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What are the differences between geometric and binomial distributions?

Geometric: Counts trials to first success | Binomial: Counts successes in a fixed number of trials.

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What are the differences between geometric and binomial distributions?
Geometric: Counts trials to *first* success | Binomial: Counts successes in a *fixed* number of trials.
Compare the focus of geometric and binomial distributions.
Geometric: Focus on the trial number of the first success. | Binomial: Focus on the number of successes.
What are the conditions for Binomial distribution?
Binary outcome, Independent trials, Number of trials fixed, Success probability fixed.
What are the conditions for Geometric distribution?
Go until first success, Independent trials, First success, Trials not fixed.
Compare the shape of Binomial and Geometric Distributions.
Geometric: Skewed Right | Binomial: Can be skewed or approximately normal depending on p and n.
Explain the concept of skewness in a geometric distribution.
Geometric distributions are always skewed right because the probability of success decreases with each trial.
Explain the meaning of the mean in a geometric distribution.
The mean (expected value) represents the average number of trials needed to achieve the first success.
Explain the meaning of standard deviation in a geometric distribution.
Standard deviation measures how much the number of trials typically varies from the mean number of trials until the first success.
What is the relationship between p and E(Y)?
The expected value is the inverse of the probability of success on each trial.
Describe the trials in geometric distribution.
Each trial is independent, meaning the outcome of one trial does not affect the outcome of any other trial.
What is the formula for the PMF of a geometric distribution?
$P(Y = k) = (1-p)^{k-1} * p$
What is the formula for the mean (expected value) of a geometric distribution?
$E(Y) = \frac{1}{p}$
What is the formula for the standard deviation of a geometric distribution?
$ฯƒ_Y = \sqrt{\frac{1-p}{p^2}}$
Given p = 0.4, what is E(Y)?
E(Y) = 1/0.4 = 2.5
Given p = 0.5, what is the standard deviation?
$ฯƒ_Y = \sqrt{\frac{1-0.5}{0.5^2}} = 1$