AP Statistics Curriculum 2007 Distrib MeanVar
Contents
General Advance-Placement (AP) Statistics Curriculum - Expectation (Mean) and Variance
Expectation (Mean)
Example
Suppose 10% of the human population carries the green-eye allele. If we choose 1,000 people randomly and let the RV X be the number of green-eyed people in the sample. Then the distribution of X is binomial distribution with n = 1,000 and p = 0.1 (denoted as \(X \sim B(1,000, 0.1)\). In a sample of 1,000 people, how many are we expecting to have this allele? Clearly the count of individuals that carry the green-eye allele will vary between different samples of 1,000 subjects. Well, now much dispersion between the samples can we expect, in terms of the number of individuals carrying this allele? These questions will be answered by computing the mean and the variance (or standard deviation) for this process.
Definition
The expected value, expectation or mean, of a discrete random variable X is defined by \(E[X]=\sum_x{xP(X=x)}.\) The expectation of a continuously-values random variable Y is analogously defined by \(E[Y]=\int{yP(y)dy}\), where the integral is over the domain of Y and P(y) is the probability density function of Y.
Variance
The variance, of a discrete random variable X is defined by \(VAR[X]=\sum_x{(x-E[X])^2P(X=x)}.\) The variance of a continuously-values random variable Y is analogously defined by \(VAR[Y]=\int{(y-E[Y])^2P(y)dy}\), where the integral is over the domain of Y and P(y) is the probability density function of Y.
Examples
A Game of Chance
Suppose we are offered to play a game of chance under these conditions: it costs us to play $1.50 and the awarded prices are {$1, $2, $3}. Assume the probabilities of winning each price are {0.6, 0.3, 0.1}, respectively. Should we play the game? What are our chances of winning/loosing? Let's let X=awarded price. Then X={1, 2, 3}.
| x | 1 | 2 | 3 |
| P(X=x) | 0.6 | 0.3 | 0.1 |
| x*P(X=x) | 0.6 | 0.6 | 0.3 |
Then the mean of this game (i.e., expected return or expectation) is computed as the weighted (by the outcome probabilities) average of all the outcome prices\[E[X] = x_1P(X=x_1) + x_2P(X=x_2)+x_3P(X=x_3) = 1\times 0.6 + 2\times 0.3 + 3\times 0.1 = 1.5\]. In other words, the expected return of this came is $1.5, which equals the entry fee, and hence the game is fair - neither the player nor the house has an advantage in this game (on the long run!) Of course, each streak of n games will produce different outcomes and may give small advantage to one side, however, on the long run, no one will make money.
You can play similar games under different conditions for the probability distribution of the prices using the SOCR Binomial Coin or Die experiments.
Children Gender Expectation Example
Suppose we conduct an (unethical!) experiment involving young couple planning to have children. Suppose, the couples are interested in the number of girls they will have, and each couple agrees to have children until one of the following 2 stopping criteria is met: (1) the couple has at least one child of each gender, or (2) the couple has at most 3 children! Let's denote the RV X ={number of Girls}. The distribution of X is given by:
| x | 0 | 1 | 2 | 3 |
| P(X=x) | 1/8 | 5/8 | 1/8 | 1/8 |
| x*P(X=x) | 0 | 5/8 | 2/8 | 3/8 |
Therefore, the expected number of girls that each couple participating in this ("odd") experiment will have is given by \(E[X] = 0 + 5/8 + 2/8 + 3/8 = 1.5\). What is the interpretation of this expectation?
References
- SOCR Home page: http://www.socr.ucla.edu
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