mean and variance of geometric distribution
Understand the Problem
The question is asking for the mean and variance of the geometric distribution, which are statistical measures that describe the characteristics of this probability distribution.
Answer
Mean: $\mu = \frac{1}{p}$, Variance: $\sigma^2 = \frac{1 - p}{p^2}$
Answer for screen readers
Mean: $\mu = \frac{1}{p}$
Variance: $\sigma^2 = \frac{1 - p}{p^2}$
Steps to Solve
- Mean of the Geometric Distribution
The mean (expected value) of a geometric distribution with success probability $p$ is calculated using the formula: $$ \mu = \frac{1}{p} $$
- Variance of the Geometric Distribution
The variance of a geometric distribution is given by the formula: $$ \sigma^2 = \frac{1 - p}{p^2} $$
- Implementation of Formulas
If for example, $p = 0.2$ (meaning the probability of success is 20%), we can compute the mean and variance.
- Mean: $$ \mu = \frac{1}{0.2} = 5 $$
- Variance: $$ \sigma^2 = \frac{1 - 0.2}{(0.2)^2} = \frac{0.8}{0.04} = 20 $$
Mean: $\mu = \frac{1}{p}$
Variance: $\sigma^2 = \frac{1 - p}{p^2}$
More Information
The geometric distribution is often used in situations involving "waiting times" for the first success in a series of independent Bernoulli trials. The mean indicates the average number of trials needed for the first success, while the variance provides insight into how much this number can vary.
Tips
- Forgetting to convert the probability $p$ into a decimal form (e.g., using 20 instead of 0.2).
- Confusing the formulas for the mean and variance.
- Not recognizing that the geometric distribution only applies to independent trials with constant probabilities.
