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Binomial distributions · Tutorial 352 of 1000

Mean of a Binomial Distribution

See why the mean of a binomial random variable is \(np\), and use it to predict and interpret the expected number of survey respondents with a specified response.

Intermediate 9 min read

What You'll Learn

  • Derive the binomial mean formula \(\mu_X=np\) using success indicators and the mean-of-a-sum rule.
  • Identify \(n\) and \(p\) in a survey context before calculating an expected count.
  • Check when a random sample without replacement can reasonably be modeled as binomial using the 10% condition.
  • Calculate expected survey counts and verify the multiplication in a second way.
  • Interpret a fractional expected count as a long-run average, not as a possible single survey result.
  • Compare expected counts when survey sample sizes and response rates differ.

From a Binomial Model to an Expected Count

In Writing Binomial Probabilities in Proper Notation, you represented a count of successes with \(X\sim B(n,p)\). The parameters tell us more than which probability command to use: they also determine the distribution’s mean. For a survey, this mean predicts the number of people expected to give a particular response under the model.

Suppose \(X\) counts respondents who say they support a proposed community project. If a model gives each respondent the same probability \(p\) of supporting it and the survey includes \(n\) respondents, then the expected number of supporters is \(np\). This is a center of the probability distribution, not a promise about the count in one particular survey.

Formula: If \(X\sim B(n,p)\), then the mean, or expected number of successes, is \(\mu_X=np\). Here \(n\) is the fixed number of trials and \(p\) is the probability of success on each trial.

The formula is useful because it avoids calculating every possible value of \(X\) and its probability. The earlier tutorial The Mean of a Discrete Random Variable defined a mean as a probability-weighted average. The binomial structure lets us simplify that calculation to one multiplication.

Why the Mean Is \(np\)

One way to see the result is to record each trial separately. For trial \(i\), define an indicator variable \(I_i\): it equals 1 if that trial is a success and 0 if it is a failure. The total number of successes is the sum of these indicators. Each indicator has mean \(p\), because it is 1 with probability \(p\) and 0 with probability \(1-p\).

As established in Mean of the Sum of Two Random Variables, the mean of a sum is the sum of the means. Applying that rule to all \(n\) indicators gives \(n\) terms, each equal to \(p\). Therefore, the mean count is \(np\). Independence is part of the binomial conditions, but the mean-of-a-sum rule itself does not require independence.

$$ \begin{aligned} X&=I_1+I_2+\cdots+I_n\\ \mu_X&=E(X)=E(I_1)+E(I_2)+\cdots+E(I_n)\\ &=p+p+\cdots+p=np. \end{aligned} $$

The same result follows from the binomial probability formula. In the probability-weighted mean, each possible count \(k\) is multiplied by \(P(X=k)\). Using the identity \(k\binom{n}{k}=n\binom{n-1}{k-1}\), the sum simplifies to \(np\). The indicator explanation is often more intuitive: each of the \(n\) trials contributes an expected \(p\) successes.

$$ \begin{aligned} \mu_X &=\sum_{k=0}^{n} k\binom{n}{k}p^k(1-p)^{n-k}\\ &=np\sum_{k=1}^{n}\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}\\ &=np\bigl(p+(1-p)\bigr)^{n-1}\\ &=np. \end{aligned} $$

The number \(np\) can be a whole number or a decimal. Since an actual count must be a whole number, a decimal mean is not a possible result of one survey. It is still a meaningful center: over many repetitions of the same modeled survey process, the average count would approach that value.

Check the Survey Model Before Using the Formula

A survey count can be modeled as binomial when the BINS conditions from the earlier tutorials are reasonable: each trial has two outcomes, trials are independent, the number of trials is fixed, and the success probability stays the same. For a random sample drawn without replacement from a population, responses are not literally independent. The 10% condition provides a common justification for treating them as approximately independent when the sample is no more than 10% of the population.

