Expected Value, Variance, and Standard Deviation for Discrete Distributions

All flashcards

Expected value of a discrete random variable

The long-term average outcome, calculated as $\mu = \sum_x xP(x)$.

Variance of a discrete random variable

The probability-weighted average of squared deviations from the mean: $\sigma^2 = \sum_x (x-\mu)^2P(x)$.

Standard deviation of a discrete random variable

The square root of the variance: $\sigma = \sqrt{\sum_x (x-\mu)^2P(x)}$.

Net profit or loss

The gain or loss after accounting for the cost of participating.

Law of Large Numbers

As the number of trials increases, relative frequency tends to approach theoretical probability.

Quiz questions

1. For a discrete random variable with mean $\mu$, which expression gives its variance?Answer: $\sum_x (x-\mu)^2P(x)$
2. If a discrete distribution has variance $0.49$, what is its standard deviation?Answer: $0.7$
3. When calculating a player's expected result for a game, what should each outcome represent?Answer: The player's net profit or loss after any participation cost.
4. From a player's perspective, what does a negative expected value indicate over many plays?Answer: A negative average result for the player.
5. Which conditions must the probabilities in a discrete probability distribution satisfy?Answer: Each is between 0 and 1, and their sum is 1.