligned} \sigma_X&=\sqrt{np(1-p)}\\ &=\sqrt{200(0.02)(0.98)}\\ &=\sqrt{3.92}\\ &\approx1.9799. \end{aligned}

Check the variance calculation: $200(0.02)=4$, and $4(0.98)=3.92$. As a reverse check of the rounded standard deviation, $1.9799^2=3.92000401$, which rounds to 3.9200; the unrounded square root squared is 3.92.

Conclude. Under this model, the number of components that meet the standard typically differs from its mean by about 1.98 components from batch to batch. The mean is $np=200(0.02)=4$, so the standard deviation describes typical variation around 4 meeting components. It does not guarantee that every batch will contain between about 2 and 6 such components.

Worked Example: A Larger Batch

A different production process models the probability that an item meets its specified standard as 0.10. A batch contains 400 items. Find the standard deviation of the number that meet the standard, and interpret it.

State. Let $Y$ be the number of items in the batch that meet the standard. The model is $Y\sim B(400,0.10)$.

Plan. One trial is checking one item, and success is meeting the standard. There are two possible outcomes for each item, the number of items is fixed at 400, and the model assumes independent outcomes with the same success probability of 0.10. These assumptions make a binomial model reasonable for the calculation.

Do. Apply the formula using $1-p=0.90$:

\begin{aligned} \sigma_Y&=\sqrt{np(1-p)}\\ &=\sqrt{400(0.10)(0.90)}\\ &=\sqrt{36}\\ &=6. \end{aligned}

Check the variance: $400(0.10)=40$, and $40(0.90)=36$. The reverse check is $6^2=36$, confirming that the standard deviation is 6 items.

Conclude. The number of items meeting the standard typically varies by about 6 items from the expected count. That expected count is $np=400(0.10)=40$, so the standard deviation describes spread around 40—not a guarantee that a particular batch will contain exactly 40, or a fixed range of 34 to 46.

Worked Example: Comparing Batch Sizes

Two hypothetical production runs use the same modeled probability, 0.04, that an item is defective. Run A contains 100 items, while Run B contains 400 items. Let $X_A$ and $X_B$ count defective items in the respective runs. Find and compare the standard deviations.

State. Under the stated models, $X_A\sim B(100,0.04)$ and $X_B\sim B(400,0.04)$.

Plan. For both runs, a trial is inspecting one item and success is a defect. Each item has two outcomes, the number of items is fixed within each run, and the model assumes independent outcomes with the same defect probability, 0.04. We can apply the formula separately to the two counts.

Do. For Run A, $1-p=0.96$:

\begin{aligned} \sigma_{X_A}&=\sqrt{100(0.04)(0.96)}\\ &=\sqrt{3.84}\\ &\approx1.9596. \end{aligned}

Check: $100(0.04)=4$, and $4(0.96)=3.84$. The rounded value squared is $1.9596^2=3.84003216$, which rounds to 3.8400.

For Run B, use the same per-item defect probability and the larger batch size:

\begin{aligned} \sigma_{X_B}&=\sqrt{400(0.04)(0.96)}\\ &=\sqrt{15.36}\\ &\approx3.9192. \end{aligned}

Check: $400(0.04)=16$, and $16(0.96)=15.36$. The rounded standard deviation, squared, is $3.9192^2=15.36012864$, which rounds to 15.360 to three decimal places, matching the variance to that precision.

Conclude. Run A’s defective count typically varies by about 1.96 items from its mean, while Run B’s typically varies by about 3.92 items. Run B has the larger standard deviation in the number of defects because it contains more items. The expected counts also differ: $100(0.04)=4$ defects in Run A and $400(0.04)=16$ in Run B. The comparison concerns counts, not defect rates.

What Changes the Standard Deviation?

The formula shows that both $n$ and $p$ matter. If $p$ stays fixed and the number of trials is multiplied by a factor, the variance is multiplied by that factor, but the standard deviation is multiplied by its square root. For instance, multiplying the batch size by four multiplies the standard deviation by two, as the comparison above illustrates.

