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5.3: Mean, Variance, and Standard Deviation of the Binomial Distribution

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    46135

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    Learning Objectives
    • Calculate the mean of a binomial distribution by multiplying the number of trials by the probability of success.
    • Determine the variance to measure the spread of outcomes, considering both success and failure probabilities.
    • Compute the standard deviation as the square root of the variance to represent the typical distance from the mean.

    In probability theory and statistics, the binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success. When analyzing this distribution, three important measures help summarize its characteristics: the mean, the variance, and the standard deviation.

    The mean, often referred to as the expected value, represents the average number of successes that can be expected over many repetitions of the binomial experiment.

    Formulas for Mean, Variance, and Standard Deviation

    Definition: Formulas

    For a Binomial distribution, the following holds.

    \(\mu\) is the expected number of successes and is computed as \(\mu=n p \).

    \(\sigma^{2}\) is the variance of the number of successes and is computed as \( \sigma^{2}=n p q\).

    \(\sigma\) is the standard deviation for the number of successes and is computed as \(\sigma=\sqrt{n p q}\).

    Where p is the probability of success and q = 1 - p.

    Example \(\PageIndex{1}\) Finding the Probability Distribution, Mean, Variance, and Standard Deviation of a Binomial Distribution

    When looking at a person’s eye color, it turns out that 1% of people in the world have green eyes ("What percentage of," 2013). Consider a group of 20 people.

    1. Find the mean.
    2. Find the variance.
    3. Find the standard deviation.
    Solution
    1. Since this is a binomial, then you can use the formula \(\mu=n p\). So \(\mu=20(0.01)=0.2\) people. You expect on average that out of 20 people, less than 1 would have green eyes.
    2. Since this is a binomial, then you can use the formula \(\sigma^{2}=n p q\).

    \(q=1-0.01=0.99\)

    \(\sigma^{2}=20(0.01)(0.99)=0.198 \)

    1. Once you have the variance, you just take the square root of the variance to find the standard deviation.

    \(\sigma=\sqrt{0.198} \approx 0.445\)

    Example \(\PageIndex{2}\)

    A calculator manufacturer produces scientific calculators with a defect probability of 0.02. Assume that a quality control department for the manufacturer gathers a batch of 800 such calculators.

    1. Find the mean.
    2. Find the variance.
    3. Find the standard deviation.
    Solution
    1. For the is problem n = 800, p = 0.02, and q = 1 - 0.02 = 0.98.

    The formula \(\mu=n p\) is used.

    \(\mu=800(0.02)=16\)

    The expected number of defective calculators in the batch of 800 is 16.

    1. The variance is computed using the formula \(\sigma^{2}=n p q\).

    \(\sigma^{2}=800(0.02)(0.98)=15.68 \)

    1. The standard deviation is computed by taking the square root of the variance.

    \(\sigma=\sqrt{15.68} \approx 3.960\)

    Exercises

    Exercise \(\PageIndex{1}\)

    A group of 15 college students launches small side businesses while in school (e.g., reselling, tutoring, online shops). Based on past data, each student has a 0.4 probability that their business will still be active after one year. Let X be the number of students whose businesses succeed.

    1. Find the mean number of successful student businesses.
    2. Find the variance.
    3. Find the standard deviation.
    Answer
    1. Mean: 6.0
    2. Variance: 3.60
    3. Standard deviation: 1.90
    Exercise \(\PageIndex{2}\)

    A group of 18 college students is surveyed about whether they choose to eat at a local restaurant instead of the campus cafeteria on a given day. Each student has a 0.5 probability of choosing the restaurant. Let X be the number of students who choose the restaurant.

    1. Find the mean number of successful student businesses.
    2. Find the variance.
    3. Find the standard deviation.
    Answer
    1. Mean: 9.0
    2. Variance: 4.50
    3. Standard deviation: 2.12

     

    Exercise \(\PageIndex{3}\)

    A college sustainability club has 16 student volunteers working in a campus garden. Each student has a 0.6 probability of showing up on a given weekend. Let X be the number of students who show up.

    1. Find the mean number of students who attend.
    2. Find the variance.
    3. Find the standard deviation
    Answer
    1. Mean: 9.6
    2. Variance: 3.84
    3. Standard deviation: 1.96
    Exercise \(\PageIndex{4}\)

    In a community survey, each selected resident has a 40% chance of responding. A researcher randomly selects 12 residents.
    Find:

    1. The mean
    2. The variance
    3. The standard deviation
    Answer
    1. Mean: 4.8
    2. Variance: 2.88
    3. Standard deviation: 1.70
    Exercise \(\PageIndex{1}\)

    A political science club surveys 25 college students about a campus policy issue. Each student has a 0.6 probability of supporting the policy. Let X be the number of students who support the policy.

     

    1. Find the mean number of students who support the policy.
    2. Find the variance.
    3. Find the standard deviation.
    Answer
    1. Mean: 15.0
    2. Variance: 6.00
    3. Standard deviation: 2.45

     

    This page titled 5.3: Mean, Variance, and Standard Deviation of the Binomial Distribution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Toros Berberyan, Tracy Nguyen, and Alfie Swan via source content that was edited to the style and standards of the LibreTexts platform.