Conditions: Before applying \(\mu_X=np\) to a survey, define one trial and what counts as success; identify the fixed sample size \(n\) and per-person probability \(p\); and check the BINS conditions. For sampling without replacement, check the 10% condition: \(n\leq 0.10N\), where \(N\) is the population size. This supports an approximate binomial model; it does not guarantee that the survey is free of response or sampling bias.

In a survey, success might mean “answers yes,” “uses the service,” or “supports the proposal.” The value of \(p\) must describe that response for one person in the modeled population. If the probability changes from person to person, or the selection process makes responses strongly dependent, a single binomial model may not be appropriate. The expected count formula depends on a suitable model, so it is important to explain why the model is reasonable rather than simply multiplying two numbers.

Worked Example: Expected Support in a Community Survey

Worked Example: Expected Support in a Community Survey

A town plans to randomly select 300 residents from a population of 15,000. A planning model estimates that each resident has probability 0.28 of supporting a new community garden. Find and interpret the expected number of supporters in the sample.

State. Let \(X\) be the number of selected residents who support the community garden. The fixed sample size is \(n=300\), and the modeled probability of support for one resident is \(p=0.28\). Thus, the proposed model is \(X\sim B(300,0.28)\).

Plan. Each selected resident either supports the garden or does not, giving two outcomes for the count. The sample size is fixed at 300. The model assumes the same support probability, 0.28, for each resident. The selection is random and is without replacement, so check the 10% condition: 10% of 15,000 is \(0.10(15{,}000)=1{,}500\), and \(300\leq1{,}500\). The sample is no more than 10% of the population, so treating the selections as approximately independent is reasonable under this model.

Do. The mean of a binomial count is \(\mu_X=np\). Substitute the sample size and support probability:

$$ \begin{aligned} \mu_X&=np\\ &=300(0.28)\\ &=84. \end{aligned} $$

As a check, \(0.28=28/100\), so \(300(28/100)=8{,}400/100=84\). The multiplication gives the same expected count.

Conclude. Under the binomial model, the expected number of residents who support the community garden is 84 out of the 300 selected. This is a model-based average across many comparable surveys; it does not mean that exactly 84 people must support it in this particular sample.

Worked Example: A Decimal Expected Count

Worked Example: A Decimal Expected Count

A library randomly surveys 120 members from a membership population of 5,000. A model estimates that a member has probability 0.07 of using the library’s new tool-lending service. Find and interpret the expected number of surveyed members who use the service.

State. Let \(X\) be the number of surveyed members who use the tool-lending service. Then \(n=120\), \(p=0.07\), and the proposed model is \(X\sim B(120,0.07)\).

Plan. For this count, a member either uses the service or does not. The survey fixes the number of members at 120, and the model assigns the same probability, 0.07, to each member. Because members are sampled without replacement, check the 10% condition: \(0.10(5{,}000)=500\), and \(120\leq500\). The sample is within 10% of the population, which supports using approximate independence for the binomial model.

Do. Apply \(\mu_X=np\):

$$ \begin{aligned} \mu_X&=np\\ &=120(0.07)\\ &=8.4. \end{aligned} $$

To check, \(0.07=7/100\), so \(120(7/100)=840/100=8.4\). This confirms the product. The value 8.4 is not a possible count in one survey, because a count must be a whole number.

Conclude. According to the model, the expected number of surveyed members who use the tool-lending service is 8.4. In repeated comparable surveys, the average number of users would be about 8.4 per sample of 120; an individual survey will have a whole-number count, such as 8 or 9, among other possibilities.

Worked Example: Comparing Expected Counts

Worked Example: Comparing Expected Counts

Two organizations plan separate surveys about a proposed recycling service. Survey A will select 400 people from a population of 50,000, and its model uses a 0.16 probability that a selected person supports the service. Survey B will select 250 people from a population of 20,000, and its model uses a 0.24 support probability. Find each expected number of supporters and compare them.