For a fixed $n$, the factor $p(1-p)$ is largest when $p=0.50$ and gets smaller as $p$ moves closer to 0 or 1. Thus, under otherwise identical binomial models, probabilities near 0.50 give more variability in the success count than probabilities near the ends of the interval. This is a feature of the formula; it does not change the need to interpret $p$ as the probability of the particular outcome defined as success.

Do not compare standard deviations without considering what is being counted. A standard deviation of 6 describes typical variation of about 6 items, not a variation of 6 percentage points. If the batches have different sizes, the count standard deviations alone do not describe variability in the same kind of quantity as a proportion or rate.

Common Mistakes and AP Exam Tips

Forgetting the failure probability. The formula uses $p(1-p)$, not just $p$. State the success probability and its complement before substituting.
Reporting the variance as the standard deviation. First calculate $np(1-p)$, then take its square root. Keep the variance and standard deviation distinct.
Using the wrong units. The variance is in squared units; the standard deviation is in the original count units, such as items.
Calling the standard deviation a range. It describes typical distance from the mean, not a maximum, minimum, or guaranteed interval. Do not claim that every batch must fall within one standard deviation of the mean.
Leaving the count undefined. Identify exactly what $X$ counts and what success means. “Defects” and “items meeting the standard” have different success definitions and probabilities.
Skipping model assumptions. A short formula does not remove the need to justify the binomial model. Name the binary outcomes, fixed trial count, independence assumption, and constant success probability.

For full-credit communication, define the random variable in context, identify $n$ and $p$, check the binomial conditions, show the substitution into $\sqrt{np(1-p)}$, and interpret the result as typical variation in the count. Include the unit “items” when describing the standard deviation.

AP Exam Tip: A complete interpretation connects the standard deviation to the mean and the setting: “The number of [successes] in a batch typically differs from its expected count of $np$ by about $\sigma_X$ items.” Avoid presenting that typical distance as a guaranteed range for one batch.

Key Takeaway

The binomial standard deviation converts the model’s trial count and success probability into a measure of typical spread for its success count. Check that the manufacturing process is reasonably modeled as binomial, calculate the variance $np(1-p)$, and take its square root before interpreting the result in items.

Key takeaway: For $X\sim B(n,p)$, $\sigma_X=\sqrt{np(1-p)}$. It describes the typical distance of the number of successes from the mean $np$, in the original count units.

Check Your Understanding

For each question, show the formula and connect the result to the count being modeled.

A batch has 250 items, each with modeled probability 0.08 of meeting a standard. Find the standard deviation of the number that meet the standard and interpret it.
A production model has $n=100$ and $p=0.20$. Find the variance and standard deviation, and state the units for each.
Explain why the formula for the binomial variance includes a factor of $n$.
Two production runs have the same defect probability, but one run has four times as many items as the other. How do their standard deviations compare? Explain using the formula.
What does it mean in context if a batch’s modeled count of conforming items has standard deviation 5? State one interpretation the number does not justify.

Standard Deviation of a Binomial Distribution

What You'll Learn

From Expected Count to Typical Variation

Why the Formula Uses \(np(1-p)\)

Check the Manufacturing Model

Worked Example: Variation in a Batch of Components

Worked Example: Variation in a Batch of Components

Worked Example: A Larger Batch

Worked Example: A Larger Batch

Worked Example: Comparing Batch Sizes

Worked Example: Comparing Batch Sizes

What Changes the Standard Deviation?

Common Mistakes and AP Exam Tips

Key Takeaway

Check Your Understanding

Standard Deviation of a Binomial Distribution

What You'll Learn

From Expected Count to Typical Variation

Why the Formula Uses \(np(1-p)\)

Check the Manufacturing Model

Worked Example: Variation in a Batch of Components

Worked Example: Variation in a Batch of Components

Worked Example: A Larger Batch

Worked Example: A Larger Batch

Worked Example: Comparing Batch Sizes

Worked Example: Comparing Batch Sizes

What Changes the Standard Deviation?

Common Mistakes and AP Exam Tips

Key Takeaway

Check Your Understanding

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