5.3: Mean, Variance, and Standard Deviation of the Binomial Distribution
- Page ID
- 46135
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- 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
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.
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.
- Find the mean.
- Find the variance.
- Find the standard deviation.
Solution
- 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.
- Since this is a binomial, then you can use the formula \(\sigma^{2}=n p q\).
\(q=1-0.01=0.99\)
\(\sigma^{2}=20(0.01)(0.99)=0.198 \)
- Once you have the variance, you just take the square root of the variance to find the standard deviation.
\(\sigma=\sqrt{0.198} \approx 0.445\)
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.
- Find the mean.
- Find the variance.
- Find the standard deviation.
Solution
- 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.
- The variance is computed using the formula \(\sigma^{2}=n p q\).
\(\sigma^{2}=800(0.02)(0.98)=15.68 \)
- The standard deviation is computed by taking the square root of the variance.
\(\sigma=\sqrt{15.68} \approx 3.960\)
Authors
"5.3: Mean, Variance, and Standard Deviation of the Binomial Distribution" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0
Attributions
"5.3: Mean and Standard Deviation of Binomial Distribution" by Kathryn Kozak is licensed under CC BY-SA 4.0