State. Let \(X_A\) count supporters in Survey A and \(X_B\) count supporters in Survey B. The models are \(X_A\sim B(400,0.16)\) and \(X_B\sim B(250,0.24)\).

Plan. For each survey, the response is support or does not support, the sample size is fixed, and the given support probability is constant within that survey’s model. For Survey A, 10% of 50,000 is \(5{,}000\), and \(400\leq5{,}000\). For Survey B, 10% of 20,000 is \(2{,}000\), and \(250\leq2{,}000\). Thus, each sample is no more than 10% of its population, supporting approximate independence under the stated random-sampling models.

Do. Calculate the mean separately for each binomial count:

$$ \begin{aligned} \mu_{X_A}&=400(0.16)=64,\\ \mu_{X_B}&=250(0.24)=60. \end{aligned} $$

Check Survey A using \(0.16=16/100\): \(400(16/100)=6{,}400/100=64\). Check Survey B using \(0.24=24/100\): \(250(24/100)=6{,}000/100=60\). Both products are confirmed.

Conclude. Survey A has an expected 64 supporters, while Survey B has an expected 60 supporters. Although Survey B’s modeled support probability is higher, Survey A’s larger sample size gives it the slightly larger expected count. These are expected counts, not guarantees about the survey results.

Common Mistakes and AP Exam Tips

The formula is short, but full-credit work still needs to connect the parameters to the situation and interpret the result correctly. Check that the count, sample size, and probability all describe the same response and survey.

  • Using the wrong probability. \(p\) is the probability of the outcome defined as success on one trial. State exactly what counts as a success before substituting it.
  • Multiplying by the population size. For the expected number in the surveyed sample, use the number of trials \(n\), not the total population \(N\). The population size matters when checking the 10% condition.
  • Treating the mean as a guaranteed result. A mean of 84 does not ensure 84 successes in one survey. Say that 84 is the expected count under the model or the long-run average for comparable surveys.
  • Rejecting a decimal mean. A mean such as 8.4 is valid even though a single survey cannot contain a fraction of a person. The mean summarizes the distribution across repetitions.
  • Skipping survey conditions. A random sample without replacement does not produce literally independent selections. Check the 10% condition, and describe independence as approximate when that condition supports it.
  • Confusing an expected count with representativeness. The formula describes the count under the probability model. It does not correct nonresponse, biased question wording, or a nonrandom selection process.

A strong response defines \(X\), states the binomial parameters, checks the conditions that support the model, shows \(\mu_X=np\), and interprets the result in context. Keep the full product until the final value; here, the products are simple enough to verify directly with equivalent fraction calculations.

AP Exam Tip: Report the answer as an expected number of successes in the stated sample. If the mean is fractional, explain that it is a long-run average, not a possible count or a prediction guaranteed for one survey.

Key Takeaway

The binomial mean follows from adding the expected contribution of each trial: each trial contributes \(p\) expected successes, and there are \(n\) trials. In a survey, first justify the binomial model, then multiply the sample size by the per-person success probability and interpret the result as an average count.

Key takeaway: For \(X\sim B(n,p)\), \(\mu_X=np\). In a survey, \(np\) is the expected number of people in the sample who give the response defined as success; it is not a guaranteed count in one survey.

Check Your Understanding

For each question, identify the count and connect the calculation to its survey context.

  1. A random survey includes 180 people, and the model assigns each person a 0.35 probability of supporting a proposal. What is the expected number of supporters?
  2. A sample of 90 is drawn without replacement from a population of 700. Does the 10% condition hold? Show the comparison.
  3. A binomial survey model has \(n=75\) and \(p=0.12\). Find the mean and explain why its value does not need to be a whole number.
  4. Explain why the mean of a binomial count is \(np\) using the expected contribution from each trial.
  5. One survey has \(n=200\), \(p=0.20\); another has \(n=150\), \(p=0.30\). Find both expected counts and compare them in context.

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